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Inverse trigonometric functions are a significant part of Class 12 Mathematics. These inverse trigonometric functions help us find the angle when the value of a function is given. This chapter in NCERT Class 12 Mathematics gives a clear idea of what are inverse trigonometric functions and how they are used to solve various problems with real life applications in determining the unknown angle in navigation, engineering, physics, etc.
This article on NCERT Solutions on Class 12 Chapter 2 Inverse Trigonometric Functions gives step-by-step solutions to all the exercises problems in the NCERT Book. These solutions are highly accurate and reliable as they are given by subject matter experts. For more practice refer to NCERT Exemplar Solutions For Class 12 Maths Chapter 2 Inverse Trigonometric Functions.
Properties of Inverse Trigonometric Functions:
Function | Domain | Range |
y = sin-1(x) | [-1, 1] | [-π/2, π/2] |
y = cos-1(x) | [-1, 1] | [0, π] |
y = cosec-1(x) | R - (-1, 1) | [-π/2, π/2] - {0} |
y = sec-1(x) | R - (-1, 1) | [0, π] - {π/2, π/2} |
y = tan-1(x) | R | (-π/2, π/2) |
y = cot-1(x) | R | (0, π) |
Self-Adjusting Trigonometric Property:
sin(sin-1(x)) = x
sin-1(sin(x)) = x
cos(cos-1(x)) = x
cos-1(cos(x)) = x
tan(tan-1(x)) = x
tan-1(tan(x)) = x
sec(sec-1(x)) = x
sec-1(sec(x)) = x
cosec-1(cosec(x)) = x
cosec(cosec-1(x)) = x
cot-1(cot(x)) = x
cot(cot-1(x)) = x
Reciprocal Relations:
sin-1(1/x) = cosec-1(x), x ≥ 1 or x ≤ -1
cos-1(1/x) = sec-1(x), x ≥ 1 or x ≤ -1
tan-1(1/x) = cot-1(x), x > 0
Even and Odd Functions:
sin-1(-x) = -sin-1(x), x ∈ [-1, 1]
tan-1(-x) = -tan-1(x), x ∈ R
cosec-1(-x) = -cosec-1(x), |x| ≥ 1
cos-1(-x) = π - cos-1(x), x ∈ [-1, 1]
sec-1(-x) = π - sec-1(x), |x| ≥ 1
cot-1(-x) = π - cot-1(x), x ∈ R
Complementary Relations:
sin-1(x) + cos-1(x) = π/2
tan-1(x) + cot-1(x) = π/2
cosec-1(x) + sec-1(x) = π/2
Sum and Difference Formulas:
tan-1(x) + tan-1(y) = tan-1((x+y)/(1-xy))
tan-1(x) - tan-1(y) = tan-1((x-y)/(1+xy))
sin-1(x) + sin-1(y) = sin-1[x√(1-y2)+y√(1-x2)]
sin-1(x) - sin-1(y) = sin-1[x√(1-y2)-y√(1-x2)]
cos-1(x) + cos-1(y) = cos-1[xy-√(1-x2)√(1-y2)]
cos-1(x) - cos-1(y) = cos-1[xy+√(1-x2)√(1-y2)]
cot-1(x) + cot-1(y) = cot-1((xy-1)/(x+y))
cot-1(x) - cot-1(y) = cot-1((xy+1)/(y-x))
Double Angle Formula:
2 tan-1(x) = sin-1(2x/(1+x2))
2 tan-1(x) = cos-1((1-x2)/(1+x2))
2 tan-1(x) = tan-1(2x/(1-x2))
2 sin-1(x) = sin-1(2x√(1+x2))
2 cos-1(x) = sin-1(2x√(1-x2))
Conversion Properties:
sin-1(x) = cos-1(√(1-x2)) = tan-1(x/√(1-x2)) = cot-1(√(1-x2)/x)
cos-1(x) = sin-1(√(1-x2)) = tan-1(√(1-x2)/x) = cot-1(x/√(1-x2))
tan-1(x) = sin-1(x/√(1-x2)) = cos-1(x/√(1+x2)) = sec-1(√(1+x2)) = cosec-1(√(1+x2)/x)
Class 12 Maths chapter 2 solutions Exercise 2.1 Page number: 26-27 Total questions: 14 |
Q1. (i) Find the principal values of the following: $\sin^{-1}\left ( \frac{-1}{2} \right )$
Answer:
Let $x = \sin^{-1}\left ( \frac{-1}{2} \right )$
$\implies \sin x = \frac{-1}{2}= -\sin(\frac{\pi}{6}) = \sin(-\frac{\pi}{6})$
We know, principle value range of $sin^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$
$\therefore$ The principal value of $\sin^{-1}\left ( \frac{-1}{2} \right )$ is $-\frac{\pi}{6},$
Q1. (ii) Find the principal values of the following: $\cos^{-1}\left(\frac{\sqrt3}{2} \right )$
Answer:
So, let us assume that $\cos^{-1}\left(\frac{\sqrt3}{2} \right ) = x$ then,
Taking inverse both sides we get;
$cos\ x = (\frac{\sqrt{3}}{2})$ , or $cos (\frac{\pi}{6}) = (\frac{\sqrt{3}}{2})$
and as we know that the principal values of $cos^{-1}$ is from [0, $\pi$ ],
Hence $cos\ x = (\frac{\sqrt{3}}{2})$ when x = $\frac{\pi}{6}$ .
Therefore, the principal value for $\cos^{-1}\left(\frac{\sqrt3}{2} \right )$ is $\frac{\pi}{6}$ .
Q1. (iii) Find the principal values of the following:$\textup{cosec}^{-1}(2)$
Answer:
Let us assume that $\textup{cosec}^{-1}(2) = x$ , then we have;
$Cosec\ x = 2$ , or
$Cosec( \frac{\pi}{6}) = 2$ .
And we know the range of principal values is $[\frac{-\pi}{2},\frac{\pi}{2}] - \left \{ 0 \right \}.$
Therefore the principal value of $\textup{cosec}^{-1}(2)$ is $\frac{\pi}{6}$ .
Q1. (iv)Find the principal values of the following:$\tan^{-1}(-\sqrt3)$
Answer:
Let us assume that $\tan^{-1}(-\sqrt3) = x$ , then we have;
$\tan x = (-\sqrt 3)$ or
$-\tan (\frac{\pi}{3}) = \tan \left ( \frac{-\pi}{3} \right ).$
and as we know that the principal value of $\tan^{-1}$ is $\left ( \frac{-\pi}{2}, \frac{\pi}{2} \right )$ .
Hence the only principal value of $\tan^{-1}(-\sqrt3)$ when $x = \frac{-\pi}{3}$ .
Q1. (v)Find the principal values of the following: $\cos^{-1}\left(-\frac{1}{2} \right )$
Answer:
Let us assume that $\cos^{-1}\left(-\frac{1}{2} \right ) =y$ then,
Easily we have; $\cos y = \left ( \frac{-1}{2} \right )$ or we can write it as:
$-\cos \left ( \frac{\pi}{3} \right ) = \cos \left ( \pi - \frac{\pi}{3} \right ) = \cos \left ( \frac{2\pi}{3} \right ).$
as we know that the range of the principal values of $\cos^{-1}$ is $\left [ 0,\pi \right ]$
Hence $\frac{2\pi}{3}$ lies in the range it is a principal solution.
Q1. (vi)Find the principal values of the following $\tan^{-1}(-1)$
Answer:
Given $\tan^{-1}(-1)$we can assume it to be equal to 'z';
$\tan^{-1}(-1) =z$ ,
$\tan z = -1$
or
$-\tan (\frac{\pi}{4}) = \tan(\frac{-\pi}{4})= -1$
And as we know the range of principal values of $\tan^{-1}$ from $\left ( \frac{-\pi}{2}, \frac{\pi}{2} \right )$ .
As only one value z = $-\frac{\pi}{4}$ lies hence we have only one principal value that is $-\frac{\pi}{4}$ .
Q1. (vii)Find the principal values of the following: $\sec^{-1}\left (\frac{2}{\sqrt3}\right)$
Answer:
Let us assume that $\sec^{-1}\left (\frac{2}{\sqrt3}\right) = z$ then,
we can also write it as; $\sec z = \left (\frac{2}{\sqrt3}\right)$ .
Or $\sec (\frac{\pi}{6}) = \left (\frac{2}{\sqrt3}\right)$ and the principal values lies between $\left [ 0, \pi \right ] - \left \{ \frac{\pi}{2} \right \}$ .
Hence we get only one principal value of $\sec^{-1}\left (\frac{2}{\sqrt3}\right)$ i.e., $\frac{\pi}{6}$ .
Q1. (viii) Find the principal values of the following: $\cot^{-1}(\sqrt3)$
Answer:
Let us assume that $\cot^{-1}(\sqrt3) = x$ , then we can write in other way,
$\cot x = (\sqrt3)$ or
$\cot (\frac{\pi}{6}) = (\sqrt3)$ .
Hence when $x=\frac{\pi}{6}$ we have $\cot (\frac{\pi}{6}) = (\sqrt3)$ .
and the range of principal values of $\cot^{-1}$ lies in $\left ( 0, \pi \right )$ .
Then the principal value of $\cot^{-1}(\sqrt3)$ is $\frac{\pi}{6}$
Q1. (ix) Find the principal values of the following: $\cos^{-1}\left(-\frac{1}{\sqrt2} \right )$
Answer:
Let us assume $\cos^{-1}\left(-\frac{1}{\sqrt2} \right ) = x$ ;
Then we have $\cos x = \left ( \frac{-1}{\sqrt 2} \right )$
or
$-\cos (\frac{\pi}{4}) = \left ( \frac{-1}{\sqrt 2} \right )$ ,
$\cos (\pi - \frac{\pi}{4}) = \cos (\frac{3\pi}{4})$ .
And we know the range of principal values of $\cos^{-1}$ is $[0,\pi]$.
So, the only principal value which satisfies $\cos^{-1}\left(-\frac{1}{\sqrt2} \right ) = x$ is $\frac{3\pi}{4}$ .
Q1. (x) Find the principal values of the following: $\textup{cosec}^{-1}(-\sqrt2)$
Answer:
Let us assume the value of $\textup{cosec}^{-1}(-\sqrt2) = y$ , then
we have $cosec\ y = (-\sqrt 2)$or
$-cosec\ (\frac{\pi}{4}) = (-\sqrt 2) = cosec\ (\frac{-\pi}{4})$ .
and the range of the principal values of $\textup{cosec}^{-1}$ lies between $\left [ \frac{-\pi}{2},\frac{\pi}{2} \right ] - \left \{ 0 \right \}$ .
hence the principal value of $\textup{cosec}^{-1}(-\sqrt2)$ is $\frac{-\pi}{4}$ .
Question:11 Find the values of the following: $\tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2} \right ) + \sin^{-1}\left(-\frac{1}{2} \right )$
Answer:
To find the values first we declare each term to some constant ;
$tan^{-1}(1) = x$ , So we have $\tan x = 1$ ;
or $\tan (\frac{\pi}{4}) = 1$
Therefore, $x = \frac{\pi}{4}$
$cos^{-1}(\frac{-1}{2}) = y$
So, we have
$\cos y = \left ( \frac{-1}{2} \right ) = -\cos \left ( \frac{\pi}{3} \right ) = \cos(\pi - \frac{\pi}{3}) = \cos \left ( \frac{2\pi}{3} \right )$ .
Therefore $y = \frac{2\pi}{3}$ ,
$\sin^{-1}(\frac{-1}{2}) = z$ ,
So we have;
$\sin z = \frac{-1}{2}$ or $-\sin (\frac{\pi}{6}) =\sin (\frac{-\pi}{6}) = \frac{-1}{2}$
Therefore $z = -\frac{\pi}{6}$
Hence we can calculate the sum:
$= \frac{\pi}{4}+\frac{2\pi}{3}-\frac{\pi}{6}$
$=\frac{3\pi + 8\pi -2\pi}{12} = \frac{9\pi}{12}=\frac{3\pi}{4}$ .
Question:12 Find the values of the following: $\cos^{-1}\left(\frac{1}{2} \right ) + 2\sin^{-1}\left(\frac{1}{2} \right )$
Answer:
Here we have $\cos^{-1}\left(\frac{1}{2} \right ) + 2\sin^{-1}\left(\frac{1}{2} \right )$
let us assume that the value of
$\cos^{-1}\left ( \frac{1}{2} \right ) = x, \:and\:\sin^{-1}\left(\frac{1}{2} \right ) = y$ ;
then we have to find out the value of+2y.
Calculation of x :
$\Rightarrow \cos^{-1}\left ( \frac{1}{2} \right ) = x$
$\Rightarrow \cos x = \frac{1}{2}$
$\Rightarrow \cos \frac{\pi}{3} = \frac{1}{2}$ ,
Hence $x = \frac{\pi}{3}$ .
Calculation of y :
$\Rightarrow \sin^{-1}\left(\frac{1}{2} \right ) = y$
$\Rightarrow \sin y = \frac{1}{2}$
$\Rightarrow \sin \frac{\pi}{6} = \frac{1}{2}$ .
Hence $y = \frac{\pi}{6}$ .
The required sum will be = $\frac{\pi}{3}+2(\frac{\pi}{6}) = \frac{2\pi}{3}$ .
Question:13 If $\sin^{-1}x = y$ then
(B) $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$
(D) $-\frac{\pi}{2} < y < \frac{\pi}{2}$
Answer:
Given if $\sin^{-1}x = y$ then,
As we know that the $\sin^{-1}$ can take values between $\left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ].$
Therefore, $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$ .
Hence answer choice (B) is correct.
Question:14 $\tan^{-1}(\sqrt3)-\sec^{-1}(-2)$ is equal to
Answer:
Let us assume the values of $\tan^{-1}(\sqrt3)$ be 'x' and $\sec^{-1}(-2)$ be 'y'.
Then we have;
$\tan^{-1}(\sqrt3) = x$ or $\tan x = \sqrt 3$ or $\tan \frac{\pi}{3} = \sqrt 3$ or
$x = \frac{\pi}{3}$ .
and $\sec^{-1}(-2) = y$or$\sec y = -2$
or $-\sec (\frac{\pi}{3}) =\sec ({\pi - \frac{\pi}{3}}) = \sec{\frac{2\pi}{3}}$
$y = \frac{2\pi}{3}$
also, the ranges of the principal values of $\tan^{-1}$ and $\sec^{-1}$ are $(\frac{-\pi}{2},\frac{\pi}{2})$ . and
$[0,\pi] - \left \{ \frac{\pi}{2} \right \}$ respectively.
$\therefore$ we have then;
$\tan^{-1}(\sqrt3)-\sec^{-1}(-2)$
$= \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3}$
Class 10 Maths chapter 2.2 solutions Exercise 2.2 Page number: 29-30 Total questions: 15 |
Question:1 Prove the following: $3\sin^{-1}x = \sin^{-1}(3x - 4x^3),\;\;x\in\left[-\frac{1}{2},\frac{1}{2} \right ]$
Answer:
Given to prove: $3\sin^{-1}x = \sin^{-1}(3x - 4x^3)$
where, $x\:\epsilon \left[-\frac{1}{2},\frac{1}{2} \right ]$ .
Take $\theta= \sin ^{-1}x$ or $x = \sin \theta$
Take R.H.S value
$\sin^{-1}(3x - 4x^3)$
= $\sin^{-1}(3\sin \theta - 4\sin^3 \theta)$
= $\sin^{-1}(\sin 3\theta)$
= $3\theta$
= $3\sin^{-1}x$ = L.H.S
Question:2 Prove the following: $3\cos^{-1} x = \cos^{-1}(4x^3 - 3x), \;\;x\in\left[\frac{1}{2},1 \right ]$
Answer:
Given to prove $3\cos^{-1} x = \cos^{-1}(4x^3 - 3x), \;\;x\in\left[\frac{1}{2},1 \right ]$ .
Take $\cos^{-1}x = \theta$ or $\cos \theta = x$ ;
Then we have;
R.H.S.
$\cos^{-1}(4x^3 - 3x)$
= $\cos^{-1}(4\cos^3 \theta - 3\cos\theta)$ $\left [ \because 4\cos^3 \theta - 3\cos\theta = \cos3 \theta \right ]$
= $\cos^{-1}(\cos3\theta)$
= $3\theta$
= $3\cos^{-1}x$ = L.H.S
Hence Proved.
Question:3 Write the following functions in the simplest form: $\tan^{-1}\frac{\sqrt{1 + x^2}- 1}{x},\;\;x\neq 0$
Answer:
We have $\tan^{-1}\frac{\sqrt{1 + x^2}- 1}{x}$
Take
$\therefore$ $\tan^{-1} \frac {\sqrt{1+x^2} - 1}{x} = \tan^{-1}\frac{\sqrt{1+\tan^2 \Theta - 1}}{\tan \Theta}$
$=\tan^{-1}(\frac{sec \Theta-1}{tan \Theta}) = \tan^{-1}\left ( \frac{1-cos \Theta}{sin \Theta} \right )$
$=\tan^{-1}\left ( \frac {2sin^2\left ( \frac{\Theta}{2} \right )}{2sin\frac{\Theta}{2}cos\frac{\Theta}{2}} \right )$
$=\tan^{-1}\left ( \tan\frac{\Theta}{2} \right ) = \frac{\Theta}{2} =\frac{1}{2}\tan^{-1}x$
$=\frac{1}{2}\tan^{-1}x$ is the simplified form.
Question:4 Write the following functions in the simplest form: $\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1 + \cos x}} \right ),\;\; 0< x < \pi$
Answer:
Given that $\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1 + \cos x}} \right ),\;\; 0< x < \pi$
We have in inside the root the term : $\frac{1-\cos x}{1 + \cos x}$
Put $1-\cos x = 2\sin^2\frac{x}{2}$ and $1+\cos x = 2\cos^2\frac{x}{2}$ ,
Then we have,
$=\tan^{-1}\left(\sqrt{\frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}}} \right )$
$=\tan^{-1}\left( \frac{\sin \frac{x}{2}}{\cos\frac{x}{2}} \right )$
$=\tan^{-1}(\tan\frac{x}{2}) = \frac{x}{2}$
Hence the simplest form is $\frac{x}{2}$
Question:5 Write the following functions in the simplest form: $\tan^{-1}\left(\frac{\cos x -\sin x }{\cos x + \sin x} \right ),\;\; \frac{-\pi}{4} < x < \frac{3\pi}{4}$
Answer:
Given $\tan^{-1}\left(\frac{\cos x -\sin x }{\cos x + \sin x} \right )$ where $x\:\epsilon\:( \frac{-\pi}{4} < x < \frac{3\pi}{4})$
So,
$=\tan^{-1}\left(\frac{\cos x -\sin x }{\cos x + \sin x} \right )$
Taking $\cos x$ common from numerator and denominator.
We get:
$=\tan^{-1}\left(\frac{1 -(\frac{\sin x}{\cos x}) }{1+(\frac{\sin x}{\cos x}) } \right )$
$=\tan^{-1}\left(\frac{1 - \tan x }{1+\tan x } \right )$
= $\tan^{-1}(1) - \tan^{-1}(\tan x)$ as, $\left [ \because \tan^{-1}x - \tan^{-1}y = \frac{x - y}{1 + xy} \right ]$
= $\frac{\pi}{4} - x$ is the simplest form.
Question:6 Write the following functions in the simplest form: $\tan^{-1} \frac{x}{\sqrt{a^2 - x^2}}, \;\; |x| < a$
Answer:
Given that $\tan^{-1} \frac{x}{\sqrt{a^2 - x^2}}, \;\; |x| < a$
Take $x = a\sin \theta$ or
$\theta = \sin^{-1}\left ( \frac{x}{a} \right )$ and putting it in the equation above;
$\tan^{-1} \frac{a\sin \theta}{\sqrt{a^2 - (a\sin \theta)^2}}$
$=\tan^{-1} \frac{a\sin \theta}{a\sqrt{1 - \sin^2 \theta}}$
$=\tan^{-1} \left ( \frac{\sin \theta}{\sqrt{\cos^2 \theta}} \right ) = \tan^{-1} \left ( \frac{\sin \theta}{{\cos \theta}} \right )$
$=\tan^{-1}\left ( \tan \theta \right )$
$=\theta = \sin^{-1}\left ( \frac{x}{a} \right )$is the simplest form.
Question:7Write the following functions in the simplest form: $\tan^{-1}\left(\frac{3a^2x -x^3}{a^3 - 3ax^2} \right ),\;\;a>0\;\;;\;\;\frac{-a}{\sqrt3} < x < \frac{a}{\sqrt3}$
Answer:
Given $\tan^{-1}\left(\frac{3a^2x -x^3}{a^3 - 3ax^2} \right )$
Here we can take $x = a\tan \theta \Rightarrow \frac{x}{a} = \tan \theta$
So, $\theta = \tan^{-1}\left ( \frac{x}{a} \right )$
$\tan^{-1}\left(\frac{3a^2x -x^3}{a^3 - 3ax^2} \right )$ will become;
$=\tan^{-1}\left(\frac{3a^2a\tan \theta -(a\tan \theta)^3}{a^3 - 3a(a\tan \theta)^2} \right ) = \tan^{-1}\left(\frac{3a^3\tan \theta -a^3\tan ^3 \theta}{a^3 - 3a^3\tan ^2 \theta} \right )$
and as $\left [ \because \left(\frac{3\tan \theta -\tan ^3 \theta}{ 1- 3\tan ^2 \theta} \right) =\tan 3\theta \right ]$ ;
$=3 \theta$
$=3 \tan^{-1}(\frac{x}{a})$
hence the simplest form is $3 \tan^{-1}(\frac{x}{a})$ .
Question:8 Find the values of each of the following: $\tan^{-1}\left[2\cos\left(2\sin^{-1}\frac{1}{2} \right ) \right ]$
Answer:
Given equation:
$\tan^{-1}\left[2\cos\left(2\sin^{-1}\frac{1}{2} \right ) \right ]$
So, solving the inner bracket first, we take the value of $\sin x^{-1} \frac{1}{2} = x.$
Then we have,
$\sin x = \frac{1}{2} = \sin \left ( \frac{\pi}{6} \right )$
Therefore, we can write $\sin^{-1} \frac{1}{2} = \frac{\pi}{6}$ .
$\tan^{-1}\left[2\cos\left(2\sin^{-1}\frac{1}{2} \right ) \right ] = \tan^{-1}\left[2\cos\left(2\times\frac{\pi}{6} \right ) \right ]$
$= \tan^{-1}\left[2\cos\left(\frac{\pi}{3} \right ) \right ] = \tan^{-1}\left[2\times\left(\frac{1}{2} \right ) \right ] = \tan^{-1}1 = \frac{\pi}{4}$ .
Question:9 Find the values of each of the following: $\tan \frac{1}{2}\left[\sin^{-1}\frac{2x}{1+x^2} + cos^{-1}\frac{1-y^2}{1+y^2} \right ],\;\;|x|<1,\;y>0$ and $xy<1$
Answer:
Taking the value $x = \tan \Theta$or$\tan^{-1}x = \Theta$ and $y = \tan \Theta$or$\tan^{-1} y = \Theta$ then we have,
= $\tan \frac{1}{2}\left[\sin^{-1}\frac{2\tan \Theta}{1+(\tan \Theta)^2} + cos^{-1}\frac{1-\tan^2 \Theta}{1+(\tan \Theta)^2} \right ]$ ,
= $\tan \frac{1}{2}\left[\sin^{-1}(\sin2\Theta) + cos^{-1} (\cos 2\Theta) \right ]$
$\because \left[\cos^{-1}(\frac{1-\tan^2 \Theta}{1+ \tan^2\Theta}) = \cos^{-1} (\cos2 \Theta) , \right ]$
$\because \left[\sin^{-1}(\frac{2\tan\Theta}{1+ \tan^2\Theta}) = \sin^{-1} (\sin2 \Theta) \right ]$
Then,
$=\tan \frac{1}{2}\left[2\tan^{-1}x + 2\tan^{-1}y \right ]$ $\because \left[\tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1- xy} \right]$
$=\tan \left [ \tan^{-1}\frac{x+y}{1-xy} \right ]$
$=\frac{x+y}{1-xy}$
Question:10 If $\tan^{-1}\frac{x-1}{x-2} + \tan^{-1}\frac{x+1}{x+2} =\frac{\pi}{4}$ , then find the value of $x$ .
Answer:
Using the identity $\tan^{-1}x+\tan^{-1} y = \tan^{-1}{\frac{x+y}{1-xy}}$ ,
We can find the value of x;
So, $\tan^{-1}\frac{x-1}{x-2} + \tan^{-1}\frac{x+1}{x+2} =\frac{\pi}{4}$
on applying,
= $\tan^{-1}{\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1- \left ( \frac{x-1}{x-2} \right )\left ( \frac{x+1}{x+2} \right )}}$
$=\tan^{-1}\frac{\frac{(x-1)(x+2)+(x-2)(x+1)}{x^2-4}}{1-\frac{x^2-1}{x^2-4}} = \tan^{-1} \left [ \frac{2x^2-4}{-3} \right ] = \frac{\pi}{4}$
$=\frac{2x^2-4}{-3} = \tan (\frac{\pi}{4})=1$
= $2x^2=1$ or $x = \pm \frac{1}{\sqrt{2}}$ ,
Hence, the possible values of x are $\pm \frac{1}{\sqrt{2}}$ .
Question:11Find the values of each of the expressions$\sin^{-1}\left (\sin\frac{2\pi}{3} \right )$
Answer:
Given $\sin^{-1}\left (\sin\frac{2\pi}{3} \right )$ ;
We know that $\sin^{-1}(\sin x) = x$
If the value of x belongs to $\left [ \frac{-\pi}{2},\frac{\pi}{2} \right ]$ then we get the principal values of $\sin^{-1}x$ .
Here, $\frac{2\pi}{3} \notin \left [ \frac{-\pi}{2},\frac{\pi}{2} \right ]$
We can write $\sin^{-1}\left (\sin\frac{2\pi}{3} \right )$ is as:
= $\sin^{-1}\left [ \sin\left ( \pi-\frac{2\pi}{3} \right ) \right ]$
= $\sin^{-1}\left [ \sin \frac{\pi}{3} \right ]$ where $\frac{\pi}{3} \epsilon \left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ]$
$\therefore \sin^{-1}\left (\sin\frac{2\pi}{3} \right )=\sin^{-1}\left [ \sin \frac{\pi}{3} \right ]=\frac{\pi}{3}$
Question:12 Find the values of each of the expressions$\tan^{-1}\left (\tan\frac{3\pi}{4} \right )$
Answer:
As we know $\tan^{-1}\left ( \tan x \right ) =x$
If $x \epsilon \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right ).$ which is the principal value range of $\tan^{-1}x$ .
So, as in $\tan^{-1}\left (\tan\frac{3\pi}{4} \right )$ ;
$\frac{3\pi}{4}\notin \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$
Hence we can write $\tan^{-1}\left (\tan\frac{3\pi}{4} \right )$ as :
$\tan^{-1}\left (\tan\frac{3\pi}{4} \right )$ = $\tan^{-1}\left (\tan\frac{3\pi}{4} \right) = \tan^{-1}\left [ \tan(\pi - \frac{\pi}{4}) \right ] = \tan^{-1}\left [ \tan (\frac{-\pi}{4}) \right ]$
Where $-\frac{\pi}{4} \epsilon \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$
and $\therefore \tan^{-1}\left (\tan\frac{3\pi}{4} \right )=\tan^{-1}\left [ \tan (\frac{-\pi}{4}) \right ]=-\frac{\pi}{4}$
Question:13 Find the values of each of the expressions $\tan\left(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2} \right )$
Answer:
Given that $\tan\left(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2} \right )$
we can take $\sin^{-1}\frac{3}{5} = x$ ,
then $\sin x = \frac{3}{5}$
or $\cos x = \sqrt{1-\sin^{2}x}= \frac{4}{5}$
$\Rightarrow \tan x = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}$
$\Rightarrow \tan^{-1}\frac{3}{4}= x$
We have similarities
$\cot^{-1} \frac{3}{2} = \tan^{-1} \frac{2}{3}$
Therefore we can write $\tan\left(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2} \right )$
$=\tan\left(\tan^{-1}\frac{3}{4}+\tan^{-1}\frac{2}{3} \right )$
$=\tan\left[\tan^{-1}\left ( \frac{\frac{3}{4}+\frac{2}{3}}{1- \frac{3}{4}.\frac{2}{3}} \right ) \right ]$ from $As, \left [ \tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1- xy} \right ]$
$=\tan \left (\tan^{-1} \frac{9+8}{12-6} \right ) = \tan \left (\tan^{-1} \frac{17}{6} \right )= \frac{17}{6}$
Question:14 $\cos^{-1}\left(\cos\frac{7\pi}{6} \right )$ is equal to
Answer:
As we know that $\cos^{-1} (cos x ) = x$ if $x\epsilon [0,\pi]$ and is principal value range of $\cos^{-1}x$ .
In this case $\cos^{-1}\left(\cos\frac{7\pi}{6} \right )$ ,
$\frac{7\pi}{6} \notin [0,\pi]$
hence we have then,
$\cos^{-1}\left(\cos\frac{7\pi}{6} \right ) =$ $\cos^{-1} \left ( \cos \frac{-7\pi}{6} \right ) = \cos^{-1}\left [ \cos\left ( 2\pi - \frac{7\pi}{6} \right ) \right ]$
$\left [ \because \cos (2\pi + x) = \cos x \right ]$
$\therefore\ we\ have \cos^{-1}\left ( \cos \frac{7\pi}{6} \right ) = \cos^{-1}\left ( \cos \frac{5\pi}{6} \right ) = \frac{5\pi}{6}$
Hence the correct answer is $\frac{5\pi}{6}$ (B).
Question:15 $\sin\left(\frac{\pi}{3} -\sin^{-1}\left(-\frac{1}{2} \right ) \right )$ is equal to
Answer:
Solving the inner bracket of $\sin\left(\frac{\pi}{3} -\sin^{-1}\left(-\frac{1}{2} \right ) \right )$ ;
$\left(\frac{\pi}{3} -\sin^{-1}\left(-\frac{1}{2} \right ) \right )$ or
Take $\sin^{-1}\left(-\frac{1}{2} \right ) = x$ then,
$\sin x =-\frac{1}{2}$ and we know the range of principal value of $\sin^{-1}x\ is\ \left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ].$
Therefore we have $\sin^{-1}\left ( -\frac{1}{2} \right ) = -\frac{\pi}{6}$ .
Hence, $\sin\left(\frac{\pi}{3} -\sin^{-1}\left(-\frac{1}{2} \right ) \right ) = \sin \left ( \frac{\pi}{3}+ \frac{\pi}{6} \right )= \sin \left ( \frac{3\pi}{6} \right ) = \sin\left ( \frac{\pi}{2} \right ) = 1$
Hence the correct answer is D.
Question:15$\tan^{-1}\sqrt3 - \cot^{-1}(-\sqrt3)$ is equal to
Answer:
We have $\tan^{-1}\sqrt3 - \cot^{-1}(-\sqrt3)$ ;
finding the value of $\cot^{-1}(-\sqrt3)$ :
Assume $\cot^{-1}(-\sqrt3) =y$ then,
$\cot y = -\sqrt 3$ and the range of the principal value of $\cot^{-1}$ is $(0,\pi)$ .
Hence, principal value is $\frac{5\pi}{6}$
Therefore $\cot^{-1} (-\sqrt3) = \frac {5\pi}{6}$
and $\tan^{-1} \sqrt3 = \frac{\pi}{3}$
so, we have now,
$\tan^{-1}\sqrt3 - \cot^{-1}(-\sqrt3)=\frac{\pi}{3} - \frac{5\pi}{6}$
$= \frac{2\pi - 5\pi}{6} = \frac{-3\pi}{6}$
or, $= \frac{ -\pi}{2}$
Hence the answer is option (B).
Class 10 Maths chapter 2 solutions Miscellaneous Exercise Page number: 31-32 Total questions: 14 |
Question 1: Find the value of the following: $\cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right )$
Answer:
If $x \epsilon [0,\pi]$ then $\cos^{-1}(\cos x) = x$ , which is principal value of $\cos^{-1} x$ .
So, we have $\cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right )$
$where \:\frac{13\pi}{6} \notin \left [ 0, \pi \right ].$
$Hence\: we \:can\: write\: \cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right ) \:as$
$=\cos^{-1}\left (\cos\left(2\pi+\frac{\pi}{6} \right ) \right )$
$=\cos^{-1}\left (\cos\left(\frac{\pi}{6} \right ) \right )$
$\frac{\pi}{6}\ \epsilon \left [ 0, \pi \right ]$
Therefore we have,
$\cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right ) = \cos^{-1}\left (\cos\left(\frac{\pi}{6} \right ) \right ) = \frac{\pi}{6}$ .
Question 2: Find the value of the following: $\tan^{-1}\left(\tan\frac{7\pi}{6} \right )$
Answer:
We have given $\tan^{-1}\left(\tan\frac{7\pi}{6} \right )$ ;
so, as we know $\tan^{-1}\left(\tan x \right ) =x \:\:if\:\:x\epsilon \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$
So, here we have $\frac{7\pi}{6} \notin \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$ .
Therefore we can write $\tan^{-1}\left(\tan\frac{7\pi}{6} \right )$ as:
$=\tan^{-1}\left(\tan \left (2\pi - \frac{5\pi}{6} \right ) \right )$ $\left [ \because \tan(2\pi - x) = -\tan x \right ]$
$=\tan^{-1}\left[-\tan \left ( \frac{5\pi}{6} \right ) \right ]$
$=\tan^{-1}\left[\tan \left (\pi- \frac{5\pi}{6} \right ) \right ]$
$=\tan^{-1}\left[\tan \left (\frac{\pi}{6} \right ) \right ]\:\:where\:\:\frac{\pi}{6} \epsilon \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$
$\therefore \tan^{-1} \left ( \tan \frac{7\pi}{6} \right ) = \tan^{-1} \left ( \tan \frac{\pi}{6} \right ) = \frac{\pi}{6}$ .
Question 3: Prove that $2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7}$
Answer:
To prove: $2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7}$ ;
$L.H.S=2\sin^{-1}\frac{3}{5}$
Assume that $\sin^{-1}\frac{3}{5} = x$
then we have $\sin x = \frac{3}{5}$ .
or $\cos x = \sqrt{1-\left (\frac{3}{5} \right )^2} = \frac{4}{5}$
Therefore we have
$\tan x = \frac{3}{4}\:\:or\:\:x = \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{3}{4}$
Now,
We can write L.H.S as
$2\sin^{-1}\frac{3}{5} = 2\tan^{-1}\frac{3}{4}$
$=\tan^{-1} \left [\frac{2\times\frac{3}{4}}{1- \left ( \frac{3}{4} \right )^2} \right ]$ as we know $\left [2\tan^{-1} x = \tan^{-1} \frac{2x}{1-x^2} \right ]$
$=\tan^{-1} \left [\frac{\frac{3}{2}}{\left ( \frac{16-9}{16} \right )} \right ] = \tan^{-1}\left ( \frac{3}{2}\times \frac{16}{7} \right )$
$=\tan^{-1} \frac{24}{7}=R.H.S$
L.H.S = R.H.S
Question:4 Prove that $\sin^{-1}\frac{8}{17} + \sin^{-1}\frac{3}{5} =\tan^{-1}\frac{77}{36}$
Answer
Taking $\sin ^{-1} \frac{8}{17} = x$
then,
$\sin x = \frac{8}{17} \Rightarrow \cos x = \sqrt{1- \left ( \frac{8}{17} \right )^2} = \sqrt {\frac{225}{289}} = \frac{15}{17}.$
Therefore we have-
$\tan^{-1} x = \frac{8}{15} \Rightarrow x = \tan^{-1} \frac{8}{15}$
$\therefore \sin ^{-1} \frac{8}{17} = \tan ^{-1} \frac{8}{15}$.............(1).
$Now, let\:\sin ^{-1} \frac{3}{5} = y$ ,
Then,
$\sin ^{-1} \frac{3}{5} = \tan ^{-1} \frac{3}{4}$.............(2).
So, we have now,
L.H.S.
$\sin^{-1}\frac{8}{17} + \sin^{-1}\frac{3}{5}$
using equations (1) and (2) we get,
$=\tan ^{-1} \frac{8}{15} + \tan^{-1} \frac{3}{4}$
$=\tan^{-1} \frac{\frac{8}{15}+ \frac{3}{4}}{1-\frac{8}{15}\times \frac{3}{4}}$ $[\because \tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1 - xy}]$
$=\tan^{-1} (\frac{32+45}{60-24})$
$=\tan^{-1} (\frac{77}{36})$
= R.H.S.
Question 5: Prove that $\cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13} = \cos^{-1}\frac{33}{65}$
Answer:
Take $\cos^{-1}\frac{4}{5} = x$ and $\cos^{-1}\frac{12}{13} = y$ and $\cos^{-1}\frac{33}{65} = z$
then we have,
$\cos x = \frac{4}{5}$
$\sin x = \sqrt {1- \left ( \frac {4}{5} \right )^2} = \frac {3}{5}$
Then we can write it as:
$\tan x = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}$ or $x= \tan^{-1} \frac{3}{4}$
$\therefore \cos ^{-1} \frac{4}{5} = \tan^{-1} \frac{3}{4}$...............(1)
Now, $\cos^{-1}\frac{12}{13} = y$
$\cos y = \frac{12}{13} \Rightarrow$ $\sin y =\frac{5}{13}$
$\therefore \tan y = \frac{5}{12} \Rightarrow y = \tan^{-1} \frac{5}{12}$
So, $\cos^{-1}\frac{12}{13} = \tan^{-1} \frac{5}{12}$...................(2)
Also, we have similarities;
$\cos^{-1}\frac{33}{65} = z$
Then,
$\cos^{-1}\frac{33}{65} = \tan^{-1} \frac{56}{33}$...........................(3)
Now, we have
L.H.S
$\cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13}$ so, using (1) and (2) we get,
$=\tan^{-1}\frac{3}{4} + \tan^{-1}\frac{5}{12}$
$=\tan^{-1}\left ( \frac{\frac{3}{4}+ \frac{5}{12}}{1-\left ( \frac{3}{4}\times \frac{5}{12} \right )} \right )$ $\because \left [ \tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1- xy} \right ]$
$=\tan^{-1}\left ( \frac{36+20}{48-15} \right )$
$=\tan^{-1}\left ( \frac{56}{33} \right )$ or we can write it as;
$=\cos^{-1}\frac{33}{65}$
= R.H.S.
Hence proved.
Question 6: Prove that $\cos^{-1}\frac{12}{13} + \sin^{-1}\frac{3}{5} = \sin^{-1}\frac{56}{65}$
Answer:
Converting all terms in tan form;
Let $\cos^{-1}\frac{12}{13} = x$ , $\sin^{-1}\frac{3}{5} = y$ and $\sin^{-1}\frac{56}{65} = z$ .
Now, converting all the terms:
$\cos^{-1}\frac{12}{13} = x$ or $\cos x = \frac{12}{13}$
We can write it in tan form as:
$\cos x = \frac{12}{13} \Rightarrow$ $\sin x = \frac{5}{13}$ .
$\therefore \tan x = \frac{5}{12} \Rightarrow x = \tan^{-1} \frac{5}{12}$
or $\cos^{-1}\frac{12}{13} = \tan^{-1} \frac{5}{12}$ ................(1)
$\sin^{-1}\frac{3}{5} = y$ or $\sin y = \frac{3}{5}$
We can write it in tan form as:
$\sin y = \frac{3}{5} \Rightarrow$ $\cos y = \frac{4}{5}$
$\therefore \tan y =\frac{3}{4} \Rightarrow y = \tan^{-1} \frac{3}{4}$
or $\sin^{-1}\frac{3}{5} = \tan^{-1} \frac{3}{4}$ ......................(2)
Similarly, for $\sin^{-1}\frac{56}{65} = z$ ;
we have $\sin^{-1}\frac{56}{65} = \tan^{-1} \frac{56}{33}$ .............(3)
Using (1) and (2) we have L.H.S
$\cos^{-1}\frac{12}{13} + \sin^{-1}\frac{3}{5}$
$= \tan^{-1} \frac{5}{12} + \tan^{-1} \frac{3}{4}$
On applying $\tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1-xy}$
We have,
$=\tan^{-1} \frac{\frac{5}{12}+\frac{3}{4}}{1-(\frac{5}{12}.\frac{3}{4})}$
$=\tan^{-1} (\frac{20+36}{48-15})$
$=\tan^{-1} (\frac{56}{33})$
$=\sin^{-1} (\frac{56}{65})$ ...........[Using (3)]
=R.H.S.
Hence proved.
Question 7: Prove that $\tan^{-1}\frac{63}{16} = \sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5}$
Answer:
Taking R.H.S;
We have $\sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5}$
Converting sin and cos terms in tan forms:
Let $\sin^{-1}\frac{5}{13} = x$ and $\cos^{-1}\frac{3}{5} = y$
now, we have $\sin^{-1}\frac{5}{13} = x$ or $\sin x = \frac{5}{13}$
$\sin x = \frac{5}{13} \:or\: \cos x =\frac{12}{13}\:or\:\tan x = \frac{5}{12}$
$\tan x = \frac{5}{12} \Rightarrow x =\tan^{-1} \frac{5}{12}$
$\therefore \sin^{-1} \frac{5}{13} = \tan^{-1} \frac{5}{12}$............(1)
Now, $\cos^{-1}\frac{3}{5} = y\Rightarrow \cos y = \frac{3}{5}$
$\cos y = \frac{3}{5} \:or\: \sin y = \frac{4}{5}\:or\:\tan y = \frac{4}{3}$
$\tan y = \frac{4}{3} \Rightarrow y = \tan^{-1} \frac{4}{3}$
$\therefore \cos^{-1}\frac{3}{5} = \tan^{-1} \frac{4}{5}$................(2)
Now, Using (1) and (2) we get,
R.H.S.
$\sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5} = \tan^{-1} \frac{5}{12} + \tan^{-1} \frac{4}{3}$
$=\tan^{-1}\left ( \frac{\frac{5}{12}+\frac{4}{3}}{1- \frac{5}{12}\times \frac{4}{3}} \right )$ as we know $\left [ \tan^{-1} x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1-xy} \right ]$
so,
$= \tan^{-1} \frac{63}{16}$
equal to L.H.S
Hence proved.
Question 8: Prove that $\tan^{-1} \sqrt{x} = \frac{1}{2}\cos^{-1}\frac{1-x}{1+x},\;\;x\in [0,1]$
Answer:
By observing the square root we will first put
$x= \tan^2 \theta$ .
Then,
we have $\tan^{-1} \sqrt{\tan^2 \theta} = \frac{1}{2}\cos^{-1}\frac{1-\tan^2 \theta}{1+\tan^2 \theta}$
or, R.H.S.
$\frac{1}{2}\cos^{-1}\frac{1-\tan^2 \theta}{1+\tan^2 \theta} = \frac{1}{2}\cos^{-1}(cos2 \theta)$
$= \frac{1}{2}\times 2\theta = \theta$ .
L.H.S.$\tan^{-1} \sqrt{\tan^2 \theta} = \tan^{-1}(\tan \theta) = \theta$
hence L.H.S. = R.H.S proved.
Question 9: Prove that $\cot^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1 - \sin x}}{\sqrt{1+\sin x} - \sqrt{1 - \sin x}} \right ) = \frac{x}{2},\;\;x\in\left(0,\frac{\pi}{4} \right )$
Answer:
Given that $\cot^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1 - \sin x}}{\sqrt{1+\sin x} - \sqrt{1 - \sin x}} \right )$
By observing we can rationalize the fraction
$\left(\frac{\sqrt{1+\sin x} + \sqrt{1 - \sin x}}{\sqrt{1+\sin x} - \sqrt{1 - \sin x}} \right )$
We get then,
$=\left(\frac{\sqrt{1+\sin x} + \sqrt{1 - \sin x}}{\sqrt{1+\sin x} - \sqrt{1 - \sin x}} \right ) = \left(\frac{(\sqrt{1+\sin x} + \sqrt{1 - \sin x})^2}{{1+\sin x} - {1 + \sin x}} \right )$
$= \left(\frac{{1+\sin x} +{1 - \sin x} + 2\sqrt{(1+\sin x)(1-\sin x)} }{{1+\sin x} - {1 + \sin x}} \right )$
$= \frac{2(1+\sqrt{1-\sin^2 x})}{2\sin x} = \frac{1+\cos x}{\sin x} = \frac{2\cos^2 \frac{x}{2}}{2\sin \frac{x}{2}\cos \frac{x}{2}}$
$= \cot \frac{x}{2}$
Therefore we can write it as;
$\cot^{-1}\left ( \cot \frac{x}{2} \right ) = \frac{x}{2}$
As L.H.S. = R.H.S.
Hence proved.
Question 10: Prove that $\tan^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} \right ) = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}x,\;\;-\frac{1}{\sqrt2}\leq x\leq 1$
[Hint: Put $x = \cos 2\theta$ ]
Answer:
By using the Hint we will put $x = \cos 2\theta$ ;
we get then,
$=\tan^{-1}\left(\frac{\sqrt{1+\cos 2\theta} - \sqrt{1-\cos2\theta}}{\sqrt{1+\cos 2\theta} + \sqrt{1-\cos 2\theta}} \right )$
$=\tan^{-1}\left(\frac{\sqrt{2\cos^2 \theta} - \sqrt{2\sin^2\theta}}{\sqrt{2\cos^2 \theta} + \sqrt{2\sin^2\theta}} \right )$
$=\tan^{-1}\left(\frac{\sqrt2{\cos \theta} - \sqrt2{\sin\theta}}{\sqrt2{\cos \theta} + \sqrt2{\sin\theta}} \right )$
$=\tan^{-1}\left(\frac{{\cos \theta} - {\sin\theta}}{{\cos \theta} + {\sin\theta}} \right )$ dividing numerator and denominator by $\cos \theta$ ,
we get,
$= \tan^{-1}\left ( \frac{1-\tan \theta}{1+\tan \theta} \right )$
$= \tan^{-1} 1 - \tan^{-1} (\tan \theta)$ using the formula $\left [ \tan^{-1}x - \tan^{-1} y = \tan^{-1} \frac{x-y}{1+xy} \right ]$
$= \frac{\pi}{4} - \theta = \frac{\pi}{4}- \frac{1}{2}\cos^{-1}x$
As L.H.S = R.H.S
Hence proved
Question 11: Solve the following equations: $2\tan^{-1}(\cos x) = \tan^{-1}(2\textup{cosec}x)$
Answer:
Given equation $2\tan^{-1}(\cos x) = \tan^{-1}(2\textup{cosec}x)$ ;
Using the formula:
$\left [ 2\tan^{-1}z = \tan^{-1} \frac{2z}{1-z^2} \right ]$
We can write
$2\tan^{-1}(\cos x) = \tan^{-1}\left [ \frac{2\cos x}{1- (\cos x )^2 }\right ]$
$\tan^{-1}\left [ \frac{2\cos x}{1- (\cos x )^2 }\right ] = \tan^{-1}\left [2cosec x \right ]$
So, we can equate;
$=\left [ \frac{2\cos x}{1- (\cos x )^2 }\right ] = \left [2cosec x \right ]$
$=\left [ \frac{2\cos x}{\sin^2 x }\right ] = \left [ \frac{2}{sinx } \right ]$
that implies that $\cos x = \sin x$ .
or $\tan x =1$ or $x = \frac{\pi}{4}$
Hence we have solution $x = \frac{\pi}{4}$ .
Question 12: Solve the following equations: $\tan^{-1} \frac{1-x}{1+x} = \frac{1}{2}\tan^{-1}x,\;(x>0)$
Answer:
Given equation is
$\tan^{-1} \frac{1-x}{1+x} = \frac{1}{2}\tan^{-1}x$ :
L.H.S can be written as;
$\tan^{-1} \frac{1-x}{1+x} = \tan^{-1}1 - \tan^{-1}x$
Using the formula $\left [ \tan^{-1}x -\tan^{-1}y = \tan^{-1} \frac{x-y}{1+xy} \right ]$
So, we have $\tan^{-1}1 - \tan^{-1}x = \frac{1}{2} \tan^{-1}x$
$\Rightarrow \tan^{-1}1= \frac{3}{2} \tan^{-1}x$
$\Rightarrow\frac{\pi}{4}= \frac{3}{2} \tan^{-1}x$
$\Rightarrow \tan^{-1}x = \frac{\pi}{6}$
$\Rightarrow x= \tan \frac{\pi}{6} = \frac{1}{\sqrt3}$
Hence the value of $x= \frac{1}{\sqrt3}$ .
Question 13: $\sin(\tan^{-1}x),\;|x|<1$ is equal to
Answer:
Let $\tan^{-1}x = y$ then we have;
$\tan y = x$ or
$y=\sin^{-1} \left ( \frac{x}{\sqrt{1+x^2}} \right ) \Rightarrow \tan^{-1} x = \sin^{-1} \left ( \frac{x}{\sqrt{1+x^2}} \right)$
$\Rightarrow \sin \left ( \tan^{-1} x \right ) = \sin\left ( \sin^{-1}\left ( \frac{x}{\sqrt{1+x^2}} \right ) \right ) = \frac{x}{\sqrt {1+x^2}}$
Hence the correct answer is D.
Question 14: $\sin^{-1}(1-x) - 2\sin^{-1}x = \frac{\pi}{2}$ then $x$ is equal to
Answer:
Given the equation: $\sin^{-1}(1-x) - 2\sin^{-1}x = \frac{\pi}{2}$
we can migrate the $\sin^{-1}(1-x)$ term to the R.H.S.
then we have;
$- 2\sin^{-1}x = \frac{\pi}{2} - \sin^{-1}(1-x)$
or $- 2\sin^{-1}x =\cos^{-1}(1-x)$............................(1)
from $\left [\because \cos^{-1}(1-x) + \sin^{-1}(1-x) = \frac{\pi}{2} \right ]$
Take $\sin^{-1}x = \Theta$ $\Rightarrow \sin \Theta = x$ or $\cos \Theta = \sqrt{1-x^2}$ .
So, we conclude that;
$\sin^{-1}x = \cos^{-1}\left ( \sqrt{1-x^2} \right )$
Therefore we can put the value of $\sin^{-1}x$ in equation (1) we get,
$- 2\cos^{-1}\left ( \sqrt{1-x^2} \right ) =\cos^{-1}(1-x)$
Putting x= sin y, in the above equation; we have then,
$\Rightarrow - 2\cos^{-1}\left ( \sqrt{1-(\sin y)^2} \right ) =\cos^{-1}(1-\sin y )$
$\Rightarrow - 2\cos^{-1}\left ( \sqrt{\cos^2 y} \right ) =\cos^{-1}(1-\sin y )$
$\Rightarrow - 2\cos^{-1}\left ( \cos y \right ) =\cos^{-1}(1-\sin y )$
$\Rightarrow \cos(-2y) = 1-\sin y$
$\Rightarrow - 2y=\cos^{-1}(1-\sin y )$
$\Rightarrow 1- 2\sin^2 y = 1-\sin y$
$\Rightarrow 2\sin^2 y - \sin y = 0$
$\Rightarrow \sin y(2 \sin y -1) = 0$
So, we have the solution;
$\sin y = 0\ or\ \frac{1}{2}$ Therefore we have $x = 0\ or\ x= \frac{1}{2}$ .
When we have $x= \frac{1}{2}$ , we can see that :
$L.H.S. = \sin ^{-1}\left ( 1 - \frac{1}{2} \right ) - 2\sin^{-1}\frac{1}{2} = - \sin^{-1}\frac{1}{2} = -\frac{\pi}{6}$
So, it is not equal to the R.H.S. $-\frac{\pi}{6} \neq \frac{\pi}{2}$
Thus we have only one solution which is x = 0
Hence the correct answer is (C).
Below are some useful links for solutions to exercises of Inverse trigonometry function of class 12:
Inverse Trigonometric Functions Class 12 Exercise 2.1
Inverse Trigonometric Functions Class 12 Exercise 2.2
Inverse Trigonometric Functions Class 12 Miscellaneous Exercise
Given below are the subject-wise exemplar solutions of class 12 NCERT:
Here are some useful links for NCERT books and the NCERT syllabus for class 12:
Use identities: Apply trigonometric identities to express inverse trignometric functions in simpler forms.
Navigation: Inverse trigonometric functions are used in calculating angles of elevation and depression, as well as directions.
Engineering: Used in signal processing, wave analysis, and electrical engineering.
Physics: To calculate angles in problems involving vectors, motion, and forces.
Architecture: Helps in designing structures by calculating angles and slopes.
Identity: Use definitions, algebraic manipulations, and trigonometric identities to prove standard properties.
Domain and Range: The domain and range for each inverse trigonometric function are explained to help students understand the constraints on values
Graphical Representation: How to sketch the graphs of inverse trigonometric functions for better visual understanding.
Solving Equations: Techniques to solve equations involving inverse trigonometric functions.
Applications: Problems involving real-life applications, including trigonometric equations and angle calculations.
There are 3 exercises in NCERT class 12 maths chapter 2
Application Date:24 March,2025 - 23 April,2025
Admit Card Date:04 April,2025 - 26 April,2025
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.
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!
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:
Consider Professional Help:
Explore Alternative Options:
Focus on NEET 2025 Preparation:
Seek Support:
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.
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!
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.
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