NCERT Exemplar Class 12 Maths Solutions Chapter 2 Inverse Trigonometric Functions 2021

NCERT Exemplar Class 12 Maths Solutions Chapter 2 Inverse Trigonometric Functions

Edited By Komal Miglani | Updated on Mar 31, 2025 03:20 AM IST | #CBSE Class 12th

Imagine trying to find an unknown angle when only its sine or cosine value is given. Have you ever wondered how calculators determine the angles for trigonometric values? This is where inverse trigonometric functions come into play. They help us reverse the process of basic trigonometric functions like sine, cosine, and tangent, allowing us to find angles from given ratios. When the trigonometric values are known, the inverse trigonometric functions allow you to find the angles. These functions are very important when it comes to calculus, solving equations, and even putting them into practice in the real world, such as physics and engineering. We will go through all the NCERT Exemplar questions and solutions in this chapter, which are aimed at developing a solid understanding of this topic to help students understand what the author is trying to assess in the exam.

Latest: JEE Main high scoring chapters | JEE Main 10 year's papers

This Story also Contains

NCERT Exemplar Class 12 Maths Solutions Chapter 2
Sub-Topics Covered in NCERT Exemplar Class 12 Maths Solutions Chapter 2
NCERT Exemplar Class 12 Maths Solutions
Importance of NCERT Exemplar Class 12 Maths Solutions Chapter 2
NCERT solutions of class 12 - Subject-wise
NCERT Notes of class 12 - Subject Wise
NCERT Books and NCERT Syllabus
NCERT Exemplar Class 12 Solutions - Subject Wise

In NCERT Solutions for Class 12 Maths Chapter 2, students will understand how to find the ranges and domains of inverse trigonometric functions. Some other key concepts that students will learn are the behavior of the function, properties of inverse trigonometric functions, and more, all to be explained in a systematic way. The NCERT Exemplar Class 12 Maths Chapter 2 Solutions will provide step-by-step guidance to ensure a strong conceptual understanding for students.

NCERT Exemplar Class 12 Maths Solutions Chapter 2

Class 12 Maths Chapter 2 exemplar solutions Exercise: 2.3
Page number: 35-41
Total questions: 55

NEET/JEE Coaching Scholarship

Get up to 90% Scholarship on Offline NEET/JEE coaching from top Institutes

Apply

JEE Main high scoring chapters and topics

As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE

Get now

Question:1

find the value of $\tan^{- 1} (\tan \frac{5 π}{6}) + \cos^{- 1} (\cos \frac{13 π}{6})$

Answer:

we know that
$\tan^{- 1} (\tan x) = x; x ϵ (- \frac{π}{2}, \frac{π}{2})$ and
$\cos^{- 1} (\cos x) = x; x ϵ (0, π)$

$∴ \tan^{- 1} (\tan \frac{5 π}{6}) + \cos^{- 1} (\cos \frac{13 π}{6})$
$= \tan^{- 1} [\tan (π - \frac{π}{6})] + \cos^{- 1} [c o s (π + \frac{7 π}{6})]$
$= \tan^{- 1} (- \tan \frac{π}{6}) + c o s^{- 1} (- \cos \frac{7 π}{6})$ [since , $\cos (π + θ) = - \cos θ$ ]
$= - \tan^{- 1} (\tan \frac{π}{6}) + π - [c o s^{- 1} (\cos \frac{7 π}{6})]$
$[s i n c e \tan^{- 1} (- x) = - \tan 1 x, x ϵ R a n d \cos^{- 1} = (- x) = π - \cos^{- 1} (x), x ϵ (- 1, 1)]$
$= - \tan^{- 1} (\tan \frac{π}{6}) + π - \cos^{- 1} [\cos (π + \frac{π}{6})]$
$= - \tan^{- 1} (\tan \frac{π}{6}) + π - \cos^{- 1} (- \cos \frac{π}{6})$ [since , $\cos (π + θ) = - \cos θ$ ]
$= - \tan^{- 1} (\tan \frac{π}{6}) + π - π + \cos^{- 1} (\cos \frac{π}{6})$
$= - \frac{π}{6} + 0 + \frac{π}{6}$
$= 0$

Question:2

Evaluate
$\cos [\cos^{- 1} (\frac{- \sqrt{3}}{2}) + \frac{π}{6}]$

Answer:

We have
$\cos [\cos^{- 1} (\frac{- \sqrt{3}}{2}) + \frac{π}{6}]$
$[S i n c e, \cos \frac{5 π}{6} = \frac{- \sqrt{3}}{2}]$
$= \cos [\cos^{- 1} (\cos \frac{5 π}{6}) + \frac{π}{6}]$
$= \cos (\frac{5 π}{6} + \frac{π}{6})$

$[s i n c e . \cos^{- 1} (\cos x) = x; x ϵ (0, π)]$
$= \cos (\frac{6 π}{6})$
$= \cos π$
=-1

Question:3

Prove that $\cot (\frac{π}{4} - 2 \cot^{- 1} 3) = 7$

Answer:

We prove that
$\cot (\frac{π}{4} - 2 \cot^{- 1} 3) = 7$
$\Rightarrow \cot (\frac{π}{4} - 2 \cot^{- 1} 3) = \cot^{- 1} 7$
$\Rightarrow 2 \cot^{- 1} 3 = \frac{π}{4} - \cot^{- 1} 7$
$\Rightarrow 2 \tan^{- 1} \frac{1}{3} = \frac{π}{4} - \tan^{- 1} \frac{1}{7}$
$\Rightarrow 2 \tan^{- 1} \frac{1}{3} + \tan^{1} \frac{1}{7} = \frac{π}{4}$
$[s i n c e, 2 \tan^{- 1} (x) = 2 t a n^{- 1} \frac{2 x}{1 - (x)^{2}}]$
$\Rightarrow \tan^{- 1} \frac{\frac{2}{3}}{1 - {(\frac{1}{3})}^{2}} + \tan^{- 1} \frac{1}{7} = \frac{π}{4}$
$\Rightarrow \tan^{- 1} (\frac{\frac{2}{3}}{\frac{8}{9}}) + \tan^{- 1} \frac{1}{7} = \frac{π}{4}$
$\Rightarrow \tan^{- 1} (\frac{3}{4}) + \tan^{- 1} \frac{1}{7} = \frac{π}{4}$
$[s i n c e, \tan^{- 1} x + t a n^{- 1} y = \tan^{- 1} \frac{x + y}{1 - x y}]$
$\Rightarrow \tan^{- 1} (\frac{\frac{3}{4} + \frac{1}{7}}{1 - \frac{3}{4}, \frac{1}{7}}) = \frac{π}{4}$
$\Rightarrow \tan^{- 1} \frac{\frac{(21 + 4)}{28}}{\frac{(28 - 3)}{28}} = \frac{π}{4}$
$\Rightarrow \tan^{- 1} \frac{25}{25} = \frac{π}{4}$
$\Rightarrow \tan^{- 1} (1) = \frac{π}{4}$
$\Rightarrow 1 = \tan \frac{π}{4}$
$\Rightarrow 1 = 1$
LHS=RHS
Hence Proved

Question:4

Find the value of $\tan^{- 1} (- \frac{1}{\sqrt{3}}) + c o t^{- 1} (\frac{1}{\sqrt{3}}) + t a n^{- 1} [s i n (\frac{- π}{2})]$

Answer:

We have
$\tan^{- 1} (- \frac{1}{\sqrt{3}}) + c o t^{- 1} (\frac{1}{\sqrt{3}}) + t a n^{- 1} [s i n (\frac{- π}{2})]$
$= \tan^{- 1} (t a n \frac{5 π}{6}) + c o t^{- 1} (c o t \frac{π}{3}) + t a n^{- 1} (- 1)$
$= \tan^{- 1} [t a n (π - \frac{5 π}{6})] + \cot^{- 1} (\cot \frac{π}{3}) + \tan^{- 1} [\tan (π - \frac{π}{4})]$
$[s i n c e, \tan^{- 1} (\tan x) = x, x ϵ (- \frac{π}{2}, \frac{π}{2}); \cot^{- 1} (c o t x) = x, x ϵ (0, π); a n d \tan^{- 1} (- x) = - \tan^{- 1} x]$
$= \tan^{- 1} (- \tan \frac{π}{6}) + \cot^{- 1} (\cot \frac{π}{3}) + \tan^{- 1} [- \tan \frac{π}{4}]$
$= - \frac{π}{6} + \frac{π}{3} - \frac{π}{4}$
$= \frac{- 2 π + 4 π - 3 π}{12}$
$= - \frac{π}{12}$

Question:5

find the value of $\tan^{- 1} (t a n \frac{2 π}{3})$

Answer:

We have
$\tan^{- 1} (\tan \frac{2 π}{3}) = \tan^{- 1} \tan (π - \frac{π}{3})$
$= \tan^{- 1} (- \tan \frac{π}{3})$
$[S i n c e, \tan^{- 1} (- x) = - \tan^{- 1} x, x ϵ R]$
$= - \tan^{- 1} t a n \frac{π}{3}$
$[S i n c e, \tan^{- 1} (\tan x) = x, x ϵ (- \frac{π}{2}, \frac{π}{2})]$
$= - \frac{π}{3}$

Question:6

show that $2 \tan^{- 1} (- 3) = - \frac{π}{2} + \tan^{- 1} (\frac{- 4}{3})$

Answer:

We have to prove ,
$2 \tan^{- 1} (- 3) = - \frac{π}{2} + t a n^{- 1} (\frac{- 4}{3})$
LHS= $2 \tan^{- 1} (- 3)$ $[S i n c e, \tan^{- 1} (- x) = - \tan^{- 1} x, x ϵ R]$
$= - [\cos^{- 1} \frac{1 - 3^{2}}{1 + 3^{2}}]$ $[S i n c e 2 \tan^{- 1} x = [\cos^{- 1} \frac{1 - x^{2}}{1 + x^{2}}], x \geq 0]$
$= - [\cos^{- 1} (\frac{- 8}{10})]$
$= - [\cos^{- 1} (\frac{- 4}{5})]$
$= - [π - \cos^{- 1} (\frac{4}{5})]$ $= - [s i n c e \cos^{- 1} (- x) = π - \cos^{- 1} x, x ϵ [- 1, 1]]$
$= - π + \cos^{- 1} (\frac{4}{5})$
$[l e t \cos^{- 1} (\frac{4}{5}) = 0 \Rightarrow \cos θ = (\frac{4}{5}) \Rightarrow \tan θ = (\frac{3}{4}) \Rightarrow θ = \tan^{- 1} (\frac{3}{4})]$
$= - π + \tan^{- 1} (\frac{3}{4}) = - π + [\frac{π}{2} - \cot^{- 1} (\frac{3}{4})]$
$= - \frac{π}{2} - \cot^{- 1} (\frac{3}{4})$
$[S i n c e, \tan^{- 1} (- x) = - \tan^{- 1} x]$
$= - \frac{π}{2} + \tan^{- 1} (\frac{- 4}{3})$
$= - \frac{π}{2} + \tan^{- 1} (- \frac{4}{3})$
=RHS
Hence Proved.

Question:7

Find the real solution of the equation:
$\tan^{- 1} \sqrt{x (x + 1)} + \sin^{- 1} \sqrt{x^{2} + x + 1} = \frac{π}{2}$

Answer:

We have , $\tan^{- 1} \sqrt{x (x + 1)} + \sin^{- 1} \sqrt{x^{2} + x + 1} = \frac{π}{2} . . . . . . . . . (i)$
Let $\sin^{- 1} \sqrt{x^{2} + x + 1} = θ$
$\Rightarrow \sin θ = \sqrt{\frac{x^{2} + x + 1}{1}}$
$\Rightarrow \tan θ = \frac{\sqrt{x^{2} + x + 1}}{\sqrt{- x^{2} - x}}$ $[S i n c e, \tan θ = \frac{\sin θ}{\cos θ}]$
$\Rightarrow θ = \tan^{- 1} \frac{\sqrt{x^{2} + x + 1}}{\sqrt{- x^{2} - x}} = \sin^{- 1} \sqrt{x^{2} + x + 1}$
On Putting the value of $θ$ in Eq. (i), We get
$\tan^{- 1} \sqrt{x (x + 1)} + \tan^{- 1} \frac{\sqrt{x^{2} + x + 1}}{\sqrt{- x^{2} - x}} = \frac{π}{2} . . . . . . . . . (i i)$
we know that,
$\tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} \frac{x + y}{1 - x y}, x y < 1$
So,(ii) becomes,
$\tan^{- 1} [\frac{\sqrt{x (x + 1)} + \frac{\sqrt{x^{2} + x + 1}}{\sqrt{- x^{2} - x}}}{1 - \sqrt{x (x + 1)} \frac{\sqrt{x^{2} + x + 1}}{\sqrt{- x^{2} - x}}}] = \frac{π}{2}$
$\Rightarrow \tan^{- 1} [\frac{\sqrt{x (x + 1)} + \frac{\sqrt{x^{2} + x + 1}}{\sqrt{- 1 (x^{2} + x)}}}{1 - \sqrt{x (x + 1)} \frac{\sqrt{x^{2} + x + 1}}{\sqrt{- 1 (x^{2} + x)}}}] = \frac{π}{2}$
$\Rightarrow \frac{x^{2} + x + \sqrt{- (x^{2} + x + 1)}}{[1 - \sqrt{- (x^{2} + x + 1) . \sqrt{(x^{2} + x)}}]} = \tan \frac{π}{2} = \frac{1}{0}$
$\Rightarrow [1 - \sqrt{- (x^{2} + x + 1)} . \sqrt{(x^{2} + x)}] = 0$
$\Rightarrow - (x^{2} + x + 1) = 1 o r x^{2} + x = 0$
$\Rightarrow x^{2} - x - 1 = 1 o r x (x + 1) = 0$
$\Rightarrow x^{2} + x + 2 = 0 o r x (x + 1) = 0$
$\Rightarrow x = \frac{- 1 \pm \sqrt{1 - (4 \times 2)}}{2} o r x = - 1$
$\Rightarrow x = 0 o r x = - 1$
For real solution , we have x=0,-1.

Question:8

Find the value of $\sin (2 \tan^{- 1} \frac{1}{3}) + \cos (t a n^{- 1} 2 \sqrt{2})$

Answer:

We have $\sin (2 \tan^{- 1} \frac{1}{3}) + \cos (t a n^{- 1} 2 \sqrt{2})$
$= \sin [\sin^{- 1} {\frac{2 \times \frac{1}{3}}{1 + {(\frac{1}{3})}^{2}}}] + \cos (\cos^{- 1} \frac{1}{3})$
$[S i n c e, \tan^{- 1} x = \cos^{- 1} \frac{1}{\sqrt{1 + x^{2}}}; 2 \tan^{- 1} (x) = 2 \tan^{- 1} \frac{2 x}{1 - {(x)}^{2}}, - 1 \leq x \leq 1 a n d \tan^{- 1} 2 \sqrt{2} = \cos^{- 1} \frac{1}{3}]$
= $\sin [\sin^{- 1} {\frac{\frac{2}{3}}{1 + \frac{1}{9}}}] + \frac{1}{3}$ $[S i n c e, \cos (\cos^{- 1} x) = x, x ϵ {- 1, 1}]$
$= \sin [\sin^{- 1} (\frac{2 \times 9}{3 \times 10})] + \frac{1}{3}$
$= \sin [\sin^{- 1} (\frac{3}{5})] + \frac{1}{3}$
$= \frac{3}{5} + \frac{1}{3} [S i n c e, \sin (\sin^{- 1} x) = x]$
$= \frac{14}{15}$

Question:9

If $2 \tan^{- 1} (\cos θ) = \tan^{- 1} (c o s e c θ),$ then show that $θ = \frac{π}{4}$ , where n is any integer.

Answer:

We have $2 \tan^{- 1} (\cos θ) = \tan^{- 1} (c o s e c θ)$
$\Rightarrow \tan^{- 1} (\frac{2 \cos θ}{1 - \cos^{2} θ}) = \tan^{- 1} (2 c o s e c θ)$ $[S i n c e 2 \tan^{- 1} x = \tan^{- 1} (\frac{2 x}{1 - x^{2}})]$
$\Rightarrow \frac{2 \cos θ}{\sin^{2} θ} = 2 c o s e c θ$
$\Rightarrow \cot θ . 2 c o s e c θ = 2 c o s e c θ$
$\Rightarrow \cot θ = 1$
$\Rightarrow \cot θ = \cot \frac{π}{4}$
$\Rightarrow θ = \frac{π}{4}$
Hence Proved

Question:10

Show that $\cos (2 \tan^{- 1} \frac{1}{7}) = \sin (4 \tan^{- 1} \frac{1}{3})$

Answer:

We have , $\cos (2 \tan^{- 1} \frac{1}{7}) = \sin (4 \tan^{- 1} \frac{1}{3})$
$\Rightarrow \cos [\cos^{- 1} (\frac{1 - {(\frac{1}{7})}^{2}}{1 + {(\frac{1}{7})}^{2}})] = s i n (2.2 \tan^{- 1} \frac{1}{3})$
$[S i n c e, 2 \tan^{- 1} x = [\cos^{- 1} \frac{1 - x^{2}}{1 + x^{2}}], x \geq 0]$
$\Rightarrow \cos [\cos^{- 1} (\frac{\frac{48}{49}}{\frac{50}{49}})] = \sin [2 (\tan^{- 1} \frac{\frac{2}{3}}{1 - {(\frac{1}{3})}^{2}})]$
$\Rightarrow \cos [\cos^{- 1} (\frac{24}{25})] = \sin (2 \tan^{- 1} \frac{3}{4})$
$\Rightarrow \cos [\cos^{- 1} (\frac{24}{25})] = \sin (\sin^{- 1} \frac{2 \times \frac{3}{4}}{1 + \frac{9}{16}})$ $[S i n c e, 2 \tan^{- 1} x = \tan^{- 1} (\frac{2 x}{1 - x^{2}})]$
$\Rightarrow \frac{24}{25} = \sin (\sin^{- 1} \frac{\frac{3}{2}}{\frac{25}{16}})$
$\Rightarrow \frac{24}{25} = \frac{48}{50}$
$\Rightarrow \frac{24}{25} = \frac{24}{25}$
Since LHS=RHS
Hence Proved

Question:11

Solve the following equation $\cos (\tan^{- 1} x) = \sin (\cot^{- 1} \frac{3}{4})$

Answer:

We have $\cos (\tan^{- 1} x) = \sin (\cot^{- 1} \frac{3}{4})$
$\Rightarrow \cos (\cos^{- 1} \frac{1}{\sqrt{x^{2} + 2}}) = \sin (\sin^{- 1} \frac{4}{5}) . . . . . . . . (i)$
Let $\tan^{- 1} x = θ_{1} \Rightarrow \tan θ_{1} = \frac{x}{1}$
$\Rightarrow \cos θ_{1} = \frac{1}{\sqrt{x^{2} + 1}} . . . . . (a)$
$\Rightarrow θ_{1} = \cos^{- 1} \frac{1}{\sqrt{x^{2} + 1}} . . . . . (c)$
And $\cot^{- 1} = θ_{2} \Rightarrow \cot^{- 1} = \frac{3}{4}$
$\Rightarrow \sin θ_{2} = \frac{4}{5} . . . . . . . (b)$
$\Rightarrow θ_{2} = \sin^{- 1} \frac{4}{5} . . . . . . . (d)$
From (c),(d);(i) becomes
$\Rightarrow \cos θ_{1} = \sin θ_{2}$
$\Rightarrow \frac{1}{\sqrt{x^{2} + 1}} = \frac{4}{5}$ [From (a),(b)]
On squarinting both Sides, we get
$\Rightarrow 16 (x^{2} + 1) = 25$
$\Rightarrow 16 x^{2} = 9$
$\Rightarrow x^{2} = {(\frac{3}{4})}^{2}$
$\Rightarrow x = \pm \frac{3}{4}$
$∴ x = \frac{3}{4}, - \frac{3}{4} .$

Question:12

Prove that $\tan^{- 1} (\frac{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}) = \frac{π}{4} + \frac{1}{2} \cos^{- 1} x^{2}$

Answer:

We have $\tan^{- 1} (\frac{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}) = \frac{π}{4} + \frac{1}{2} c o s^{- 1} x^{2}$
LHS= $\tan^{- 1} (\frac{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}) . . . . . . . . (i)$
$[x^{2} = \cos 2 θ = \cos^{2} θ + \sin^{2} θ = 1 - 2 \sin^{2} θ = 2 \cos^{2} θ - 1]$
$\Rightarrow \cos^{- 1} x^{2} = 2 θ$
$\Rightarrow θ = \frac{1}{2} \cos^{- 1} x^{2}$
$∴ \sqrt{1 + x^{2}} = \sqrt{1 + \cos 2 θ}$
$\Rightarrow \sqrt{1 + 2 \cos^{2} θ - 1} = \sqrt{2} \cos θ$
And $\sqrt{1 - x^{2}} = \sqrt{1 - \cos 2 θ}$
$\sqrt{1 - 1 + 2 \sin^{2} θ} = \sqrt{2} \sin θ$
$∴ L H S = \tan^{- 1} (\frac{\sqrt{2} \cos θ + \sqrt{2} \sin θ}{\sqrt{2} \cos θ - \sqrt{2} \sin θ})$
$= \tan^{- 1} (\frac{\cos θ + \sin θ}{\cos θ - \sin θ})$
$= \tan^{- 1} (\frac{1 + \tan θ}{1 - \tan θ})$
$= \tan^{- 1} {\frac{\tan (\frac{π}{4}) + \tan θ}{\tan (\frac{π}{4}) - \tan θ}}$
$= \tan^{- 1} [\tan (\frac{π}{4} + θ)]$ $[S i n c e, \tan (x + y) = \frac{\tan x + \tan y}{1 - \tan x . \tan y}]$
$= \frac{π}{4} + θ$
$= \frac{π}{4} + \frac{1}{2} \cos^{- 1} x^{2}$
=RHS
LHS=RHS
Hence Proved

Question:13

Find the simplified from of $\cos^{- 1} (\frac{3}{5} \cos x + \frac{4}{5} \sin x)$ , where $x ϵ [\frac{- 3 π}{4}, \frac{π}{4}]$

Answer:

Let $\cos y = \frac{3}{5}$
$\Rightarrow \sin y = \frac{4}{5}$
$\Rightarrow y = \cos^{- 1} \frac{3}{5} = \sin^{- 1} \frac{4}{5} = \tan^{- 1} \frac{4}{3}$
$∴ \cos^{- 1} [\cos y . \cos x + \sin y . \sin x]$
$[s i n c e, \cos (A - B) = \cos A . \cos B + \sin A . \sin B]$
$= \cos^{- 1} [\cos (y - x)]$
$[s c i n e, \cos (\cos^{- 1} x) = x, x ϵ {- 1, 1}]$
=y-x
$[s c i n e, y = \tan^{- 1} \frac{4}{3}]$
$= \tan^{- 1} \frac{4}{3} - x$

Question:14

Prove that $\sin^{- 1} \frac{8}{17} + \sin^{- 1} \frac{3}{5} = \sin^{- 1} \frac{77}{85}$

Answer:

we have $\sin^{- 1} \frac{8}{17} + \sin^{- 1} \frac{3}{5} = \sin^{- 1} \frac{77}{85}$
$L H S = \sin^{- 1} \frac{8}{17} + \sin^{- 1} \frac{3}{5}$

$l e t \sin^{- 1} \frac{8}{17} = θ_{1}$
$\Rightarrow \sin θ_{1} = \frac{8}{17}$
$\Rightarrow \tan θ_{1} = \frac{8}{15} \Rightarrow θ_{1} = \tan^{- 1} \frac{8}{15}$
And, $\sin \frac{3}{5} = θ_{2} \Rightarrow \sin^{- 1} \frac{3}{5}$
$\Rightarrow \tan θ_{2} = \frac{3}{4} \Rightarrow θ_{2} = \tan^{- 1} \frac{3}{4}$
$= \tan^{- 1} \frac{8}{15} + \tan^{- 1} \frac{3}{4}$
$= \tan^{- 1} [\frac{\frac{8}{15} + \frac{3}{4}}{1 - \frac{8}{15} \times \frac{3}{4}}]$ $[S i n c e, \tan^{- 1} x + \tan^{- 1} y = t a n^{- 1} (\frac{x + y}{1 - x y})]$
$= \tan^{- 1} [\frac{\frac{77}{60}}{\frac{36}{60}}]$
$= \tan^{- 1} (\frac{77}{36})$
Let $= θ_{3} = t a n^{- 1} (\frac{77}{36}) \Rightarrow \tan θ_{3} = \frac{77}{36}$
$\Rightarrow \sin θ_{3} = \frac{77}{\sqrt{5929 + 1296}} = \frac{77}{85}$
$∴ θ_{3} = \sin^{- 1} (\frac{77}{85})$
$= \sin^{- 1} (\frac{77}{85}) = R H S$
Hence proved

Question:15

Show that $\sin^{- 1} \frac{5}{13} + \cos^{- 1} \frac{3}{5} = \tan^{- 1} \frac{63}{16}$

Answer:

Solving LHS, $\sin^{- 1} \frac{5}{13} + \cos^{- 1} \frac{3}{5}$
$L e t \sin^{- 1} \frac{5}{13} = x$
$\Rightarrow \sin x = \frac{5}{13}$
$A n d \cos^{2} x = 1 - \sin^{2} x$
$\Rightarrow 1 - \frac{25}{169} = \frac{144}{169}$
$\Rightarrow \cos x = \sqrt{\frac{144}{169}} = \frac{12}{13}$
$∴ \tan x = \frac{\sin x}{\cos x} = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12}$
$\Rightarrow \tan x = \frac{5}{12} . . . . . . . . . . (i)$
Again , let $\cos^{- 1} \frac{3}{5} = y$
$\Rightarrow \cos y = \frac{3}{5}$
$∴ \sin y = \sqrt{1 - \cos^{2} y}$
$\Rightarrow \sin y = \sqrt{1 - {(\frac{3}{5})}^{2}}$
$\Rightarrow \sin y = \sqrt{\frac{16}{25}} = \frac{4}{5}$
$\Rightarrow \tan y = \frac{\sin y}{\cos y} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3} . . . . . . . . . (i i)$
We know that, $\tan (x + y) = \frac{\frac{5}{12} + \frac{4}{3}}{1 - \frac{5}{12} \frac{4}{3}}$ [from (i),(ii)]
$\Rightarrow \tan (x + y) = \frac{\frac{15 + 48}{36}}{\frac{36 - 20}{36}}$
$\Rightarrow \tan (x + y) = \frac{\frac{63}{36}}{\frac{16}{36}}$
$\Rightarrow (x + y) = \tan^{- 1} \frac{63}{16}$ $\Rightarrow (x + y) = \tan^{- 1} \frac{63}{16} = R H S$
Since , LHS=RHS
Hence Proved.

Question:16

Prove that , $\tan^{- 1} \frac{1}{4} + \tan^{- 1} \frac{2}{9} = \sin^{- 1} \frac{1}{\sqrt{5}}$

Answer:

Solving LHS, $\tan^{- 1} \frac{1}{4} + \tan^{- 1} \frac{2}{9}$
Let $\tan^{- 1} \frac{1}{4} = x$
$\Rightarrow \tan x = \frac{1}{4}$
Squaring both sides,
$\Rightarrow \tan^{2} x = \frac{1}{16}$
$\Rightarrow \sec^{2} x - 1 = \frac{1}{16}$
$\Rightarrow \sec^{2} x = \frac{17}{16}$
$\Rightarrow \frac{1}{\cos^{2} x} = \frac{17}{16}$
$\Rightarrow \cos^{2} x = \frac{16}{17}$
$\Rightarrow \cos x = \frac{4}{\sqrt{17}}$
$S i n c e, \sin^{2} x = 1 - \cos^{2} x$
$\Rightarrow \sin^{2} x = 1 - \frac{16}{17} = \frac{1}{17}$
$\Rightarrow \sin x = \frac{1}{\sqrt{17}}$
Again,
Let $\tan^{- 1} \frac{2}{9} = y$
$\Rightarrow \tan y = \frac{2}{9}$
Squaring both sides,
$\Rightarrow \tan^{2} y = \frac{4}{81}$
$\Rightarrow \sec^{2} y - 1 = \frac{4}{81}$
$\Rightarrow \sec^{2} y = \frac{85}{81}$
$\Rightarrow \frac{1}{\cos^{2} y} = \frac{85}{81}$
$\Rightarrow \cos^{2} y = \frac{81}{85}$
$\Rightarrow \cos y = \frac{9}{\sqrt{85}}$
Since, $\sin^{2} y = 1 - \cos^{2} y$
$\Rightarrow \sin^{2} = 1 - \frac{81}{85} = \frac{4}{85}$
$\Rightarrow \sin x = \frac{2}{\sqrt{85}}$
We know that, $\sin (x + y) = \sin x . \sin y + \cos x . \sin y$
$\Rightarrow \sin (x + y) = \frac{1}{\sqrt{17}} . \frac{9}{\sqrt{85}} + \frac{4}{\sqrt{17}} . \frac{2}{\sqrt{85}}$
$\Rightarrow \sin (x + y) = \frac{17}{\sqrt{17} . \sqrt{85}}$
$\Rightarrow \sin (x + y) = \frac{\sqrt{17}}{\sqrt{17} . \sqrt{5}}$
$\Rightarrow \sin (x + y) = \frac{1}{\sqrt{5}}$
$\Rightarrow x + y = \sin^{- 1} \frac{1}{\sqrt{5}} = R H S$
Since , LHS=RHS
Hence Proved

Question:17

Find the value of $4 \tan^{- 1} \frac{1}{5} - t a n^{- 1} \frac{1}{239}$

Answer:

We have, $4 t a n^{- 1} \frac{1}{5} - \tan^{- 1} \frac{1}{239}$
$= 2 \times (2 \tan^{- 1} \frac{1}{5}) - \tan^{- 1} \frac{1}{239}$
$= 2 [\tan^{- 1} \frac{\frac{2}{5}}{1 - {(\frac{1}{5})}^{2}}] - \tan^{- 1} \frac{1}{239}$ $[s i n c e, 2 \tan^{- 1} x = \tan^{- 1} \frac{2 x}{1 - {(x)}^{2}}]$

$= 2 [\tan^{- 1} \frac{\frac{2}{5}}{\frac{24}{25}}] - \tan^{- 1} \frac{1}{239}$
$= 2 \tan^{- 1} \frac{5}{12} - \tan^{- 1} \frac{1}{239}$
$= [\tan^{- 1} \frac{\frac{5}{6}}{1 - {(\frac{5}{12})}^{2}}] - \tan^{- 1} \frac{1}{239}$ $[s i n c e, 2 \tan^{- 1} x = \tan^{- 1} \frac{2 x}{1 - {(x)}^{2}}]$
$= \tan^{- 1} \frac{\frac{5}{6}}{1 - \frac{25}{144}} - \tan^{- 1} \frac{1}{139}$
$= \tan^{- 1} (\frac{144 \times 5}{119 \times 6}) - \tan^{- 1} \frac{1}{239}$
$= \tan^{- 1} \frac{120}{119} - \tan^{- 1} \frac{1}{239}$
$= \tan^{- 1} \frac{\frac{120}{119} - \frac{1}{239}}{1 + \frac{120}{119} . \frac{1}{239}} [s i n c e, \tan^{- 1} x - \tan^{- 1} y = \tan^{- 1} (\frac{x - y}{1 + x y})]$
$= \tan^{- 1} [\frac{28680 - 119}{28441 + 120}]$
$= \tan^{- 1} \frac{28561}{28561}$
$= \tan^{- 1} (1)$
$= \tan^{- 1} (\tan \frac{π}{4})$
$= \frac{π}{4}$
Hence, $4 t a n^{- 1} \frac{1}{5} - \tan^{- 1} \frac{1}{239} = \frac{π}{4}$

Question:18

Show that $\tan (\frac{1}{2} \sin^{- 1} \frac{3}{4}) = \frac{4 - \sqrt{7}}{3}$ and, justify why the other value $\frac{4 + \sqrt{7}}{3}$ is ignored.

Answer:

Solving LHS,
$= \tan (\frac{1}{2} \sin^{- 1} \frac{3}{4})$
$L e t \frac{1}{2} \sin^{- 1} \frac{3}{4} = θ$
$\Rightarrow \sin^{- 1} \frac{3}{4} = 2 θ$
$\Rightarrow \frac{3}{4} = \sin 2 θ$
$\Rightarrow \sin 2 θ = \frac{3}{4}$
$\Rightarrow \frac{2 \tan θ}{1 + \tan^{2} θ} = \frac{3}{4}$
$\Rightarrow 3 + 3 \tan^{2} θ = 8 \tan θ$
$\Rightarrow 3 \tan^{2} θ - 8 \tan θ = 3$
$L e t \tan θ = y$
$∴ 3 y^{2} + 8 y + 3 = 0$
$\Rightarrow y = \frac{8 \pm \sqrt{64 - 4 \times 3 \times 3}}{2 \times 3}$ $\Rightarrow= \frac{8 \pm \sqrt{28}}{6}$
$\Rightarrow y = \frac{2 (4 \pm \sqrt{7})}{2 \times 3}$
$\Rightarrow \tan θ = \frac{(4 \pm \sqrt{7})}{3}$
$\Rightarrow θ = \tan^{- 1} \frac{(4 \pm \sqrt{7})}{3}$
{ but as we can see , $\frac{4 + \sqrt{7}}{3} > 1$ , since $m a x [\tan (\frac{1}{2} \sin^{- 1} \frac{3}{4})] = 1$ }
$\tan (\frac{1}{2} \sin^{- 1} \frac{3}{4}) = \frac{4 - \sqrt{7}}{3} = R H S$
Note: Scince $- \frac{π}{2} \leq s i n^{- 1} \frac{3}{4} \leq \frac{π}{2}$
$\Rightarrow - \frac{π}{4} \leq \frac{1}{2} s i n^{- 1} \frac{3}{4} \leq \frac{π}{4}$
$∴ \tan (- \frac{π}{4}) \leq \tan (\frac{1}{2} s i n^{- 1} \frac{3}{4}) \leq \tan (\frac{π}{4})$
$\Rightarrow - 1 \leq \tan (\frac{1}{2} \sin^{- 1} \frac{3}{4}) \leq 1$

Question:19

If a₁,a₂,a₃,.................a_n is an arithmetic progression with common difference d, then evaluate the following expression.
$\tan [\tan^{- 1} (\frac{d}{1 + a_{1} a_{2}}) + \tan^{- 1} (\frac{d}{1 + a_{2} a_{3}}) + \tan^{- 1} (\frac{d}{1 + a_{3} a_{4}}) + . . . . . . . . . . + \tan^{- 1} (\frac{d}{1 + a_{n - 1} a_{n}})]$

Answer:

We have $a_{1} = a, a_{2} = a + d, a_{3} = a + 2 d . . . . . . .$
And, $d = a_{2} - a_{1} = a_{3} - a_{2} = a_{4} - a_{3} = . . . . . . = a_{n} - a_{n - 1}$
Given that,
$\tan [\tan^{- 1} (\frac{d}{1 + a_{1} a_{2}}) + \tan^{- 1} (\frac{d}{1 + a_{2} a_{3}}) + \tan^{- 1} (\frac{d}{1 + a_{3} a_{4}}) + . . . . . . . . . . . . + \tan^{- 1} (\frac{d}{1 + a_{n - 1} a_{n}})]$
$= \tan^{- 1} [\tan^{- 1} (\frac{a_{2} - a_{1}}{1 + a_{1} a_{2}}) + \tan^{- 1} (\frac{a_{3} - a_{2}}{1 + a_{2} a_{3}}) + . . . . . . . . . + \tan^{- 1} (\frac{a_{n} - a_{n - 1}}{1 + a_{n - 1} a_{n}})]$
$= \tan [(\tan^{- 1} a_{2} - \tan^{- 1} a_{1}) + (\tan^{- 1} a_{3} - \tan^{- 1} a_{2}) + . . . . . . . . . . . . . . . . + (\tan^{- 1} a_{n} - \tan^{- 1} a_{n - 1})]$
$= \tan [\tan^{- 1} a_{n} - \tan^{- 1} a_{1}]$
$[S c i n c e, \tan^{- 1} x - \tan^{- 1} y = \tan^{- 1} (\frac{x - y}{1 + x y})]$
$= \tan [\tan^{- 1} (\frac{a_{n} - a_{1}}{1 + a_{1} a_{n}})]$
$[s c i n c e, \tan (\tan^{- 1} x) = x]$
$= \frac{a_{n} - a_{1}}{1 + a_{1} a_{n}}$

Question:20

Which of the following in the principal value branch of $\cos^{- 1} x$
A. $[- \frac{π}{2}, \frac{π}{2}]$
B. $(0, - π)$
C. $[0, π]$
D. $(0, π) - \frac{π}{2}$

Answer:

Answer : (c)
We know that the principal value branch of $\cos^{- 1}$ is $[0, π]$

Question:21

Which of the following in the principal value branch of $c o s e c^{- 1} x$ .
$A . (- \frac{π}{2}, \frac{π}{2})$
$B . [0, π] - {\frac{π}{2}}$
$C . [- \frac{π}{2}, \frac{π}{2}]$
$D . [- \frac{π}{2}, \frac{π}{2}] - {0}$

Answer:

Answer :(D)

We know that the principal value branch of $c o s e c^{- 1} x$ is $[- \frac{π}{2}, \frac{π}{2}] - (0)$

Question:22

If $3 \tan^{- 1} x + \cot^{- 1} x = π$ , then x equals to
A. 0
B. 1
C. -1
D. 1/2

Answer:

Answer : B
Given That, $3 \tan^{- 1} x + \cot^{- 1} x = π$
$\Rightarrow 2 \tan^{- 1} x + \tan^{- 1} x + \cot^{- 1} x = π$
$\Rightarrow 2 \tan^{- 1} x = π - \frac{π}{2}$ $[S c i n c e, \tan^{- 1} x + \cot^{- 1} x = \frac{π}{2}]$
$\Rightarrow \tan^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{2}$ $[S c i n c e, 2 t a n^{- 1} x = \tan^{- 1} \frac{2 x}{1 - x^{2}}]$
$\Rightarrow \frac{2 x}{1 - x^{2}} = \tan \frac{π}{2}$
$\Rightarrow \frac{2 x}{1 - x^{2}} = \tan \frac{1}{0}$
Cross multiplying
$\Rightarrow 1 - x^{2} = 0$
$\Rightarrow x^{2} = \pm 1$
Here only x=1 satifies the given equation.
Note:- Here ,Putting x=-1 in the given equation we get,
$3 \tan^{- 1} (- 1) + c o t^{- 1} (- 1) = π$
$3 \tan^{- 1} [\tan (\frac{- π}{4})] + c o t^{- 1} [\cot (\frac{- π}{4})] = π$
$3 \tan^{- 1} [- \tan (\frac{π}{4})] + c o t^{- 1} [- \cot (\frac{π}{4})] = π$
$3 \tan^{- 1} [\tan (\frac{π}{4})] + π - c o t^{- 1} [\cot (\frac{π}{4})] = π$
$- 3 \times \frac{π}{4} + π - \frac{π}{4} = π$
$- π + π = π$
$0 \neq π$
Hence , x=-1 does not satisfy the given equation.

Question:23

The value of $\sin^{- 1} [c o s (\frac{33 π}{5})]$ is
$A . \frac{3 π}{5}$
$B . \frac{- 7 π}{5}$
$C . \frac{π}{10}$
$D . \frac{- π}{10}$

Answer:

Answer :(D)
We have
$\sin^{- 1} [\cos (\frac{33 π}{5})]$
$= \sin^{- 1} [\cos (6 π + \frac{3 π}{5})]$
$= \sin^{- 1} [\cos (\frac{3 π}{5})]$ $[S i n c e, \cos (2 n π + θ) = \cos θ]$
$= \sin^{- 1} [\cos (\frac{π}{2} + \frac{π}{10})]$
$= \sin^{- 1} [\sin (- \frac{π}{10})] [S i n c e, \sin^{- 1} (x) = - \sin^{- 1} x]$
$= - \frac{π}{10} [S i n c e, \sin^{- 1} (\sin x) = - x, x ϵ (- \frac{π}{2}, \frac{π}{2})]$

Question:24

The domain of the function $\cos^{- 1} (2 x - 1)$ is
A.[0,1]
B.[-1,1]
C.(-1,1)
D.[0, $π$ ]

Answer:

Answer:(A)
We Have $f (x) = c o s^{- 1} (2 x - 1)$
Scince $- 1 \leq 2 x - 1 \leq 1$
$\Rightarrow 0 \leq 2 x \leq 2$
$\Rightarrow 0 \leq x \leq 1$
$∴ x ϵ [0, 1]$

Question:25

The domain of the function defined by $f (x) = \sin^{- 1} \sqrt{x - 1}$ is
A.[1,2]
B.[-1,1]
C.[0,1]
D. None of these

Answer:

Answer: (A)
$f (x) = \sin^{- 1} \sqrt{x - 1}$
$\Rightarrow 0 \leq x - 1 \leq 1 [S i n c e, \sqrt{x - 1} \geq 0 a n d - 1 \leq \sqrt{x - 1} \leq 1]$
$\Rightarrow 1 \leq x \leq 2$
$∴ x ϵ [1, 2]$

Question:26

If $\cos (s i n^{- 1} \frac{2}{5} + c o s^{- 1} x) = 0,$ then x is equal to
$A . \frac{1}{5}$
$B . \frac{2}{5}$
C.0
D.1

Answer:

Answer: (B)
Given, $\cos (S i n^{- 1} \frac{2}{5} + c o s^{- 1} x) = 0$
Let $S i n^{- 1} \frac{2}{5} + c o s^{- 1} x = θ$
So $\cos θ = 0. . . . . . . (1)$
Principal value $\cos^{- 1} x$ is $[0, π]$ .......(2)
Also , we know that $\cos \frac{π}{2} = 0. . . . . . (3)$
From (1) ,,(2), and (3) we have
$θ = \frac{π}{2}$
But $θ = \sin^{- 1} \frac{2}{5} + \cos^{- 1} x$
So,
$\sin^{- 1} \frac{2}{5} + \cos^{- 1} x = \frac{π}{2}$
We know that $\sin^{- 1} x + \cos^{- 1} x = \frac{π}{2} f o r a l l x ϵ [- 1, 1]$
As $\sin^{- 1} \frac{2}{5} + \cos^{- 1} x = \frac{π}{2}$
so $x = \frac{2}{5}$

Question:27

The value of sin (2tan^–1 (.75)) is equal to
A. 0.75
B. 1.5
C. 0.96
D. sin 1.5

Answer:

Answer :(c)
sin (2tan^–1 (.75))
Let, tan^–1 (.75) = θ
$\Rightarrow \tan^{- 1} (\frac{3}{4}) = θ$
$\Rightarrow \tan θ = \frac{3}{4}$
As, $\tan θ = \frac{3}{4}$ so
$\sin θ = \frac{3}{5}, \cos θ = \frac{4}{5} . . . . . . (1)$
Now,
sin (2tan^–1 (.75)) = sin 2θ
= 2 sin θ cos θ
$= 2 (\frac{3}{5}) (\frac{4}{5})$
$= \frac{24}{25}$
So, sin (2tan^–1(.75)) = 0.96.

Question:28

The Value of $\cos^{- 1} \cos \frac{3 π}{2}$ is equal to
$A . \frac{π}{2}$
$B . \frac{3 π}{2}$
$C . \frac{5 π}{2}$
$D . \frac{7 π}{2}$

Answer:

We have $\cos^{- 1} \cos \frac{3 π}{2}$
We know that,
$\cos \frac{3 π}{2} = 0$
So, $\cos^{- 1} \cos \frac{3 π}{2} = \cos^{- 1} 0$
Let $\cos^{- 1} 0 = θ$
⇒ cos θ = 0
Principal value of cos^-1 x is [0, π]
For, cos θ = 0
so, $θ = \frac{π}{2}$

Question:29

The value of the expression $2 \sec^{- 1} 2 + \sin^{- 1} (\frac{1}{2})$ is
$A . \frac{π}{6}$
$B . \frac{5 π}{6}$
$C . \frac{7 π}{6}$
D.1

Answer:

Answer :(B)
We have,
Principal value of sin^-1 x is $(- \frac{π}{2}, \frac{π}{2})$
Principal value of sec^-1 x is [0, π] $- {\frac{π}{2}}$
Let $\sin^{- 1} \frac{1}{2} = A$
$\Rightarrow \sin A = \frac{1}{2}$
$\Rightarrow A = \frac{π}{6}$
So, $\Rightarrow \sin^{- 1} \frac{1}{2} = \frac{π}{6}$ … (1)

Let sec^-1 2 = B
⇒ sec B = 2
$\Rightarrow B = \frac{π}{3}$
So, 2 sec^-1 2 = 2B
$\Rightarrow 2 \sec^{- 1} 2 = \frac{2 π}{3}$ ...(2)
So, the value of $2 \sec^{- 1} 2 + \sin^{- 1} \frac{1}{2}$ from (1) and (2) is
$2 \sec^{- 1} 2 + \sin^{- 1} \frac{1}{2} = \frac{2 π}{3} + \frac{π}{6}$
$= \frac{4 π}{6} + \frac{π}{6}$
$= \frac{5 π}{6}$
So, $2 \sec^{- 1} 2 + \sin^{- 1} \frac{1}{2} = \frac{5 π}{6}$

Question:30

If tan^–1 x + tan^–1y = 4π/5, then cot^–1x + cot^–1 y equals
$A . \frac{π}{5}$
$B . \frac{2 π}{5}$
$C . \frac{3 π}{5}$
$D . π$

Answer:

Answer :(A)
We know that,
$\tan^{- 1} x + \cot^{- 1} x = \frac{π}{2}$
We have,
tan^–1 x + tan^–1 y = 4π/5 … (1)
Let, cot^–1x + cot^–1 y = k … (2)
Adding (1) and (2) –
$\tan^{- 1} x + \tan^{- 1} y + \cot^{- 1} x + \cot^{- 1} y = \frac{4 π}{5} + k$ ...(3)
Now, tan^–1 A + cot^–1 A = π/2 for all real numbers.
So, (tan^–1 x + cot^–1 x) + (tan^–1y + cot^–1 y) = π … (4)
From (3) and (4), we get,
$\frac{4 π}{5} + k = π$
$\Rightarrow k = π - \frac{4 π}{5}$
$\Rightarrow k = \frac{π}{5}$

Question:31

If $\sin^{- 1} \frac{2 a}{1 + a^{2}} + \cos^{- 1} \frac{1 - a^{2}}{1 + a^{2}} = t a n^{- 1} \frac{2 x}{1 - x^{2}}$ where a, $x ϵ [0, 1]$ then the value of x is
A. 0
B. a/2
C. a
D. $\frac{2 a}{1 - a^{2}}$

Answer:

Answer:(D)
We have
$s i n^{- 1} \frac{2 a}{1 + a^{2}} + c o s^{- 1} \frac{1 - a^{2}}{1 + a^{2}} = \tan^{- 1} \frac{2 x}{1 - x^{2}}$
we know that
$2 \tan^{- 1} p = \sin^{- 1} \frac{2 p}{1 + p^{2}} . . . . . . . . . (1)$
$A l s o, 2 \tan^{- 1} p = \cos^{- 1} \frac{1 - p^{2}}{1 + p^{2}} . . . . . . . . . (2)$
$A l s o, 2 \tan^{- 1} p = \tan^{- 1} \frac{2 p}{1 - p^{2}} . . . . . . . . . (3)$
From (1) and (2) we have,
L.H.S-
$\sin^{- 1} \frac{2 a}{1 + a^{2}} + \cos^{- 1} \frac{1 - a^{2}}{1 + a^{2}} = 2 \tan^{- 1} a + 2 \tan^{- 1} a$
$\Rightarrow \sin^{- 1} \frac{2 a}{1 + a^{2}} + \cos^{- 1} \frac{1 - a^{2}}{1 + a^{2}} = 4 \tan^{- 1} a$
From (3) R.H.S
$\tan^{- 1} \frac{2 x}{1 - x^{2}} = 2 \tan^{- 1} x$
So, we have 4 tan^-1 a = 2 tan^-1 x
⇒ 2 tan^-1 a = tan^-1 x
But from (3) $2 \tan^{- 1} a = \tan^{- 1} \frac{2 a}{1 - a^{2}}$
So $\tan^{- 1} \frac{2 a}{1 - a^{2}} = \tan^{- 1} x$
$x = \frac{2 a}{1 - a^{2}}$

Question:32

The value of $\cot \cos^{- 1} \frac{7}{25} i s$
A. 25/24
B.25/7
C.24/25
D.7/24

Answer:

Answer :(d)
We have to find $\cot \cos^{- 1} \frac{7}{25}$
Let $\cos^{- 1} \frac{7}{25} = A$
$\Rightarrow \cos^{- 1} = \frac{7}{25}$
Also, $\cot A = \cot \cos^{- 1} \frac{7}{25}$
As, $\sin A = \sqrt{1 - \cos^{2} A}$
So $\sin A = \sqrt{1 - {(\frac{7}{5})}^{2}}$
$\Rightarrow \sin A = \sqrt{1 - \frac{49}{625}}$
$\Rightarrow \sin A = \sqrt{\frac{625 - 49}{625}}$
$\Rightarrow \sin A = \sqrt{\frac{576}{625}}$
$\Rightarrow \sin A = \frac{24}{25}$
We need to find cot A
$\cot A = \frac{\cos A}{\sin A}$
$\Rightarrow \cot A = \frac{\frac{7}{25}}{\frac{24}{25}}$
$\Rightarrow \cot A = \frac{7}{24}$
So $\cot \cos^{- 1} \frac{7}{25} = \frac{7}{24}$

Question:33

The value of the expression $\tan \frac{1}{2} \cos^{- 1} \frac{2}{\sqrt{5}}$ is $[H i n t : \tan \frac{θ}{2} = \sqrt{\frac{1 - \cos θ}{1 + \cos θ}}]$
$A . 2 + \sqrt{5}$
$B . \sqrt{5} - 2$
$C . \frac{2 + \sqrt{5}}{2}$
$D . \sqrt{5} + 2$
Answer:

Answer:(B)
We need to find , $\tan \frac{1}{2} \cos^{- 1} \frac{2}{\sqrt{5}}$
Let, $\cos^{- 1} \frac{2}{\sqrt{5}} = A$
$\Rightarrow \cos A = \frac{2}{\sqrt{5}}$
Also we need to find $\tan \frac{A}{2}$
We know that $\tan \frac{θ}{2} = \sqrt{\frac{(1 - \cos θ)}{1 + \cos θ}}$
so, $\tan^{- 1} \frac{A}{2} = \sqrt{\frac{(1 - \cos A)}{1 + \cos A}}$
$\Rightarrow \tan^{- 1} \frac{A}{2} = \sqrt{\frac{1 - \frac{2}{\sqrt{5}}}{1 + \frac{2}{\sqrt{5}}}}$
$\Rightarrow \tan^{- 1} \frac{A}{2} = \sqrt{\frac{\frac{\sqrt{5} - 2}{\sqrt{5}}}{\frac{\sqrt{5} + 2}{\sqrt{5}}}}$
$\Rightarrow \tan^{- 1} \frac{A}{2} = \sqrt{\frac{\sqrt{5} - 2}{\sqrt{5} + 2}}$
on rationalizing,
$\Rightarrow \tan^{- 1} \frac{A}{2} = \sqrt{\frac{(\sqrt{5} - 2) (\sqrt{5} + 2)}{(\sqrt{5} + 2) (\sqrt{5} + 2)}}$
$\Rightarrow \tan^{- 1} \frac{A}{2} = \sqrt{\frac{{(\sqrt{5})}^{2} - 2^{2}}{({\sqrt{5}}^{2} + 2^{2})}}$
$\Rightarrow \tan^{- 1} \frac{A}{2} = \sqrt{\frac{5 - 4}{({\sqrt{5}}^{2} + 2^{2})}}$
$\Rightarrow \tan^{- 1} \frac{A}{2} = \sqrt{\frac{1}{({\sqrt{5}}^{2} + 2^{2})}}$
$\Rightarrow \tan^{- 1} \frac{A}{2} = \frac{1}{\sqrt{5} + 2}$
Again rationalizing
$\Rightarrow \tan^{- 1} \frac{A}{2} = \frac{1 (\sqrt{5} - 1)}{(\sqrt{5} + 2) (\sqrt{5} - 2)}$
$\Rightarrow \tan^{- 1} \frac{A}{2} = \frac{(\sqrt{5} - 2)}{({\sqrt{5}}^{2} - 2^{2})}$
$\Rightarrow \tan^{- 1} \frac{A}{2} = \frac{(\sqrt{5} - 2)}{(5 - 4)}$
$\Rightarrow \tan^{- 1} \frac{A}{2} = \sqrt{5} - 2$

Question:34

If |x| ≤ 1, then $2 \tan^{2} x + \sin^{- 1} \frac{2 x}{1 + x^{2}}$ is equal to
A. 4 tan–1 x
B. 0
C. $\frac{π}{2}$
D. π

Answer:

Answer(A)
We need to find, $2 \tan^{- 1} x + \sin^{- 1} \frac{2 x}{1 + x^{2}}$
We know that
$2 \tan^{- 1} p = \sin^{- 1} \frac{2 x}{1 + x^{2}}$
So,
$2 \tan^{1} x + \sin^{- 1} \frac{2 x}{1 + x^{2}} = 2 \tan^{1} x + 2 \tan^{- 1} x$
= $4 \tan^{1} x$

Question:35

If cos^–1α + cos^–1β + cos^–1γ = 3π, then α(β + γ) + β (γ + α) + γ (α + β) equals
A. 0
B. 1
C. 6
D. 12

Answer:

Answer :(C)
Given, cos^–1α + co^s–1β + cos^–1γ = 3π … (1)
Principal value of cos^-1 x is [0, π]
So, maximum value which cos^-1 x can have is π.
So, if (1) is correct then all the three terms i.e,
cos^–1α, cos^–1β, cos^–1γ should be equal to π
So, cos^–1α = π
cos^–1β = π
cos^–1γ = π
So, α = β = γ = -1
So, α(β + γ) + β (γ + α) + γ (α + β)
= (-1)(-1-1) + (-1)(-1-1) + (-1)(-1-1)
= 3(-1)(-2)
= 6

Question:36

The number of real solutions of equarion $\sqrt{1 + \cos x} = \sqrt{2} \cos^{- 1} (\cos x) i n [\frac{π}{2}, π]$ is
A. 0
B. 1
C. 2
D. Infinite

Answer:

Answer:(A)
We have , $\sqrt{1 + \cos 2 x} = \sqrt{2} \cos^{- 1} (\cos x)$ , x is in $[\frac{π}{2}, π]$
R.H.S
$\sqrt{2} \cos^{- 1} (\cos x) = \sqrt{2} x$
So, $\sqrt{1 + \cos 2 x} = \sqrt{2} x$
Squaring both side , we get,
$(1 + \cos 2 x) = 2 x^{2}$
$\Rightarrow \cos 2 x = 2 x^{2} - 1$
Now plotting cos 2x and 2x²-1, we get,
a36
As , there is no point of intersection in $[\frac{π}{2}, π]$ , so therre is no
solution of the given equation in $[\frac{π}{2}, π]$

Question:37

If $\cos^{- 1} x > \sin^{- 1} x$ , then
$A . \frac{1}{\sqrt{2}} < x \leq 1$
$B . 0 \leq x < \frac{1}{\sqrt{2}}$
$C . - 1 \leq x < \frac{1}{\sqrt{2}}$
D.x>0

Answer:

Answer :(C)
Plotting cos^-1 x and sin^-1 x, we get,
a37
As, graph of cos^-1x is above graph of sin^-1 x in $[- 1, \frac{1}{\sqrt{2}})$ .
So, cos^–1x > sin^–1 x for all x in $[- 1, \frac{1}{\sqrt{2}})$ .

Question:38

Fill in the blanks The principle value of $\cos^{- 1} (- \frac{1}{2})$ is ___________.

Answer:

The principal value of $\cos^{- 1} (- \frac{1}{2})$ is $\frac{2 π}{3}$ .
Principal value cos^-1 x is [0, $π$ ]
Let, $\cos^{- 1} (- 1) = θ$
$\Rightarrow \cos θ = - \frac{1}{2}$
As, $\cos \frac{2 π}{3} = - \frac{1}{2}$
So, $θ = \frac{2 π}{3}$

Question:39

Fill in the blanks The value of $s i n^{- 1} (s i n \frac{3 π}{5})$ is_______.

Answer:

The value of $\sin^{- 1} (\sin \frac{3 π}{5})$ is $\frac{2 π}{5}$
Principal value of $\sin^{- 1}$ is $[- \frac{π}{2}, \frac{π}{2}]$
now, $\sin^{- 1} (\sin \frac{3 π}{5})$ should be in the given range
$\frac{3 π}{5}$ is outside the range $[- \frac{π}{2}, \frac{π}{2}]$
As, sin (π – x) = sin x
So, $\sin^{- 1} (\sin \frac{3 π}{5}) = \sin^{- 1} (\sin (π - \frac{3 π}{5}))$
$= \sin^{- 1} (\sin \frac{2 π}{5})$
$= \sin^{- 1} (\sin \frac{2 π}{5}) = \frac{2 π}{5}$

Question:40

Fill in the blanks
If cos (tan^–1x + cot^–1 √3) = 0, then value of x is _________.

Answer:

$\begin{aligned} If \cos (\tan^{- 1} x + \cot^{- 1} \sqrt{3}) = 0, then value of x is \sqrt{3} . \\ Given, \cos (\tan^{- 1} x + \cot^{- 1} \sqrt{3}) = 0 \\ \Rightarrow \tan^{- 1} x + \cot^{- 1} \sqrt{3} = \frac{π}{2} \\ We know that, \tan^{- 1} x + \cot^{- 1} x = \frac{π}{2} \\ So, x = \sqrt{3} \end{aligned}$

Question:40

Fill in the blanks
If cos (tan^–1x + cot^–1 √3) = 0, then value of x is _________.

Answer:

If cos (tan^–1x + cot^–1 √3) = 0, then value of x is $\sqrt{3}$
Given, cos (tan^–1x + cot^–1 $\sqrt{3}$ ) = 0
$\Rightarrow \tan^{- 1} x + \cot^{- 1} \sqrt{3} = \frac{π}{2}$
we know that, $\Rightarrow \tan^{- 1} x + \cot^{- 1} x = \frac{π}{2}$
so, x= $\sqrt{3}$

Question:41

fill in blanks the set of value of $\sec^{- 1} (\frac{1}{2})$ is______________.
Answer:

Fill in the blanks the set of value of $\sec^{- 1} (\frac{1}{2})$ is $ϕ$
Domain of sec^-1 x is R – (-1,1).
As, $- \frac{1}{2}$ is outside domain of sec^-1 x.
Which means there is no set of value of $\sec^{- 1} \frac{1}{2}$
So, the solution set of $\sec^{- 1} \frac{1}{2}$ is null set or $ϕ$

Question:42

Fill in the blanks
The principal value of tan^–1 √3 is _________.

Answer:

The Principal value of $\tan^{- 1} \sqrt{3}$ is $\frac{π}{3}$
Principal value of tan^-1 x is $(- \frac{π}{2}, \frac{π}{2})$
Let, $\tan^{- 1} (\sqrt{3}) = θ$
$\Rightarrow \tan θ = \sqrt{3}$
As $\Rightarrow \tan \frac{π}{3} = \sqrt{3}$
so, $\Rightarrow θ = \sqrt{3}$

Question:43

The value of $\cos^{- 1} (\cos \frac{14 π}{3})$

Answer:

The value of $\cos^{- 1} (\cos \frac{14 π}{3})$ is $\frac{2 π}{3}$
We needd, $\cos^{- 1} (\cos \frac{14 π}{3})$
Principal value of cos^-1 x is [0,π]
Also, cos (2nπ + θ) = cos θ for all n ? N
$\cos \frac{14 π}{3} = \cos (4 π + \frac{2 π}{3})$
$\Rightarrow \cos \frac{14 π}{3} = \cos \frac{2 π}{3}$
$S o, \cos^{- 1} (\cos \frac{14 π}{3}) = \cos^{- 1} (\cos \frac{2 π}{3})$
$\Rightarrow \cos^{- 1} (\cos \frac{14 π}{3}) = \frac{2 π}{3}$

Question:44

Fill in the blanks
The value of cos (sin^–1x + cos^–1 x), |x| ≤ 1 is ________.

Answer:

The value of cos (sin^–1 x + cos^–1 x) for |x| ≤ 1 is 0.
cos (sin^–1 x + cos^–1 x), |x| ≤ 1
We know that, (sin^–1 x + cos^–1 x), |x| ≤ 1 is $\frac{π}{2}$
So, $\cos (\sin^{- 1} x + \cos^{- 1} x) = \cos \frac{π}{2}$
= 0

Question:45

The value of expression $\tan (\frac{\sin^{- 1} x + \cos^{- 1} x}{2}),$ when $x = \frac{\sqrt{3}}{2}$ is___________.

Answer:

The value of expression $\tan (\frac{\sin^{- 1} x + \cos^{- 1} x}{2})$ When $X = \frac{\sqrt{3}}{2}$ is 1
$\tan (\frac{\sin^{- 1} x + \cos^{- 1} x}{2})$ When $X = \frac{\sqrt{3}}{2}$
We know that, (sin^–1 x + cos^–1 x) for all |x| ≤ 1 is $\frac{π}{2}$
As, $x = \frac{\sqrt{3}}{2}$ lies in domain
So $\tan (\frac{\sin^{- 1} x + \cos^{- 1} x}{2})$ = $\tan \frac{π}{4}$
=1

Question:46

Fill in the blanks if $y = 2 \tan^{- 1} x + \sin^{- 1} \frac{2 x}{1 + x^{2}}$ for all x, then ______<y<_____.

Answer:

Fill in the blanks if $y = 2 \tan^{- 1} x + \sin^{- 1} \frac{2 x}{1 + x^{2}}$ for all x, then $- 2 π < y < 2 π$
$y = 2 \tan^{- 1} x + \sin^{- 1} \frac{2 x}{1 + x^{2}}$
We know that,
$2 \tan^{- 1} p = \sin^{- 1} \frac{2 x}{1 + x^{2}}$
so
$2 \tan^{1} x + \sin^{- 1} \frac{2 x}{1 + x^{2}} = 2 \tan^{1} x + 2 \tan^{- 1} x$
=4 tan^-1 x
So, y = 4 tan^-1 x
As, principal value of tan-1 x is $(- \frac{π}{2}, \frac{π}{2})$
So, $4 \tan^{- 1} x ϵ (- 2 π, 2 π)$
Hence, -2π < y < 2π

Question:47

The result $\tan^{- 1} x - \tan^{- 1} (\frac{x - y}{1 + x y})$ is true when value of xy is _________.

Answer:

The result $\tan^{- 1} x - \tan^{- 1} (\frac{x - y}{1 + x y})$ is true when value of xy is > -1.
We have,
$\tan^{- 1} x - \tan^{- 1} = \tan^{- 1} \frac{x - y}{1 + x y}$
Principal range of tan^-1a is $(- \frac{π}{2}, \frac{π}{2})$
Let tan^-1x = A and tan^-1y = B … (1)
So, A,B $ϵ (- \frac{π}{2}, \frac{π}{2})$
We know that, $\tan (A - B) = \frac{\tan A - \tan B}{1 - \tan A \tan B}$ … (2)
From (1) and (2), we get,
Applying, tan^-1 both sides, we get,
$\tan^{- 1} \tan (A - B) = \tan^{- 1} \frac{x - y}{1 - x y}$
As, principal range of tan^-1a is $(- \frac{π}{2}, \frac{π}{2})$
So, for tan^-1tan(A-B) to be equal to A-B,
A-B must lie in $(- \frac{π}{2}, \frac{π}{2})$ – (3)
Now, if both A,B < 0, then A, B $ϵ (- \frac{π}{2}, 0)$
∴ A $ϵ (- \frac{π}{2}, 0)$ and -B $ϵ (0, \frac{π}{2})$
So, A – B $ϵ (- \frac{π}{2}, \frac{π}{2})$
So, from (3),
tan^-1tan(A-B) = A-B
$\Rightarrow \tan^{- 1} x - \tan^{- 1} y = \tan^{- 1} \frac{x - y}{z + x y}$
Now, if both A,B > 0, then A, B $ϵ (0, \frac{π}{2})$
∴ A $ϵ (0, \frac{π}{2})$ and -B $ϵ (- \frac{π}{2}, 0)$
So, A – B $ϵ (- \frac{π}{2}, \frac{π}{2})$
So, from (3),
tan^-1tan(A-B) = A-B
$\Rightarrow \tan - 1 x - \tan - 1 y = \tan^{- 1} \frac{x - y}{z + x y}$
Now, if A > 0 and B < 0,
Then, A $ϵ (0, \frac{π}{2})$ and B $ϵ (0, \frac{π}{2})$
∴ A $ϵ (0, \frac{π}{2})$ and -B $ϵ (0, \frac{π}{2})$
So, A – B $ϵ$ (0,π)
But, required condition is A – B $ϵ$ $(- \frac{π}{2}, \frac{π}{2})$
As, here A – B $ϵ$ (0,π), so we must have A – B $ϵ$ $(0, \frac{π}{2})$
$A - B < \frac{π}{2}$
$A < \frac{π}{2} + B$
Applying tan on both sides,
$\tan A < \tan (\frac{π}{2} + B)$
As, $\tan (\frac{π}{2} + α) = - \cot α$
So, tan A < - cot B
Again, $\cot α = \frac{1}{\tan α}$
So, $\tan A < \frac{1}{\tan B}$
⇒ tan A tan B < -1
As, tan B < 0
xy > -1
Now, if A < 0 and B > 0,
Then, A $ϵ$ $(- \frac{π}{2}, 0)$ and B $ϵ$ $(0, \frac{π}{2})$
∴ A $ϵ$ $(- \frac{π}{2}, 0)$ and -B $ϵ$ $(- \frac{π}{2}, 0)$
So, A – B $ϵ$ (-π,0)
But, required condition is A – B $ϵ$ $(- \frac{π}{2}, \frac{π}{2})$
As, here A – B $ϵ$ (0,π), so we must have A – B $ϵ$ $(- \frac{π}{2}, 0)$
$\Rightarrow A - B > - \frac{π}{2}$
$\Rightarrow A > B - \frac{π}{2}$
Applying tan on both sides,
$\tan A > \tan (B - \frac{π}{2})$
As, $\tan (α - \frac{π}{2}) = - \cot α$
So, tan B > - cot A
Again, $\cot α \frac{1}{\tan α}$
So, $\tan B > - \frac{1}{\tan A}$
⇒ tan A tan B > -1
⇒xy > -1

Question:48

Fill in the blanks
The value of cot^–1(–x) for all x ? R in terms of cot^–1 x is _______.

Answer:

The value of cot^–1(–x) for all x ? R in terms of cot^–1 x is
π – cot^-1 x.
Let cot^–1(–x) = A
⇒ cot A = -x
⇒ -cot A = x
⇒ cot (π – A) = x
⇒ (π – A) = cot^-1 x
⇒ A = π – cot-1 x
So, cot^–1(–x) = π – cot-1 x

Question:49

State True or False for the statement
All trigonometric functions have inverse over their respective domains.

Answer:

True.
It is well known that all trigonometric functions have inverse over their respective domains.

Question:50

State True or False for the statement
The value of the expression (cos^–1x)²is equal to sec² x.

Answer:

As, cos^-1 x is not equal to sec x. So, (cos^–1x)² is not equal to sec² x.

Question:51

State True or False for the statement
The domain of trigonometric functions can be restricted to any one of their branch (not necessarily principal value) in order to obtain their inverse functions.

Answer:

As, all trigonometric and their corresponding inverse functions are periodic so, we can obtain the inverse of a trigonometric ratio in any branch in which it is one-one and onto.

Question:52

State True or False for the statement
The least numerical value, either positive or negative of angle θ is called the principal value of the inverse trigonometric function.

Answer:

True
We know that the smallest value, either positive or negative of angle θ is called principal value of the inverse trigonometric function

Question:53

State True or False for the statement
The graph of an inverse trigonometric function can be obtained from the graph of their corresponding trigonometric function by interchanging x and y axes.

Answer:

True.
Graph of any inverse function can be obtained by interchanging x and y-axis in the graph of the corresponding function. If (p, q) are two points on f(x) then (q, p) will be on f^-1(x).

Question:54

State True or False for the statement
The minimum value of n for which $\tan^{- 1} \frac{n}{π} > \frac{π}{4}, n ϵ N$ is valid is 5.

Answer:

false
$\tan^{- 1} \frac{n}{π} > \frac{π}{4}$
As , tan is an increasing function so applying tan on both side
we get,
$\tan (\tan^{- 1} \frac{n}{π}) > \tan \frac{π}{4}$
As, $\tan (\tan^{- 1} \frac{n}{π}) = \frac{n}{π}$ and $\tan \frac{π}{4} = 1$
so $\frac{n}{π} > 1$
⇒ n > π
⇒ n > 3.14
As, n is a natural number, so least value of n is 4.

Question:55

State True or False for the statement

The principal value of $\sin^{- 1} [\cos (\sin^{- 1} \frac{1}{2})]$ is $\frac{π}{3}$

Answer:

True
Principal value of sin^-1 x is $[- \frac{π}{2}, \frac{π}{2}]$
Principal value of cos^-1 x is [0, π]
We have, $\sin^{- 1} [\cos [\sin^{- 1} (\frac{1}{2})]]$
As, $\sin \frac{π}{6} = \frac{1}{2}$ so
$\sin^{- 1} [\cos [\sin^{- 1} (\frac{1}{2})]] = \sin^{- 1} [\cos [\sin^{- 1} (\sin \frac{π}{6})]]$
$\Rightarrow \sin^{- 1} [\cos [\sin^{- 1} (\frac{1}{2})]] = \sin^{- 1} [\cos [\frac{π}{6}]]$
As, $\cos \frac{π}{6} = \frac{\sqrt{3}}{2}$ so,
$\Rightarrow \sin^{- 1} [\cos [\sin^{- 1} (\frac{1}{2})]] = \sin^{- 1} [\sin [\frac{π}{3}]]$
$\Rightarrow \sin^{- 1} [\cos [\sin^{- 1} (\frac{1}{2})]] = \frac{π}{3}$

Sub-Topics Covered in NCERT Exemplar Class 12 Maths Solutions Chapter 2

The sub-topics that are covered in this chapter of inverse trigonometric functions are:

Introduction
Basic concepts
Properties of inverse trigonometric functions

JEE Main Highest Scoring Chapters & Topics

Just Study 40% Syllabus and Score upto 100%

Download EBook

NCERT Exemplar Class 12 Maths Solutions

NCERT Exemplar Class 12 Mathematics Chapter 1: Relations and Functions

NCERT Exemplar for Class 12 Mathematics Chapter 2: Inverse Trigonometric Functions

NCERT Exemplar Class 12 Mathematics Chapter 3: Matrices

NCERT Exemplar Class 12 Mathematics Chapter 4: determinants

NCERT Exemplar Class 12 Mathematics Chapter 5: Continuity and Differentiability

NCERT Exemplar Class 12 Mathematics Chapter 6: Applications of derivatives

NCERT Exemplar Class 12 Mathematics Chapter 7: Integrals

NCERT Exemplar Class 12 Mathematics Chapter 8: Applications of Integrals

NCERT Exemplar Class 12 Mathematics Chapter 9: Differential equations

NCERT Exemplar Class 12 Mathematics Chapter 10: Vector Algebra

NCERT Exemplar Class 12 Mathematics Chapter 11: Three Dimensional Geometry

NCERT Exemplar Class 12 Mathematics Chapter 12: Linear Programming

NCERT Exemplar Class 12 Mathematics Chapter 13: Probability

Importance of NCERT Exemplar Class 12 Maths Solutions Chapter 2

Class 12 Math NCERT Exemplar solutions chapter 2, students will get detailed answers to the questions in the NCERT book after every topic. Understanding and grasping this chapter can help one aim for a better score in their school exams, boards and their entrance exams.

In NCERT Exemplar solutions for Class 12 Math chapter 2, cover properties and graphical representations of inverse trigonometric functions.
One will learn about the necessity of studying inverse trigonometric functions and their properties. It covers the basic details about inverse trigonometric functions.
These solutions provide plenty of questions to practice.

JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

Get now

NCERT Solutions for Class 12 Maths: Chapter Wise

NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions

NCERT solutions for class 12 maths chapter 3 Matrices

NCERT solutions for class 12 maths chapter 4 Determinants

NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability

NCERT solutions for class 12 maths chapter 6 Application of Derivatives

NCERT solutions for class 12 maths chapter 7 Integrals

NCERT solutions for class 12 maths chapter 8 Application of Integrals

NCERT solutions for class 12 maths chapter 9 Differential Equations

NCERT solutions for class 12 maths chapter 10 Vector Algebra

NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry

NCERT solutions for class 12 maths chapter 12 Linear Programming

NCERT solutions for class 12 maths chapter 13 Probability

NCERT solutions of class 12 - Subject-wise

Here are the subject-wise links for the NCERT solutions of class 12:

NCERT Notes of class 12 - Subject Wise

Given below are the subject-wise NCERT Notes of class 12 :

NCERT Books and NCERT Syllabus

Here are some useful links for NCERT books and the NCERT syllabus for class 12:

NCERT Exemplar Class 12 Solutions - Subject Wise

Given below are the subject-wise exemplar solutions of class 12 NCERT:

NCERT Exemplar Class 12 Chemistry Solutions

NCERT Exemplar Class 12 Physics Solutions

NCERT Exemplar Class 12 Biology Solutions

Frequently Asked Questions (FAQs)

1. What are the important topics of this chapter?

Introduction to Inverse Trigonometric Functions, The Basic Concepts of Inverse Trigonometric Functions and Properties of Inverse Trigonometric Functions are important topics of this chapter.

2. Are these solutions helpful for board examinations?

Yes, the NCERT exemplar Class 12 Maths chapter 2 solutions are helpful for you to prepare for board exams.

3. How many questions are there in this chapter?

There is only 1 exercise in this chapter with 55 problem solving questions.

4. Are these solutions helpful for competitive examinations?

Yes, NCERT exemplar solutions for Class 12 Maths chapter 2 cover syllabus for exams are very reliable for preparing for competitive entrance exams like NEET and JEE Main.

Also Read

NCERT Book Class 12 English 2025-26 Free PDF Download

NCERT Book Class 12 Physics 2025-26 FREE PDF Download

NCERT Book Class 12 Maths 2025-26 PDF Free Download

NCERT Books for Class 12 Biology 2025-26: Download Free PDF

NCERT Books for Class 12 Hindi 2025-26; Download Chapter Wise PDF here

NCERT Books for Class 12 - Download All Subjects PDFs Here

NCERT Syllabus 2025-26 for Class 6-12 - Download All Subjects PDF Free

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry

NCERT Solutions for Class 12 Physics Chapter 7 - Alternating Current

NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12 - Integrals

NCERT Solutions for Exercise 4.1 Class 11 Maths Chapter 4 - Principle of Mathematical Induction

Dual nature of radiation and matter Class 12th Notes - Free NCERT Class 12 Physics Chapter 11 Notes - Download PDF

Quadratic Equations Class 10th Notes - Free NCERT Class 10 Maths Chapter 4 Notes - Download PDF

NCERT Class 12th Maths Chapter 5 Continuity and Differentiability Notes

NCERT Class 12 Physics Chapter 4 Notes, Moving Charges and Magnetism Class 12 Chapter 4 Notes

NCERT Class 12 Physics Chapter 3 Notes, Current Electricity Class 12 Chapter 3 Notes

Atoms Class 12th Notes - Free NCERT Class 12 Physics Chapter 12 Notes - Download PDF

NCERT Exemplar Class 12 Physics Solutions Chapter 8 Electromagnetic Waves

Articles

CBSE Class 12 Preparation Tips for Computer Science 2024-25

May 17, 2025

Himachal Pradesh 12th Class Result 2025 Declared, Check hpbose.org 12th Results Here

May 17, 2025

HP Board 12th Result 2025 Out; Download Link, Check HPBOSE Class 12th Results Here

May 17, 2025

HPBOSE HP Board Result 2025 Out: Check Class 10th, 12th Results @hpbose.org

May 17, 2025

HP Board 12th Revaluation & Rechecking 2025, Check Dates, Fees, Apply Online at hpbose.org

May 17, 2025

CBSE Class 12 English Preparation Tips 2025 - Tips to Score Maximum Marks

May 17, 2025

CBSE Sports Calendar 2025-26: Cluster, Zonal, and National level Games Dates

May 16, 2025

CBSE Sports Registration 2025-26, Games Login, Portal, Schedule, Venue

May 16, 2025

Upcoming School Exams

National Institute of Open Schooling 10th examination

Admit Card Date:02 April,2025 - 19 May,2025

Result

Application Process

Exam Pattern

National Institute of Open Schooling 12th Examination

Admit Card Date:02 April,2025 - 19 May,2025

Application Process

Exam Pattern

Admit Card

Andhra Pradesh Board of Intermediate Education Examination

Admit Card Date:06 May,2025 - 20 May,2025

Exam Pattern

Mock Test

Admit Card

Andhra Pradesh Board of Intermediate Education (Year 1) Exam

Admit Card Date:06 May,2025 - 20 May,2025

Result

Admit Card

Maharashtra Higher Secondary School Certificate Examination

Application Date:07 May,2025 - 17 May,2025

Exam Pattern

Result

Mock Test

View All School Exams

Certifications By Top Providers

Full-Stack Web Development

Via Masai School

Digital Banking Business Model

Via State Bank of India

Google Artificial Intelligence for JavaScript Developers with TensorFlow JavaScript

Via Google

Introduction to Psychology

Via San Jose State University, San Jose

Banking 101 A Guide for Beginners in the Banking Sector

Via EduBridge

Forensic Science DNA Analysis

Via University of Cambridge, Cambridge

1114 courses

818 courses

394 courses

294 courses

Harvard University, Cambridge

98 courses

University of Michigan, Ann Arbor

84 courses

IT & Software

Teaching and Academics

Programming and Development

Health, Fitness and Medicine

Business and Management

Engineering and Architecture

General Management

Financial Management

Environmental Studies

History

Psychology

Artificial Intelligence

Explore Top Universities Across Globe

University of Essex, Colchester

Wivenhoe Park Colchester CO4 3SQ

University College London, London

Gower Street, London, WC1E 6BT

The University of Edinburgh, Edinburgh

Old College, South Bridge, Edinburgh, Post Code EH8 9YL

University of Bristol, Bristol

Beacon House, Queens Road, Bristol, BS8 1QU

University of Delaware, Newark

Newark, DE 19716

University of Nottingham, Nottingham

University Park, Nottingham NG7 2RD

ACT Vocabulary Words You Must Know in 2025

21 min ⋆ May 09, 2025 06:05 AM IST

Tips 5 ACT Preparation Tips 2025 - Section wise Strategies to Score High

7 min ⋆ May 09, 2025 05:05 AM IST

ACT Practice Test 2025 - Download PDF with Answers and Explanations

5 min ⋆ May 09, 2025 05:05 AM IST

MBBS Fees in Bangladesh for Indian Students 2025: Medical Colleges

5 min ⋆ Apr 15, 2025 05:04 AM IST

New German Student Visa Process: Online Portal at digital.diplo.de for Applications

5 min ⋆ May 09, 2025 03:05 AM IST

LSAT Abroad Sample Papers 2025, Question Paper - Download PDF Here!

4 min ⋆ May 09, 2025 02:05 AM IST

Related E-books & Sample Papers

List of Competitive Exams for Class 9 to 12 Students

418+ Downloads

CBSE Class 11, 12 English (Core) Syllabus 2025-26

110+ Downloads

CBSE Class 11, 12 Biology Syllabus 2025-26

222+ Downloads

CBSE Class 11, 12 Physics Syllabus 2025-26

632+ Downloads

CBSE Class 11, 12 Maths Syllabus 2025-26

318+ Downloads

CBSE Class 11, 12 Applied Maths Syllabus 2025-26

35+ Downloads

Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Can I change my board from CBSE to CHSE Odisha in Class 12th?

Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.

Read Complete Answer

Manasvini Gupta 30 January,2025

quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer

Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

Read Complete Answer

Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

Read Complete Answer

Yuvan 30 August,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer

Ashvadha 12 July,2024

View All

Stay up-to date with CBSE Class 12th News

Ask

Question

Student Community: Where Questions Find Answers

Ask and get expert answers on exams, counselling, admissions, careers, and study options.

Download Careers360 App

All this at the convenience of your phone

Regular Exam Updates
Best College Recommendations
College & Rank predictors
Detailed Books and Sample Papers
Question and Answers

Scan and download the app

Central Board of Secondary Education Class 12th Examination

JEE Test Series

Applications Open

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

Exams

Top Schools

Products & Resources

NCERT Study Material

Exams

Colleges

Predictors

Resources

Exams

Colleges & Courses

Predictors

Resources

Exams

Colleges

Predictors

Resources

Exams

Colleges

Predictors & E-Books

Resources

Exams

Predictors & Articles

Colleges

Resources

Exams

Colleges

Resources

Exams

Resources

Top Courses & Careers

Colleges

Exams

Colleges

Resources

Quick Links

Exams

Colleges

Resources

Exams

Colleges

Resources

Diploma Colleges

Exams

Resources

Upcoming Events

Other Exams

Exams

Colleges

Top Countries

Resources

Exams

Colleges

Upcoming Events

Resources

Engineering Preparation

Medical Preparation

Online Courses

Products

Top Streams

Specializations

Resources

Top Providers

NCERT Exemplar Class 12 Maths Solutions Chapter 2 Inverse Trigonometric Functions

NCERT Exemplar Class 12 Maths Solutions Chapter 2

NEET/JEE Coaching Scholarship

JEE Main high scoring chapters and topics

Sub-Topics Covered in NCERT Exemplar Class 12 Maths Solutions Chapter 2

NCERT Exemplar Class 12 Maths Solutions

Importance of NCERT Exemplar Class 12 Maths Solutions Chapter 2

JEE Main Important Mathematics Formulas

NCERT solutions of class 12 - Subject-wise

NCERT Notes of class 12 - Subject Wise

NCERT Books and NCERT Syllabus

NCERT Exemplar Class 12 Solutions - Subject Wise

Frequently Asked Questions (FAQs)

Also Read

Articles

Upcoming School Exams

Certifications By Top Providers

Explore Top Universities Across Globe