NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions (Updated For 2023-24)

Q: What are important topics of the chapter Trigonometric Functions ?

In the chapter 3 class 11 maths Trigonometric functions, identities of trigonometric functions, trigonometric functions of some standard angles, trigonometric functions of sum and difference of two angles and trigonometric equations are the important topics of this chapter.

Q: How does the NCERT solutions are helpful ?

NCERT solutions will help the students, if the stuck while solving the NCERT problems. T rigonometry class 11 solutions provides practice problems for students that help them to command and get indepth understanding. finally class 11 maths trigonometry help to build strong foundation for higher classes. For ease Interested students can study trigonometric functions class 11 pdf both online and offline. therefore students can practice class 11 maths chapter 3 solutions.

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NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

Edited By Ramraj Saini | Updated on Sep 21, 2023 09:10 PM IST

Trigonometric Functions Class 11 Questions And Answers

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions are provided here. Trigonometry class 11 has various real-time applications. It is used to solve height and distance problems. This chapter gives an introduction to basic properties and identities of trigonometric functions and questions based on the same are answered in class 11 trigonometric functions NCERT solutions. Students can practice NCERT Solutions that are prepared by Careers360 expert team according to CBSE latest syllabus 2023. The chapter 3 maths class 11 trigonometric identities are used in multiple other chapters of mathematics and throughout the NCERT physics for both class 11 and 12 also. So it is very important to memories and understand the basic identities and properties of trigonometric functions. Trigonometric functions class 11 NCERT solutions will help you for the same.

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Trigonometric Functions Class 11 Questions And Answers
Trigonometric Functions Class 11 Questions And Answers PDF Free Download
Trigonometric Functions Class 11 Solutions - Important Formulae
Trigonometric Functions Class 11 NCERT Solutions (Intext Questions and Exercise)
Trigonometric Functions Class 11 Solutions - Topics
NCERT Solutions for Class 11 Mathematics - Chapter Wise
Key Features of NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions
NCERT Solutions for Class 11 - Subject Wise
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NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

Command on this NCERT syllabus is required to understand the concepts of inverse trigonometric functions which will be taught in class 12. You should practice more in order to get clarity of the concepts. First, try to solve NCERT textbook exercise problems. If you getting difficulties in doing so, you can take help from solutions of NCERT for class 11 maths chapter 3 trigonometric functions. Check all NCERT solutions at a single place which will be helpful when you are not able to solve the NCERT questions. Here you will get NCERT solutions for class 11 also.

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Trigonometric Functions Class 11 Questions And Answers PDF Free Download

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Trigonometric Functions Class 11 Solutions - Important Formulae

Angle Conversion:

Radian Measure = π/180 × Degree Measure

Degree Measure = 180/π × Radian Measure

Trigonometric Ratios:

sin θ = P / H
cos θ = B / H
tan θ = P / B
cosec θ = H / P
sec θ = H / B
cot θ = B / P

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Reciprocal Trigonometric Ratios:

sin θ = 1 / cosec θ
cosec θ = 1 / sin θ
cos θ = 1 / sec θ
sec θ = 1 / cos θ
tan θ = 1 / cot θ
cot θ = 1 / tan θ

Trigonometric Ratios of Complementary Angles:

sin (90° – θ) = cos θ
cos (90° – θ) = sin θ
tan (90° – θ) = cot θ
cot (90° – θ) = tan θ
sec (90° – θ) = cosec θ
cosec (90° – θ) = sec θ

Periodic Trigonometric Ratios:

sin(π/2-θ) = cos θ
cos(π/2-θ) = sin θ
sin(π-θ) = sin θ
cos(π-θ) = -cos θ
sin(π+θ) = -sin θ
cos(π+θ) = -cos θ
sin(2π-θ) = -sin θ
cos(2π-θ) = cos θ

Trigonometric Identities:

sin^² θ + cos^² θ = 1
cosec^² θ – cot^² θ = 1
sec^² θ – tan^² θ = 1

Product to Sum Formulas:

sin x sin y = 1/2 [cos(x–y) − cos(x+y)]
cos x cos y = 1/2[cos(x–y) + cos(x+y)]
sin x cos y = 1/2[sin(x+y) + sin(x−y)]
cos x sin y = 1/2[sin(x+y) – sin(x−y)]

Sum to Product Formulas:

sin x + sin y = 2 sin [(x+y)/2] cos [(x-y)/2]
sin x – sin y = 2 cos [(x+y)/2] sin [(x-y)/2]
cos x + cos y = 2 cos [(x+y)/2] cos [(x-y)/2]
cos x – cos y = -2 sin [(x+y)/2] sin [(x-y)/2]

General Trigonometric Formulas:

sin (x+y) = sin x × cos y + cos x × sin y
cos(x+y) = cosx × cosy − sinx × siny
cos(x–y) = cosx × cosy + sinx × siny
sin(x–y) = sinx × cosy − cosx × siny

Sum and Difference Formulas for tan:

tan (x+y) = (tan x + tan y) / (1 − tan x tan y)
tan (x−y) = (tan x − tan y) / (1 + tan x tan y)

Double Angle Formulas for tan:

tan 2θ = (2 tan θ) / (1 - tan²θ)

Triple Angle Formulas for sin, cos, and tan:

sin 3θ = 3sin θ – 4sin^³θ
cos 3θ = 4cos^³θ – 3cos θ
tan 3θ = [3tan θ – tan^³θ] / [1 − 3tan^²θ]

Free download NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions for CBSE Exam.

Trigonometric Functions Class 11 NCERT Solutions (Intext Questions and Exercise)

NCERT trigonometric functions class 11 questions and answers - Exercise: 3.1

Question:1 Find the radian measures corresponding to the following degree measures:

(i) $25 \degree$
(ii) $-47 \degree$ $30'$
(iii) $240\degree$
(iv) $520\degree$

Answer:

It is solved using relation between degree and radian

(i) $25\degree$
We know that $180\degree$ = $\pi$ radian

So, $1\degree = \frac{\pi }{180}$ radian

$25\degree = \frac{\pi }{180}\times 25$ radian $=\frac{5\pi }{36}$ radian
(ii) $-47\degree30'$
We know that
$-47\degree30' = -47\frac{1}{2}degree = -\frac{95}{2}\degree$

Now, we know that $180\degree = \pi \Rightarrow 1\degree = \frac{\pi}{180}$ radian

So, $-\frac{95}{2}\degree = \frac{\pi}{180}\times \left (-\frac{95}{2} \right )$ radian $\Rightarrow \frac{-19\pi}{72}$ radian
(iii) $240\degree$
We know that

$180\degree = \pi \Rightarrow 1\degree = \frac{\pi}{180}$ radian

So, $240\degree = \frac{\pi}{180}\times 240 \Rightarrow \frac{4\pi}{3}$ radian
(iv) $520\degree$
We know that

$180\degree = \pi \Rightarrow 1\degree = \frac{\pi}{180}$ radian

So, $520\degree \Rightarrow \frac{\pi}{180}\times 520$ radian $\Rightarrow \frac{26\pi}{9}$ radian

Question:2 Find the degree measures corresponding to the following radian measures. (Use $\small \pi =\frac{22}{7}$ )

$\small (i) \frac{11}{16}$
$\small (ii) -4$
$\small (iii) \frac{5\pi }{3}$
$\small (iv) \frac{7\pi }{6}$

Answer:

(1) $\frac{11}{16}$

We know that
$\pi$ radian $= 180\degree \Rightarrow 1 radian = \frac{180}{\pi} degree$

So, $\frac{11}{16} radian = \frac{180}{\pi}\times \frac{11}{16}degree$ (we need to take $\pi = \frac{22}{7}$ )

$\frac{11}{16}radian = \frac{180\times 7}{22}\times \frac{11}{16}degree \Rightarrow \frac{315}{8}degree$

(we use $1\degree = 60'$ and 1' = 60'')

Here 1' represents 1 minute and 60" represents 60 seconds
Now,

$\frac{315}{8}degree=39\frac{3}{8}degree\\ \\ =39\degree +\frac{3\times 60}{8}minutes \Rightarrow 39\degree +22' + \frac{1}{2}minutes \Rightarrow 39\degree +22' +30''\\ \\ \Rightarrow \frac{315}{8}degree = 39\degree22'30''$

(ii) -4
We know that

$\pi$ radian $= 180\degree \Rightarrow 1 radian = \frac{180}{\pi} degree$ (we need to take $\pi = \frac{22}{7}$ )

So, -4 radian = $\frac{-4\times 180}{\pi} \Rightarrow \frac{-4\times 180\times 7}{22} \Rightarrow \frac{-2520}{11}degree$

(we use $1\degree = 60'$ and 1' = 60'')

$\Rightarrow \frac{-2520}{11}degree = -229\frac{1}{11}degree =-229\degree + \frac{1\times 60}{11}minutes \\ \\ \Rightarrow -229\degree + 5' + \frac{5}{11}minutes = -229\degree +5' +27''\\ \\ -\frac{2520}{11} = -229\degree5'27''$

(iii) $\frac{5\pi}{3}$

We know that
$\pi$ radian $= 180\degree \Rightarrow 1 radian = \frac{180}{\pi} degree$ (we need to take $\pi = \frac{22}{7}$ )

So, $\frac{5\pi}{3}radian = \frac{180}{\pi}\times \frac{5\pi}{3}degree = 300\degree$
(iv) $\frac{7\pi}{6}$

We know that
$\pi$ radian $= 180\degree \Rightarrow 1 radian = \frac{180}{\pi} degree$ (we need to take $\pi = \frac{22}{7}$ )

So, $\frac{7\pi}{6}radian = \frac{180}{\pi}\times \frac{7\pi}{6} = 210\degree$

Question:3 A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

Answer:
Number of revolutions made by the wheel in 1 minute = 360
$\therefore$ Number of revolutions made by the wheel in 1 second = $\frac{360}{60} = 6$
( $\because$ 1 minute = 60 seconds)
In one revolutions wheel will cover $2\pi$ radian
So, in 6 revolutions it will cover = $6\times 2\pi = 12\pi$ radian

$\therefore$ In 1 the second wheel will turn $12\pi$ radian

Question:4 Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm (use $\small \pi =\frac{22}{7}$ )

Answer:

We know that
$l = r\Theta$ ( where l is the length of the arc, r is the radius of the circle and $\Theta$ is the angle subtended)

here r = 100 cm
and l = 22 cm
Now,
$\Theta = \frac{l}{r} = \frac{22}{100}radian$

We know that
$\pi radian = 180\degree\\ \\So, 1radian = \frac{180}{\pi}degree\\ \\ \therefore \frac{22}{100}radian = \frac{180}{\pi}\times\frac{22}{100}degree\Rightarrow \frac{180\times7}{22}\times\frac{22}{100} = \frac{63}{5}degree \\ \\ So, \\ \\\frac{63}{5}degree = 12\frac{3}{5}degree = 12\degree + \frac{3\times60}{5}minute = 12\degree + 36'\\ \\ \therefore \frac{63}{5}degree = 12 \degree36'$
So,
Angle subtended at the centre of a circle $\Theta = 12\degree36'$

Question:5 In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.

Answer:

Given :- radius (r)of circle = $\frac{Diameter}{2} = \frac{40cm}{2} = 20 cm$
length of chord = 20 cm

We know that
$\theta = \frac{l}{r}$ (r = 20cm , l = ? , $\theta$ = ?)

Now,
1654678632225 AB is the chord of length 20cm and OA and OB are radii of circle 20 cm each
The angle subtended by OA and OB at centre = $\theta$
$\because$ OA = OB = AB
$\therefore$ $\Delta$ OAB is equilateral triangle
So, each angle equilateral is $60\degree$
$\therefore$ $\theta = 60\degree$ $= \frac{\pi}{3}radian$
Now, we have $\theta$ and r
So,
$l = r\theta = 20\times\frac{\pi}{3}=\frac{20\pi}{3}$
$\therefore$ the length of the minor arc of the chord (l) = $\frac{20\pi}{3}$ cm

Question:6 If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.

Answer:

Given:-
$\theta_1 = 60\degree\\ \theta_2 = 75\degree\\$ and $l_1 = l_2$

We need to find the ratio of their radii $\frac{r_1}{r_2} = ?$

We know that arc length $l = r \theta$
So,
$l_1 = r_1 \theta_1\\l_2 = r_2\theta_2$
Now,
$\frac{l_1}{l_2}=\frac{ r_1 \theta_1}{ r_2\theta_2}$ ( $l_1 = l_2$ )
So,
$\frac{ r_1 }{ r_2}= \frac{\theta_2}{\theta_1} = \frac {75}{60} = \frac{5}{4}$ is the ratio of their radii

Question:7 Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of length
(i) 10 cm
(ii) 15 cm
(iii) 21 cm

Answer:

(i) We know that

$l = r \theta$
Now,
r = 75cm
l = 10cm

So,
$\theta = \frac{l}{r} = \frac{10}{75} = \frac{2}{15}radian$

(ii) We know that

$l = r \theta$
Now,
r = 75cm
l = 15cm

So,
$\theta = \frac{l}{r} = \frac{15}{75} = \frac{1}{5}radian$

(iii) We know that

$l = r \theta$
Now,
r = 75cm
l = 21cm

So,
$\theta = \frac{l}{r} = \frac{21}{75} = \frac{7}{25}radian$

NCERT class 11 maths chapter 3 question answer - Exercise: 3.2

Question:1 Find the values of other five trigonometric functions $\small \cos x = -\frac{1}{2}$ , x lies in third quadrant.

Answer:

Solution
$\cos x = -\frac {1}{2}$
$\because \sec x = \frac{1}{\cos x} = \frac{1}{-\frac {1}{2}} = -2$
x lies in III quadrants. Therefore sec x is negative

$\sin ^{2}x +\cos^{2}x = 1 \\ \sin^{2}x = 1 - \cos^{2}x\\ \sin^{2}x = 1 -\left ( -\frac{1}{2} \right )^{2}\\ \sin^{2}x = 1 - \frac{1}{4} = \frac{3}{4}\\ \sin x = \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2}$
x lies in III quadrants. Therefore sin x is negative
$\therefore \sin x= - \frac{\sqrt{3}}{2}$

$\because cosec \ x = \frac {1}{\sin x}= \frac{1}{- \frac{\sqrt{3}}{2}} =- \frac{2}{\sqrt{3}}$

x lies in III quadrants. Therefore cosec x is negative

$\tan x = \frac{\sin x}{\cos x} = \frac {-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \sqrt{3}$
x lies in III quadrants. Therefore tan x is positive

$\cot x = \frac{1}{\tan x} = \frac{1}{\sqrt{3}}$
x lies in III quadrants. Therefore cot x is positive

Question:2 Find the values of other five trigonometric functions $\small \sin x = \frac{3}{5}$ x lies in second quadrant.

Answer:

Solution

$\sin x = \frac {3}{5}$

$cosec \ x = \frac{1}{\sin x}=\frac {1}{\frac {3}{5}} = \frac {5}{3}$
x lies in the second quadrant. Therefore cosec x is positive

$\sin^{2}x + \cos ^{2}x = 1\\ \cos ^{2}x = 1 - \sin ^{2}x\\ \cos ^{2}x = 1 - \left ( \frac{3}{5} \right )^{2}\\ \cos ^{2}x = 1 - \frac {9}{25} = \frac {16}{25}\\ \cos x = \sqrt{\frac {16}{25}} = \pm \frac {4}{5}$
x lies in the second quadrant. Therefore cos x is negative
$\therefore \cos x = - \frac {4}{5}$

$\sec x = \frac {1}{\cos x} = \frac{1}{- \frac {4}{5}} = -\frac {5}{4}$
x lies in the second quadrant. Therefore sec x is negative

$\tan x = \frac {\sin x}{\cos x} = \frac {\frac{3}{5}}{-\frac{4}{5}} = -\frac {3}{4}$
x lies in the second quadrant. Therefore tan x is negative

$\cot x = \frac {1}{\tan x} = \frac {1}{-\frac {3}{4}} = -\frac{4}{3}$
x lies in the second quadrant. Therefore cot x is negative

Question:3 Find the values of other five trigonometric functions $\small \cot x = \frac{3}{4}$ , x lies in third quadrant.

Answer:

Solution

$\cot x= \frac {3}{4}$

$\tan x = \frac{1}{\cot x}= \frac{1}{\frac {3}{4}} = \frac {4}{3}$
$1 + \tan ^ {2}x = \sec ^{2}x\\ 1+\frac{4^2}{3^2} = \sec ^{2}x\\ \\ 1 + \frac {16}{9} = \sec ^{2}x\\ \frac {25}{9} = \sec ^{2}x\\ \sec x = \sqrt {\frac {25}{9}} = \pm \frac {5}{3}$
x lies in x lies in third quadrant. therefore sec x is negative
$\sec x = -\frac{5}{3}$

$\cos x = \frac {1}{\sec x} = \frac {1}{-\frac{5}{3}} = -\frac{3}{5}$
$\sin ^{2 }x+ \cos ^{2}x = 1\\ \sin ^{2 }x = 1 - \cos ^{2}x\\ \sin ^{2 }x = 1 -\left ( -\frac{3}{5} \right )^{2}\\ \sin ^{2 }x = 1 - \frac {9}{25}\\ \sin ^{2 }x = \frac{16}{25}\\ \sin x = \sqrt {\frac{16}{25}} = \pm \frac{4}{5}$
x lies in x lies in third quadrant. Therefore sin x is negative
$\sin x = -\frac {4}{5}$
$cosec x = \frac {1}{\csc} = \frac {1}{-\frac{4}{5}} = - \frac{5}{4}$

Question:4 Find the values of other five trigonometric functions $\small \sec x = \frac{13}{5}$ , x lies in fourth quadrant.

Answer:

Solution
$\sec x = \frac {13}{5}$
$\cos x = \frac {1}{\sec x} = \frac{1}{\frac {13}{5}} = \frac {5}{13}$
$\sin^{2}x + \cos^{2}x = 1\\ \sin^{2}x = 1 - \cos^{2}x\\ \sin^{2}x = 1 - \frac {5}{13}\\ \sin^{2}x = 1 - \frac {25}{169} = \frac {144}{169}\\ \sin x = \sqrt { \frac {144}{169}} = \pm \frac {12}{13}$
lies in fourth quadrant. Therefore sin x is negative
$\sin x =- \frac {12}{13}$
$\csc x = \frac {1}{\sin x} = \frac {1}{-\frac {12}{13}} = -\frac {13}{12}$
$\tan x = \frac {\sin x}{\cos x} = \frac {-\frac{12}{13}}{\frac{5}{13}} = -\frac {12}{5}$
$\cot x = \frac {1}{\tan x} = \frac {1}{-\frac{12}{5}} = -\frac{5}{12}$

Question:5 Find the values of the other five trigonometric functions $\small \tan x = -\frac{5}{12}$ , x lies in second quadrant.

Answer:
$\tan x = -\frac {5}{12}$
$\cot x = \frac {1}{\tan x} = \frac {1}{-\frac{5}{12}} = -\frac {12}{5}$
$1 + \tan^{2}x = \sec^{2}x\\ 1 + \left ( -\frac{5}{12} \right )^{2} = \sec^{2}x\\ 1 + \frac {25}{144} = \sec^{2}x\\ \\ \frac {169}{144} = \sec^{2}x\\ \sec x = \sqrt {\frac {169}{144}} = \pm \frac {13}{12}$
x lies in second quadrant. Therefore the value of sec x is negative
$\sec x = - \frac {13}{12}$
$\cos x = \frac{1}{\sec x}= \frac{1}{-\frac{13}{12}} = -\frac {12}{13}$
$\sin^{2}x + \cos^{2}x = 1\\ \sin^{2}x = 1 - \cos^{2}x\\ \sin^{2}x = 1 - \left ( -\frac{12}{13} \right )^{2}\\ \sin^{2}x = 1 - \frac{144}{169}\\ \sin^{2}x = \frac {25}{169}\\ \sin x = \sqrt {\frac{25}{169}} = \pm \frac{5}{13}$
x lies in the second quadrant. Therefore the value of sin x is positive
$\sin x = \frac {5}{13}$
$\csc = \frac {1}{\sin x} = \frac {1}{\frac {5}{13}} = \frac {13}{5}$

Question:6 Find the values of the trigonometric functions $\small \sin 765\degree$

Answer:
We know that values of sin x repeat after an interval of $2\pi\ or\ 360\ degree$

$\sin765\degree = \sin (2\times360\degree + 45\degree ) = \sin45\degree\\ sin45\degree = \frac {1}{\sqrt{2}}$

Question:7 Find the values of the trigonometric functions $\small cosec \ (-1410\degree)$

Answer:

We know that value of cosec x repeats after an interval of $2\pi \ or \ 360\degree$
$cosec (-1410\degree) = cosec (-1410\degree + 360\degree\times4)\\ cosec\ 30\degree = 2$

$cosec(-1410\degree)= - cosec(1410\degree)\\= -cosec(4 \times 360\degree - 30\degree)= - cosec(-30\degree) = 2$

Question:8 Find the values of the trigonometric functions $\small \tan \frac{19\pi }{3}$

Answer:

We know that tan x repeats after an interval of $\pi$ or 180 degree
$\tan (\frac{19\pi}{3}) = \tan (6\pi+\frac{\pi}{3})= \tan \frac{\pi}{3} =\tan 60\degree = \sqrt{3}$

Question:9 Find the values of the trigonometric functions $\sin\left ( -\frac{11\pi}{3} \right )$

Answer:

We know that sin x repeats after an interval of $2\pi or 360\degree$
$\sin \left ( -\frac{11\pi}{3} \right ) = \sin \left ( -4\pi +\frac{\pi}{3} \right ) = \sin \frac{\pi}{3} = \frac {\sqrt{3}}{2}$

Question:10 Find the values of the trigonometric functions $\small \cot \left ( -\frac{15\pi }{4} \right )$

Answer:

We know that cot x repeats after an interval of $\pi or 180\degree$
$\cot \left ( -\frac{15\pi}{4} \right ) = \cot \left ( -4\pi +\frac {\pi}{4} \right ) = \cot \left ( \frac{\pi}{4} \right ) = 1$

NCERT class 11 maths chapter 3 question answer - Exercise: 3.3

Question:1 Prove that $\small \sin ^{2} \left ( \frac{\pi }{6} \right ) + \cos ^{2}\left ( \frac{\pi }{3} \right ) - \tan ^{2}\left ( \frac{\pi }{4} \right ) = -\frac{1}{2}$

Answer:

We know the values of sin (30 degree), cos (60 degree) and tan (45 degree). That is:

$\sin \left ( \frac{\pi}{6} \right ) = \left ( \frac{1}{2} \right )\\ \\ \cos \left ( \frac{\pi}{3} \right ) = \left ( \frac{1}{2} \right )\\ \\ \tan \left ( \frac{\pi}{4} \right ) = 1$
$\sin^{2}\frac{\pi}{6}+\cos^{2}\frac{\pi}{3}-\tan^{2}\frac{\pi}{4}=$ $\left ( \frac{1}{2} \right )^{2}+ \left ( \frac {1}{2} \right )^{2}-1^{2}$

$= \frac{1}{4}+\frac{1}{4}-1= -\frac{1}{2}$
= R.H.S.

Question:2 Prove that $\small 2\sin ^{2}\left ( \frac{\pi }{6} \right ) + cosec ^{2}\left ( \frac{7\pi }{6} \right )\cos ^{2}\frac{\pi }{3} = \frac{3}{2}$

Answer:

The solutions for the given problem is done as follows.

$\sin\frac{\pi}{6} = \frac {1}{2}\\ \\ cosec\frac{7\pi}{6} = cosec\left ( \pi + \frac{\pi}{6} \right ) = -cosec \frac{\pi}{6}=-2\\ \\ \cos \frac{\pi}{3} = \frac{1}{2}$
$2\sin^{2}\frac{\pi}{6} +cosec^{2}\frac{7\pi}{6}\cos^{2}\frac{\pi}{3} = 2\left ( \frac{1}{2} \right )^{2}+\left ( -2 \right )^{2}\left ( \frac{1}{2} \right )^{2}\\ \\ \Rightarrow 2\times\frac{1}{4} + 4\times\frac{1}{4} = \frac {1}{2} + 1= \frac{3}{2}$
R.H.S.

Question:3 Prove that $\small \cot ^{2}\left ( \frac{\pi }{6} \right ) + \csc \left ( \frac{5\pi }{6} \right ) + 3\tan ^{2}\left ( \frac{\pi }{6} \right ) = 6$

Answer:

We know the values of cot(30 degree), tan (30 degree) and cosec (30 degree)

$\cot \frac{\pi}{6} = \sqrt{3}\\ \\ cosec\frac{5\pi}{6} = cosec\left ( \pi - \frac{\pi}{6} \right )=cosec\frac{\pi}{6} = 2\\ \\ \tan\frac{\pi}{6}= \frac{1}{\sqrt{3}}$

$\cot^{2}\frac{\pi}{6} + cosec\frac{5\pi}{6} +3\tan^{2}\frac{\pi}{6} = \left ( \sqrt(3) \right )^{2} + 2 + 3\times\left ( \frac{1}{\sqrt{3}} \right )^{2}\\ \\ \Rightarrow 3+2+1 = 6$
R.H.S.

Question:4 Prove that $\small 2\sin ^{2}\left ( \frac{3\pi }{4} \right ) + 2\cos ^{2}\left ( \frac{\pi }{4} \right ) + 2\sec ^{2}\left ( \frac{\pi }{3} \right ) = 10$

Answer:

$\sin \frac{3\pi}{4} = \sin\left ( \pi-\frac{\pi}{4} \right ) = \sin \frac{\pi}{4}= \frac{1}{\sqrt{2}}\\ \\ \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}\\ \\ \sec\frac{\pi}{3}= 2$
Using the above values

$2\sin^{2}\frac{3\pi}{4} +2\cos^{2}\frac{\pi}{4}+2\sec^{2}\frac{\pi}{3} = 2\times\left ( \frac{1}{\sqrt{2}} \right )^{2}+2\times\left ( \frac{1}{\sqrt{2}} \right )^{2}+2\left ( 2 \right )^{2}\\ \\ \Rightarrow 1+1+8=10$
R.H.S.

Question:5(i) Find the value of $\small (i) \sin 75\degree$

Answer:

$\sin 75\degree = \sin(45\degree + 30\degree)$
We know that
(sin(x+y)=sinxcosy + cosxsiny)
Using this idendity

$\sin 75\degree = \sin(45\degree + 30\degree) = \sin45\degree\cos30\degree + \cos45\degree\sin30\degree\\ \\ \Rightarrow \frac{1}{\sqrt{2}}\times\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}\times\frac{1}{2}\\ \\ \Rightarrow \frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}} = \frac{\sqrt{3}+1}{2\sqrt{2}}$

Question:5(ii) Find the value of
$\small (ii) \tan 15\degree$

Answer:

$\tan 15\degree = \tan (45\degree - 30\degree)$
We know that,

$\left [ \tan(x-y)= \frac{\tan x - \tan y}{1+\tan x\tan y} \right ]$
By using this we can write

$\tan (45\degree - 30\degree)= \frac{\tan 45\degree - tan30\degree}{1+\tan45\degree\tan30\degree}\\ \\ \Rightarrow \frac{1-\frac{1}{\sqrt{3}}}{1+1\left ( \frac{1}{\sqrt{3}} \right )} = \frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}} = \frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{\left ( \sqrt{3}-1 \right )^{2}}{\left ( \sqrt{3}+1 \right )\left ( \sqrt{3} -1\right )}=\frac{3+1-2\sqrt{3}}{\left ( \sqrt{3} \right )^{2}-\left ( 1 \right )^{2}}\\ \\ \Rightarrow \frac {4-2\sqrt{3}}{3-1}=\frac{2\left ( 2-\sqrt{3} \right )}{2}= 2-\sqrt{3}$

Question:6 Prove the following: $\small \cos \left ( \frac{\pi }{4}-x \right )\cos \left ( \frac{\pi }{4}-y \right ) - \sin \left ( \frac{\pi }{4} -x\right )\sin \left ( \frac{\pi }{4}-y \right ) =\sin (x+y)$

Answer:

$\cos\left ( \frac{\pi}{4}-x \right )\cos\left ( \frac{\pi}{4}-y \right ) - \sin\left ( \frac{\pi}{4}-x \right )\sin\left ( \frac{\pi}{4}-y \right )$

Multiply and divide by 2 both cos and sin functions
We get,

$\frac{1}{2}\left [2 \cos\left ( \frac{\pi}{4}-x \right )\cos\left ( \frac{\pi}{4}-y \right ) \right ] + \frac{1}{2}\left [- 2\sin\left ( \frac{\pi}{4}-x \right )\sin\left ( \frac{\pi}{4}-y \right ) \right ]$

Now, we know that

2cosAcosB = cos(A+B) + cos(A-B) -(i)
-2sinAsinB = cos(A+B) - cos(A-B) -(ii)
We use these two identities

In our question A = $\left (\frac{\pi}{4}-x \right )$

B = $\left (\frac{\pi}{4}-y \right )$
So,

$\frac{1}{2}\left [ \cos \left \{ \left ( \frac{\pi}{4}-x \right) +\left ( \frac{\pi}{4}-y \right ) \right \} + \cos \left \{ \left ( \frac{\pi}{4}-x \right) -\left ( \frac{\pi}{4}-y \right ) \right \} \right ] +\\ \\ \frac{1}{2}\left [ \cos \left \{ \left ( \frac{\pi}{4}-x \right) +\left ( \frac{\pi}{4}-y \right ) \right \} - \cos \left \{ \left ( \frac{\pi}{4}-x \right) +\left ( \frac{\pi}{4}-y \right ) \right \} \right ]$

$\Rightarrow 2 \times \frac{1}{2} \left [ \cos \left \{ \left ( \frac{\pi}{4}-x \right )+\left ( \frac{\pi}{4}-y \right ) \right \} \right ]$

$= \cos \left [ \frac{\pi}{2}-(x+y) \right ]$

As we know that

$(\cos \left ( \frac{\pi}{2} - A \right ) = \sin A)$
By using this

$= \cos \left [ \frac{\pi}{2}-(x+y) \right ]$ $=\sin(x+y)$

R.H.S

Question:7 Prove the following $\small \frac{\tan \left ( \frac{\pi }{4}+x \right )}{\tan \left ( \frac{\pi }{4} -x\right )} = \left ( \frac{1+\tan x}{1-\tan x} \right )^{2}$

Answer:

As we know that

$(\tan (A +B ) = \frac {\tan A + \tan B}{1- \tan A\tan B})$ and $\tan (A-B) = \frac {\tan A - \tan B }{1+ \tan A \tan B}$

So, by using these identities

$\frac{\tan \left ( \frac{\pi}{4}+x \right )}{\tan \left ( \frac{\pi}{4}-x \right )} = \frac{\frac{\tan \frac {\pi}{4} + \tan x}{1- \tan \frac{\pi}{4}\tan x}} {\frac{\tan \frac {\pi}{4} - \tan x}{1+ \tan \frac{\pi}{4}\tan x}} =\frac{ \frac {1+\tan x }{1- \tan x}} { \frac {1-\tan x }{1+ \tan x}} = \left ( \frac{1 + \tan x}{1 - \tan x} \right )^{2}$
R.H.S

Question:8 Prove the following $\small \frac{\cos (\pi +x)\cos (-x)}{\sin (\pi -x)\cos \left ( \frac{\pi }{2}+x \right )} = \cot ^{2} x$

Answer:

As we know that,
$\cos(\pi+x) = -\cos x$ , $\sin (\pi - x ) = \sin x$ , $\cos \left ( \frac{\pi}{2} + x\right ) = - \sin x$
and
$\cos (-x) = \cos x$

By using these our equation simplify to

$\frac{\cos x \times -\cos x}{sin x \times - \sin x} = \frac{- \cos^{2}x}{-\sin^{2}x} = \cot ^ {2}x$ $(\because \cot x = \frac {\cos x}{\sin x})$
R.H.S.

Question:9 Prove the following $\small \cos \left ( \frac{3\pi }{2} +x\right )\cos (2\pi +x)\left [ \cot \left ( \frac{3\pi }{2}-x \right ) +\cot (2\pi +x)\right ] = 1$

Answer:

We know that

$\cos \left ( \frac{3\pi}{2}+x \right ) = \sin x\\ \\ \cos (2\pi +x)= \cos x\\ \\ \cot\left ( \frac{3\pi}{2} -x\right ) = \tan x\\ \\ \cot (2\pi + x) = \cot x$

So, by using these our equation simplifies to

$\cos \left ( \frac{3\pi }{2} +x\right )\cos (2\pi +x)\left [ \cot \left ( \frac{3\pi }{2}-x \right ) +\cot (2\pi +x)\right ] \\=\sin x\cos x [\tan x + \cot x] = \sin x\cos x [\frac {\sin x}{\cos x} + \frac{\cos x}{\sin x}]\\ \\ \Rightarrow \sin x\cos x\left [ \frac{\sin^{2}x+\cos^{2}x}{\sin x\cos x } \right ] =\sin^{2}x+\cos^{2}x = 1$ R.H.S.

Question:10 Prove the following $\small \sin (n+1)x\sin(n+2)x + \cos(n+1)x\cos(n+2)x =\cos x$

Answer:

Multiply and divide by 2

$= \frac {2\sin(n+1)x \sin(n+2)x + 2\cos (n+1)x\cos(n+2)x}{2}$

Now by using identities

-2sinAsinB = cos(A+B) - cos(A-B)
2cosAcosB = cos(A+B) + cos(A-B)

$\frac{\left \{ -\left (\cos(2n+3)x - \cos (-x) \right ) + \left ( \cos(2n+3) +\cos(-x) \right )\right \}}{2}\\ \\ \left ( \because \cos(-x) = \cos x \right )\\ \\ = \frac{2\cos x}{2} = \cos x$

R.H.S.

Question:11 Prove the following $\small \cos \left ( \frac{3\pi }{4}+x \right ) - \cos\left ( \frac{3\pi }{4} -x\right ) = -\sqrt{2} \sin x$

Answer:

We know that

[ cos(A+B) - cos (A-B) = -2sinAsinB ]

By using this identity

$\cos \left ( \frac {3\pi}{4}+x \right ) - \cos \left ( \frac {3\pi}{4}-x \right ) = -2\sin\frac{3\pi}{4}\sin x = -2\times \frac{1}{\sqrt{2}}\sin x\\ \\ = -\sqrt{2}\sin x$ R.H.S.

Question:12 Prove the following $\small \sin^{2}6x - \sin^{2}4x = \sin2x\sin10x$

Answer:

We know that
$a^{2} - b^{2} = (a+b)(a-b)$

So,
$\sin^{2}6x - \sin^{2}4x =(\sin6x + \sin4x)(\sin6x - \sin4x)$

Now, we know that

$\sin A + \sin B = 2\sin \left ( \frac{A+B}{2} \right )\cos\left ( \frac{A-B}{2} \right )\\ \\ \sin A - \sin B = 2\cos \left ( \frac{A+B}{2} \right )\sin\left ( \frac{A-B}{2} \right )$
By using these identities
sin6x + sin4x = 2sin5x cosx
sin6x - sin4x = 2cos5x sinx

$\Rightarrow \sin^{2}6x - \sin^{2}4x = (2\cos5x\sin5x)(2\sin x\cos x)$

Now,

2sinAcosB = sin(A+B) + sin(A-B)
2cosAsinB = sin(A+B) - sin(A-B)

by using these identities

2cos5x sin5x = sin10x - 0
2sinx cosx = sin2x + 0

hence
$\sin^{2}6x-\sin^{2}4x = \sin2x\sin10x$

Question:13 Prove the following $\small \cos^{2}2x - \cos^{2}6x = \sin4x\sin8x$

Answer:

As we know that

$a^{2}-b^{2} =(a-b)(a+b)$

$\therefore \cos^{2}2x -\cos^{2}6x = (\cos2x-\cos6x)(\cos2x+\cos6x)$
Now
$\cos A - \cos B = -2\sin\left ( \frac{A+B}{2} \right )\sin\left ( \frac{A-B}{2} \right )\\ \\ \cos A + \cos B = 2\cos\left ( \frac{A+B}{2} \right )\cos\left ( \frac{A-B}{2} \right )$
By using these identities

cos2x - cos6x = -2sin(4x)sin(-2x) = 2sin4xsin2x ( $\because$ sin(-x) = -sin x
cos(-x) = cosx)
cos2x + cos 6x = 2cos4xcos(-2x) = 2cos4xcos2x

So our equation becomes

R.H.S.

Question:14 Prove the following $\small \sin2x +2\sin4x + \sin6x = 4\cos^{2}x\sin4x$

Answer:

We know that

$\sin A+ \sin B = 2\sin \left ( \frac{A+B}{2} \right )\cos\left ( \frac{A-B}{2} \right )$
We are using this identity
sin2x + 2sin4x + sin6x = (sin2x + sin6x) + 2sin4x

sin2x + sin6x = 2sin4xcos(-2x) = 2sin4xcos(2x) ( $\because$ cos(-x) = cos x)

So, our equation becomes
sin2x + 2sin4x + sin6x = 2sin4xcos(2x) + 2sin4x
Now, take the 2sin4x common
sin2x + 2sin4x + sin6x = 2sin4x(cos2x +1) ( $\because \cos2x = 2\cos^{2}x - 1$ )
=2sin4x( $2\cos^{2}x - 1$ +1 )
=2sin4x( $2\cos^{2}x$ )
= $4\sin4x\cos^{2}x$
R.H.S.

Question:15 Prove the following $\small \cot4x(\sin5x + \sin3x) = \cot x(\sin5x - \sin3x)$

Answer:

We know that
$\sin x + \sin y = 2\sin\left ( \frac{x+y}{2} \right )\cos\left (\frac{x-y}{2} \right )$
By using this , we get

sin5x + sin3x = 2sin4xcosx

$\frac{\cos4x}{\sin4x}\left ( 2\sin4x\cos x \right ) = 2\cos4x\cos x\\ \\$

now nultiply and divide by sin x

$\\\ \\ \frac{2\cos4x\cos x \sin x}{\sin x } \ \ \ \ \ \ \ \ \ \ \\ \\ =\cot x (2\cos4x\sin x) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left ( \because \frac{\cos x}{\ sin x} = \cot x \right )\\ \\$

Now we know that

$\\ 2\cos x\sin y = \sin(x+y) - \sin(x-y)\\ \\$

By using this our equation becomes

$\\ \\=\cot x (\sin5x - sin3x)\\$
R.H.S.

Question:16 Prove the following $\small \frac{\cos 9x - \cos 5x}{\sin17x - \sin3x} = -\frac{\sin2x}{\cos10x}$

Answer:

As we know that

$\\ \cos x - \cos y = -2\sin\frac{x+y}{2}\sin\frac{x-y}{2 }\\ \\ \cos 9x - \cos 5x = -2\sin 7x \sin2x \\ \\ \sin x - \sin y = 2\cos\frac{x+y}{2}\sin\frac{x-y}{2 }\\ \\ \sin 17x - \sin 3x = 2\cos10x \sin7x\\ \\ \frac{\cos 9x - \cos 5x}{\sin 17x - \sin 3x} =\frac{-2\sin 7x \sin2x}{2\cos10x \sin7x} = -\frac{\sin 2x}{\cos10x}$
R.H.S.

Question:17 Prove the following $\small \frac{\sin5x + \sin3x}{\cos5x + \cos3x} = \tan4x$

Answer:

We know that

$\\ \sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}\\and\\ \\ \cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} \\$

We use these identities

$\\ \sin5x + \sin3x = 2\sin4x\cos x\\ \cos5 x + \cos 3x = 2\cos4x\cos x \\ \\ \frac{\sin5x + \sin3x}{\cos5 x + \cos 3x} = \frac{ 2\sin4x\cos x}{2\cos4x\cos x} = \frac{\sin4x}{\cos 4x} = \tan 4x$
R.H.S.

Question:18 Prove the following $\small \frac{\sin x - \sin y}{\cos x+\cos y} = \tan \frac{(x-y)}{2}$

Answer:

We know that
$\sin x - \sin y = 2\cos\frac{x+y}{2 }\sin\frac{x-y}{2}\\and \\ \\ \cos x +\cos y = 2\cos\frac{x+y}{2 }\cos\frac{x-y}{2}\\$

We use these identities

$\\ We \ use \ these \ identities\\ \\ \frac{\sin x - \sin y}{\cos x +\cos y} =\frac{2\cos\frac{x+y}{2 }\sin\frac{x-y}{2}}{ 2\cos\frac{x+y}{2 }\cos\frac{x-y}{2}} = \frac{\sin\frac{x-y}{2}}{\cos\frac{x-y}{2}} = \tan \frac{x-y}{2}$

R.H.S.

Question:19 Prove the following $\small \frac{\sin x + \sin 3x}{\cos x + \cos3x} = \tan2x$

Answer:

We know that

$\\ \sin x + \sin y = 2\sin\frac{x+y}{2}\cos\frac{x-y}{2}\\and\\ \\ \cos x + \cos y = 2\cos\frac{x+y}{2}\cos\frac{x-y}{2}\\ \\ We \ use \ these \ equations \\ \\ \sin x + \sin3x = 2\sin2x\cos(-x) = 2\sin2x\cos x \ \ \ \ \ (\because \cos(-x) = \cos x)\\ \\ \cos x + \cos3x = 2\cos2x\cos(-x) =2\cos2x\cos x \ \ \ \ \ (\because \cos(-x) = \cos x)\\ \\ \frac{\sin x + \sin3x}{\cos x + \cos3x} = \frac {2\sin2x\cos x}{2\cos2x\cos x}= \frac{\sin2x}{\cos2x} = \tan2x$ R.H.S.

Question:20 Prove the following $\small \frac{\sin x - \sin 3x}{\sin^{2}x-\cos^{2}x} = 2\sin x$

Answer:
We know that

$\sin3x = 3\sin x - 4\sin^{3}x \ \ \ , \ \ \cos^{2}-\sin^{2}x = \cos2x \\and \\ \cos2x = 1 - 2\sin^{2}x \\$

We use these identities

$\sin x - \sin3x = \sin x - (3\sin x - 4\sin^{3}x) = 4\sin^{3}x - 2\sin x\\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2\sin x (2\sin^{2}x - 1)\\ \\ \cos^{2}x- \sin^{2} = \cos2x\\ \cos2x = 1 - 2\sin^{2}x$

$\sin x - \sin3x = \sin x - (3\sin x - 4\sin^{3}x) = 4\sin^{3}x - 2\sin x\\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2\sin x (2\sin^{2}x - 1)\\ \\ \sin^{2}-\cos^{2}x = - \cos2x \ \ \ \ \ \ \ \ \ \ (\cos2x = 1 - 2\sin^{2}x)\\ \sin^{2}-\cos^{2}x = -( 1 - 2\sin^{2}x) = 2\sin^(2)x - 1\\ \\ \frac{\sin x - \sin3x}{\sin^{2}-\cos^{2}x } = \frac{ 2\sin x (2\sin^{2}x - 1)}{ 2\sin^(2)x - 1} = 2\sin x$

$\sin x - \sin3x = \sin x - (3\sin x - 4\sin^{3}x) = 4\sin^{3}x - 2\sin x\\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2\sin x (2\sin^{2}x - 1)\\ \\ \sin^{2}-\cos^{2}x = - \cos2x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because \cos2x = 1 - 2\sin^{2}x)\\ \sin^{2}-\cos^{2}x = -( 1 - 2\sin^{2}x) = 2\sin^(2)x - 1\\ \\ \frac{\sin x - \sin3x}{\sin^{2}-\cos^{2}x } = \frac{ 2\sin x (2\sin^{2}x - 1)}{ 2\sin^{2}x - 1} = 2\sin x$
R.H.S.

Question:21 Prove the following $\small \frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} = \cot 3x$

Answer:

We know that

$\cos x + \cos y = 2\cos\frac{x+y}{2}\cos\frac{x-y}{2}\\ and \\ \sin x + \sin y = 2\sin\frac{x+y}{2}\cos\frac{x-y}{2}$
We use these identities

$\frac{(\cos4x + \cos2x) + \cos3x}{(\sin4x+\sin2x)+\sin3x} = \frac{2\cos3x\cos x + \cos3x}{2\sin3x\cos x+\sin3x} = \frac{2\cos3x(1+\cos x)}{2\sin3x(1+\cos x)}\\ \ \ \\ \ \ \ \ \ \ \ = cot 3x$

=RHS

Question:22 prove the following $\small \cot x \cot2x - \cot2x\cot3x - \cot3x\cot x =1$

Answer:

cot x cot2x - cot3x(cot2x - cotx)
Now we can write cot3x = cot(2x + x)

and we know that

$cot(a+b) = \frac{\cot a \cot b - 1}{\cot a + \cot b}$
So,
$cotx\ cot2x-\frac{\cot 2x \cot x - 1}{\cot 2x + \cot x}(cot2x+cotx)$
= cotx cot2x - (cot2xcotx -1)
= cotx cot2x - cot2xcotx +1
= 1 = R.H.S.

Question:23 Prove that $\small \tan4x = \frac{4\tan x(1-\tan^{2}x)}{1-6 \tan^{2}x+\tan^{4}x}$

Answer:

We know that

$tan2A=\frac{2\tan A}{1 - \tan^{2}A}$

and we can write tan 4x = tan 2(2x)
So, $tan4x=\frac{2\tan 2x}{1 - \tan^{2}2x}$ = $\frac{2( \frac{2\tan x}{1 - \tan^{2}x})}{1 - (\frac{2\tan x}{1 - \tan^{2}x})^{2}}$

= $\frac{2 (2\tan x)(1 - \tan^{2}x)}{(1-\tan x)^{2} - (4\tan^{2} x)}$

= $\frac{(4\tan x)(1 - \tan^{2}x)}{(1)^{2}+(\tan^{2} x)^{2} - 2 \tan^{2} x - (4\tan^{2} x)}$

= $\frac{(4\tan x)(1 - \tan^{2}x)}{1^{2}+\tan^{4} x - 6 \tan^{2} x }$ = R.H.S.

Question:24 Prove the following $\small \cos4x = 1 - 8\sin^{2}x\cos^{2}x$

Answer:

We know that
$cos2x=1-2\sin^{2}x$
We use this in our problem
cos 4x = cos 2(2x)
= $1-2\sin^{2}2x$
= $1-2(2\sin x \cos x)^{2}$ $(\because \sin2x = 2\sin x \cos x)$
= $1-8\sin^{2}x\cos^{2}x$ = R.H.S.

Question:25 Prove the following $\small \cos6x = 32\cos^{6}x -48\cos^{4}x + 18\cos^{2}x-1$

Answer:

We know that
cos 3x = 4 $\cos^{3}x$ - 3cos x
we use this in our problem
we can write cos 6x as cos 3(2x)
cos 3(2x) = 4 $\cos^{3}2x$ - 3 cos 2x
= $4(2\cos^{2}x - 1)^{3}$ - $3(2\cos^{2}x - 1)$ $(\because \cos 2x = 2\cos^{2}x - 1)$
= $4[(2cos^{2}x)^{3} -(1)^{3} -3(2cos^{2}x)^{2}(1) + 3(2cos^{2}x)(1)^{2}]$ $-6\cos^{2}x + 3$ $(\because (a-b)^{3} = a^{3} - b^{3} - 3a^{2}b+ 3ab^{2})$
= 32 $cos^{6}x$ - 4 - 48 $cos^{4}x$ + 24 $cos^{2}x$ - $6\cos^{2}x + 3$
= 32 $cos^{6}x$ - 48 $cos^{4}x$ + 18 $cos^{2}x$ - 1 = R.H.S.

NCERT class 11 maths chapter 3 question answer - Exercise: 3.4

Question:1 Find the principal and general solutions of the following equations: $\tan x= \sqrt{3}$

Answer:

It is given that given
$\tan x= \sqrt{3}$
Now, we know that $\tan\frac{\pi}{3}= \sqrt3$ and $\tan\frac{4\pi}{3}= \tan \left ( \pi+\frac{\pi}{3} \right )=\sqrt3$

Therefore,
the principal solutions of the equation are $x = \frac{\pi}{3},\frac{4\pi}{3}$
Now,
The general solution is $\tan x =\tan \frac{\pi}{3}$

$x =n{\pi} + \frac{\pi}{3}$ where $n \ \epsilon \ Z$ and Z denotes sets of integer

Therefore, the general solution of the equation is $x =n{\pi} + \frac{\pi}{3}$ where $n \ \epsilon \ Z$ and Z denotes sets of integer

Question:2 Find the principal and general solutions of the following equations: $\small \sec x = 2$

Answer:

We know that value of $\sec\frac{\pi}{3} = 2$ and $\sec\frac{5\pi}{3} = \sec\left ( 2\pi -\frac{\pi}{3} \right ) = \sec\frac{\pi}{3} = 2$

Therefore the principal solutions are x = $\frac{\pi}{3} and \frac{5\pi}{3}$
$\sec x = \sec\frac{\pi}{3}$
We know that value of sec x repeats after an interval of $2\pi$
So, by this we can say that

the general solution is x = $2n\pi \pm \frac{\pi}{3}$ where n $\epsilon$ Z

Question:3 Find the principal and general solutions of the following equations: $\small \cot x = - \sqrt{3}$

Answer:

we know that $\ cot\frac{\pi}{6} = \sqrt{3}$ and we know that $\ \cot\frac{5\pi}{6} = \cot\left ( \pi -\frac{\pi}{6} \right ) = -cot\frac{\pi}{6} = -\sqrt{3}$

Similarly , the value for $\ \cot\frac{11\pi}{6} = \cot\left ( 2\pi -\frac{\pi}{6} \right ) = -cot\frac{\pi}{6} = -\sqrt{3}$
Therefore, principal solution is x = $\frac{5\pi}{6} \ and \ \frac{11\pi}{6}$

We also know that the value of cot x repeats after an interval of $\pi$
There the general solution is x = $n\pi \pm \frac{5\pi}{6} \ where \ n \ \epsilon \ Z$

Question:4 Find the principal and general solutions of the following equations: $\small cosec x = -2$

Answer:

We know that
$cosec \frac{\pi}{6} = 2$

$cosec (\pi + \frac{\pi}{6}) = -cosec\frac{\pi}{6} = -2$ and also $cosec (2\pi - \frac{\pi}{6}) = cosec\frac{11\pi}{6} = -2$
So,
$cosec x= cosec\frac{7\pi}{6}$ and $cosec x= cosec\frac{11\pi}{6}$

So, the principal solutions are $x = \frac{7\pi}{6} \ and \ \frac{11\pi}{6}$

Now,
$cosec x= cosec\frac{7\pi}{6}$

$\sin x = \sin\frac{7\pi}{6}$ $\left ( \because \sin x = \frac{1}{cosec x} \right )$

$x = n\pi + (-1)^{n}\frac{7\pi}{6}$
Therefore, the general solution is

$x = n\pi + (-1)^{n}\frac{7\pi}{6}$

where $n \ \epsilon \ Z$

Question:5 Find the general solution for each of the following equation $\small \cos 4x = \cos 2x$

Answer:

cos4x = cos2x
cos4x - cos2x = 0
We know that
$\cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$
We use this identity
$\therefore$ cos 4x - cos 2x = -2sin3xsinx
$\Rightarrow$ -2sin3xsinx = 0 $\Rightarrow$ sin3xsinx=0
So, by this we can that either
sin3x = 0 or sinx = 0
3x = $n\pi$ x = $n\pi$
x = $\frac{n\pi}{3}$ x = $n\pi$

Therefore, the general solution is

$x=\frac{n\pi}{3}\ or\ n\pi \ where \ n\in Z$

Question:6 Find the general solution of the following equation $\small \cos 3x + \cos x -\cos 2x = 0$

Answer:

We know that
$\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} \\ and \\ \cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$
We use these identities
(cos3x + cosx) - cos2x = 2cos2xcosx -cos2x = 0
= cos2x(2cosx-1) = 0
So, either
cos2x = 0 or $cosx=\frac{1}{2}$
$2x=(2n+1)\frac{\pi}{2}$ $cosx =\cos\frac{\pi}{3}$
$x=(2n+1)\frac{\pi}{4}$ $x =2n\pi \pm \frac{\pi}{3}$

$\therefore$ the general solution is

$x=(2n+1)\frac{\pi}{4}$ $\ or \ 2n\pi \pm \frac{\pi}{3}$

Question:7 Find the general solution of the following equation $\small \sin 2x + \cos x = 0$

Answer:

sin2x + cosx = 0
We know that
sin2x = 2sinxcosx
So,
2sinxcosx + cosx = 0
cosx(2sinx + 1) = 0
So, we can say that either

cosx = 0 or 2sinx + 1 = 0
$x=(2n+1)\frac{\pi}{2}$ $sinx =\sin\frac{7\pi}{6}$
$x=n\pi +(-1)^{n}\frac{7\pi}{6}$
Therefore, the general solution is

$x=(2n+1)\frac{\pi}{2}$ $or$ $n\pi +(-1)^{n}\frac{7\pi}{6} \ where \ n\in Z$

Question:8 Find the general solution of the following equation $\small \sec^{2}2x = 1 - \tan2x$

Answer:

We know that
$\sec^{2}x = 1 + \tan^{2}x$
So,
$1 + \tan^{2}2x = 1 -\tan2x$
$\tan^{2}2x + \tan2x = 0\\ \\ \tan2x(\tan2x+1) = 0$
either
tan2x = 0 or tan2x = -1 ( $\tan x = \tan \left ( \pi - \frac{\pi}{4} \right ) = \tan\frac{3\pi}{4}$ )
2x = $n\pi$ $2x=n\pi + \frac{3\pi}{4}$
$x=\frac{n\pi}{2}$ $x=\frac{n\pi}{2} + \frac{3\pi}{8}$
Where n $\epsilon$ Z

Question:9 Find the general solution of the following equation $\small \sin x + \sin 3x + \sin 5x = 0$

Answer:

We know that
$\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$
We use this identity in our problem
$\sin 5x + \sin x = 2\sin\frac{5x+x}{2}\cos\frac{5x-x}{2} =2\sin3x\cos2x$
Now our problem simplifeis to
$2\sin3x\cos2x+ \sin3x$ = 0
take sin3x common
$\sin3x(2\cos2x+ 1) = 0$
So, either
sin3x = 0 or $\cos2x = -\frac{1}{2}$ $\left ( \cos2x = -\cos\frac{\pi}{3} = \cos\left ( \pi - \frac{\pi}{3} \right ) = \cos\frac{2\pi}{3} \right )$
$3x = n\pi$ $2x = 2n\pi \pm \frac{2\pi}{3}$
$x = \frac{n\pi}{3}$ $x = n\pi \pm \frac{\pi}{3}$
Where $n \ \epsilon \ Z$

Class 11 maths chapter 3 NCERT solutions - Miscellaneous Exercise

Question:1 Prove that $\small 2\cos\frac{\pi }{13}\cos\frac{9\pi }{13}+\cos\frac{3\pi }{13}+\cos\frac{5\pi }{13}=0$

Answer:

We know that

cos A+ cos B = $2\cos(\frac{A+B}{2})\cos(\frac{A-B}{2})$

we use this in our problem

$\small 2\cos\frac{\pi }{13}\cos\frac{9\pi }{13}+2\cos\frac{(\frac{3\pi }{13}+\frac{5\pi}{13})}{2}\cos\frac{(\frac{3\pi}{13}-\frac{5\pi }{13})}{2}$

$\small 2\cos\frac{\pi }{13}\cos\frac{9\pi }{13}+2\cos\frac{4\pi }{13}\cos\frac{-\pi}{13}$ ( we know that cos(-x) = cos x )

$\small 2\cos\frac{\pi }{13}\cos\frac{9\pi }{13}+2\cos\frac{4\pi }{13}\cos\frac{\pi}{13}$
$\small 2\cos\frac{\pi }{13}(\cos\frac{9\pi }{13}+\cos\frac{4\pi }{13})$
again use the above identity

$\small 2\cos\frac{\pi }{13}(2\cos(\frac{\frac{9\pi }{13}+\frac{4\pi }{13}}{2})\cos(\frac{\frac{9\pi }{13}-\frac{4\pi }{13}}{2})$
$\small 2\cos\frac{\pi }{13}2\cos\frac{\pi }{2}\cos\frac{5\pi }{26}$
we know that

$\small \cos\frac{\pi }{2}$ = 0
So,
$\small 2\cos\frac{\pi }{13}2\cos\frac{\pi }{2}\cos\frac{5\pi }{26}$ = 0 = R.H.S.

Question:2 Prove that $\small (\sin 3x + \sin x)\sin x + (\cos 3x - \cos x )\cos x = 0$

Answer:

We know that
$sin3x=3\sin x - 4\sin^{3}x$
and
$cos3x=4\cos^{3}x - 3\cos x$
We use this in our problem
$\small (\sin 3x + \sin x)\sin x + (\cos 3x - \cos x )\cos x$
= $(3\sin x - 4\sin^{3}x+ sin x) sinx$ + $(4\cos^{3}x - 3\cos x- cos x)cos x$
= (4sinx - 4 $\small \sin^{3}x$ )sinx + (4 $\small \cos^{3}x$ - 4cos x)cosx
now take the 4sinx common from 1st term and -4cosx from 2nd term
= 4 $\small \sin^{2}x$ (1 - $\small \sin^{2}x$ ) - 4 $\small \cos^{2}x$ (1 - $\small \cos^{2}x$ )
= 4 $\small \sin^{2}x$ $\small \cos^{2}x$ - 4 $\small \cos^{2}x$ $\small \sin^{2}x$ $\small \because \ \ \ \cos^{2}x = 1 - \sin^2x\\ and\\ \sin^{2}x = 1 -\cos^{2}x$
= 0 = R.H.S.

Question:3 Prove that $\small (\cos x + \cos y)^{2} + (\sin x - \sin y)^{2} = 4 \cos^{2}\left ( \frac{x+y}{2} \right )$

Answer:

We know that $(a+b)^{2} = a^{2} + 2ab + b^{2}$
and
$(a-b)^{2} = a^{2} - 2ab + b^{2}$
We use these two in our problem

$(\sin x-\sin y)^{2} = \sin^{2}x - 2\sin x\sin y + \sin^{2}y$
and
$(\cos x+\cos y)^{2} = \cos^{2}x + 2\cos x\cos y + \cos^{2}y$

$\small (\cos x + \cos y)^{2} + (\sin x - \sin y)^{2}$ = $\cos^{2}x + 2\cos x\cos y + \cos^{2}y$ + $\sin^{2}x - 2\sin x\sin y + \sin^{2}y$
= 1 + 2cosxcosy + 1 - 2sinxsiny $\left ( \because \sin^{2}x + \cos^{2}x = 1\ and \ \sin^{2}y + \cos^{2}y = 1 \right )$
= 2 + 2(cosxcosy - sinxsiny)
= 2 + 2cos(x + y)
= 2(1 + cos(x + y) )
Now we can write
$cos(x + y) =2cos^{2}\frac{(x + y)}{2} - 1$ $\left ( \because \cos2x = 2cos^{2}x - 1 \ \Rightarrow \cos x = 2\cos^{2}\frac{x}{2} - 1\right )$

= $2(1 + 2cos^{2}\frac{(x + y)}{2} - 1)$
$=4cos^{2}\frac{(x + y)}{2}$

= R.H.S.

Question:4 Prove that $\small (\cos x-\cos y)^{2} + (\sin x - \sin y)^{2} = 4\sin^{2}\left ( \frac{x-y}{2} \right )$

Answer:

We know that $(a+b)^{2} = a^{2} + 2ab + b^{2}$
and
$(a-b)^{2} = a^{2} - 2ab + b^{2}$
We use these two in our problem

$(\sin x-\sin y)^{2} = \sin^{2}x - 2\sin x\sin y + \sin^{2}y$
and
$(\cos x-\cos y)^{2} = \cos^{2}x - 2\cos x\cos y + \cos^{2}y$

$\small (\cos x - \cos y)^{2} + (\sin x - \sin y)^{2}$ = $\cos^{2}x - 2\cos x\cos y + \cos^{2}y$ + $\sin^{2}x - 2\sin x\sin y + \sin^{2}y$
= 1 - 2cosxcosy + 1 - 2sinxsiny $\left ( \because \sin^{2}x + \cos^{2}x = 1\ and \ \sin^{2}y + \cos^{2}y = 1 \right )$
= 2 - 2(cosxcosy + sinxsiny)
= 2 - 2cos(x - y) $\small (\because \cos(x-y) =\cos x \cos y + \sin x \sin y)$
= 2(1 - cos(x - y) )
Now we can write
$cos(x + y) = 1 -2sin^{2}\frac{(x + y)}{2}$ $\left ( \because \cos2x = 1 - 2\sin^{2}x \ \Rightarrow \cos x = 1 - 2\sin^{2}\frac{x}{2} \right )$

$2(1 - cos(x - y) ) = 2(1 - ( 1 -2sin^{2}\frac{(x + y)}{2}))$

$= 4sin^{2}\frac{(x - y)}{2}$ = R.H.S.

Question:5 Prove that $\small \sin x + \sin 3x + \sin 5x + \sin 7x = 4\cos x\cos2x \sin4x$

Answer:
we know that
$sinA + sinB =2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$
We use this identity in our problem
If we notice we need sin4x in our final result so it is better if we made a combination of sin7x and sin x , sin3x and sin5x tp get sin4x

$(sin7x + sinx) + (sin5x + sin3x) = 2\sin\frac{7x+x}{2}\cos\frac{7x-x}{2}$ $+2\sin\frac{5x+3x}{2}\cos\frac{5x-3x}{2}$
$=$ $2\sin4x\cos3x + 2\sin4x\cos x$
take 2sin4x common
= 2sin4x(cos3x + cosx)
Now,
We know that
$cosA + cosB =2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$
We use this
$cos3x + cosx =2\cos\frac{3x+x}{2}\cos\frac{3x-x}{2}$
= $2\cos2x\cos x$
= 2sin4x( $2\cos2x\cos x$ )
= 4cosxcos2xsin4x = R.H.S.

Question:6 Prove that $\small \frac{(\sin 7x + \sin 5x) + (\sin9x + \sin 3x)}{(\cos7x + \cos5x) + (\cos9x + \cos3x)} = \tan6x$

Answer:

We know that

$sinA + sinB = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$
and
$cosA + cosB =2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$

We use these two identities in our problem

sin7x + sin5x = $2\sin\frac{7x+5x}{2}\cos\frac{7x-5x}{2}$ = $2\sin6x\cos x$

sin 9x + sin 3x = $2\sin\frac{9x+3x}{2}\cos\frac{9x-3x}{2}$ = $2\sin6x\cos 3x$

cos 7x + cos5x = $2\cos\frac{7x+5x}{2}\cos\frac{7x-5x}{2}$ = $2\cos6x\cos x$

cos 9x + cos3x = $2\cos\frac{9x+3x}{2}\cos\frac{9x-3x}{2}$ = $2\cos6x\cos 3x$

$\small \frac{(\sin 7x + \sin 5x) + (\sin9x + \sin 3x)}{(\cos7x + \cos5x) + (\cos9x + \cos3x)}$ = $\small \frac{(2\sin 6x\cos x) + (2\sin6x \cos3x)}{(2\cos6x cos x) + (2\cos6x cos3x)}$

= $\small \frac{2\sin6x(\cos x + \cos3x)}{2\cos6x (cos x + cos3x)} = \tan6x$ = R.H.S. $\small \left ( \because \frac{\sin x}{\cos x} = \tan x\right )$

Question:7 Prove that $\small \sin3x + \sin2x - \sin x = 4\sin x \cos\frac{x}{2}\cos\frac{3x}{2}$

Answer:

We know that
$cosA + cosB = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$
$sinA - sinB = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$

we use these identities
$sin3x - sinx = 2\cos\frac{3x+x}{2}\sin\frac{3x-x}{2}$

$= 2\cos2x\sin x$

sin2x + $2\cos2x\sin x$ = 2sinx cosx + $2\cos2x\sin x$
take 2 sinx common
$2sinx ( cosx + cos2x) = 2sinx(2\cos\frac{2x+x}{2}\cos\frac{2x-x}{2})$

$= 2sinx(2\cos\frac{3x}{2}\cos\frac{x}{2})$
$= 4sinx\cos\frac{3x}{2}\cos\frac{x}{2}$

= R.H.S.

Question:8 Find $\small \sin\frac{x}{2} , \cos\frac{x}{2} , and \tan\frac{x}{2}$ in $\small \tan x = - \frac{4}{3}$ , x in quadrant II

Answer:

tan x = $-\frac{4}{3}$
We know that ,
$\sec^{2}x = 1 + \tan^{2}x$
$= 1 +\left ( -\frac{4}{3} \right )^{2}$
$= 1 + \frac{16}{9}$ = $\frac{25}{9}$
$sec x = \sqrt{\frac{25}{9}}$ = $\pm\frac{5}{3}$
x lies in II quadrant thats why sec x is -ve
So,

$sec x =-\frac{5}{3}$
Now, $cos x = \frac{1}{\sec x}$ = $-\frac{3}{5}$
We know that,
$cos x = 2\cos^{2}\frac{x}{2}- 1$ ( $\because \cos2x = 2\cos^{2}x - 1 \Rightarrow \cos x = 2\cos^{2}\frac{x}{2} - 1$ )
$-\frac{3}{5}+ 1 = 2$ $\cos^{2}\frac{x}{2}$

= $\frac{-3+5}{5}$ = $2\cos^{2}\frac{x}{2}$

$\frac{2}{5}$ = $2\cos^{2}\frac{x}{2}$
$\cos^{2}\frac{x}{2}$ = $\frac{1}{5}$
$\cos\frac{x}{2}$ = $\sqrt{\frac{1}{5}}$ = $\pm\frac{1}{\sqrt5}$
x lies in II quadrant so value of $\cos\frac{x}{2}$ is +ve

$\cos\frac{x}{2}$ = $\frac{1}{\sqrt5} = \frac{\sqrt5}{5}$
we know that
$cos x =1 - 2\sin^{2}\frac{x}{2}$

$2\sin^{2}\frac{x}{2}$ = 1 - $(-\frac{3}{5})$ = $\frac{8}{5}$

$\sin^{2}\frac{x}{2} = \frac{4}{5}\\ \\=\sin\frac{x}{2} = \sqrt{ \frac{4}{5}} = \pm \frac{2}{\sqrt{5}}$
x lies in II quadrant So value of sin x is +ve

$\sin\frac{x}{2} = \frac{2}{\sqrt{}5} = \frac{2\sqrt5}{5}$

$\tan \frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \frac{\frac{2\sqrt5}{5}}{\left ( \frac{\sqrt5}{5} \right )} = 2$

Question:9 Find $\small \sin\frac{x}{2} , \cos\frac{x}{2} , and \tan\frac{x}{2}$ in $\small \cos x = -\frac{1}{3}$ , x in quadrant III

Answer:

$\pi < x < \frac{3\pi}{2}\\ \\ \frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$

We know that
cos x = $2\cos^{2}\frac{x}{2} - 1$
$2\cos^{2}\frac{x}{2} =$ cos x + 1
= $\left ( -\frac{1}{3} \right )$ + 1 = $\left ( \frac{-1+3}{3} \right )$ = $\frac{2}{3}$

$\cos\frac{x}{2} = \sqrt{ \frac{1}{3}} = \pm \frac{1}{\sqrt3}$

$\cos\frac{x}{2} = - \frac{1}{\sqrt3} = - \frac{\sqrt3}{3}$
Now,
we know that
cos x = $1 - 2\sin^{2}\frac{x}{2}$
$2\sin^{2}\frac{x}{2} = 1 - \cos x$
= 1 - $\left ( -\frac{1}{3} \right )$ = $\frac{3+1}{3}$ = $\frac{4}{3}$

$2\sin^{2}\frac{x}{2} = \frac{4}{3} \\ \\ \sin^{2}\frac{x}{2} = \frac{2}{3}\\ \sin\frac{x}{2} = \sqrt{ \frac{2}{3}} = \pm \sqrt{ \frac{2}{3}} = \frac{\sqrt6}{3}$
Because $\sin\frac{x}{2}$ is +ve in given quadrant

$\tan\frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \frac{\frac{\sqrt6}{3}}{\frac{-\sqrt3}{3}} = - \sqrt2$

Question:10 Find $\small \sin\frac{x}{2} , \cos\frac{x}{2} , and \tan\frac{x}{2}$ in $\small \sin x = \frac{1}{4}$ ,x in quadrant II

Answer:

$\frac{\pi}{2} < x < \pi\\ \\ \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$ all functions are positive in this range
We know that
$\cos^{2}x = 1 - \sin^{2}x$
= 1 - $\left ( \frac{1}{4} \right )^{2}$ = $1 - \frac{1}{16}$ = $\frac{15}{16}$

cos x = $\sqrt\frac{15}{16} = \pm \frac{\sqrt15}{4} = - \frac{\sqrt15}{4}$ (cos x is -ve in II quadrant)

We know that
cosx = $2\cos^{2}\frac{x}{2} - 1$
$2\cos^{2}\frac{x}{2} = \cos x + 1 = -\frac{\sqrt15}{4} + 1 = \frac{-\sqrt15+4}{4}$

$\cos^{2}\frac{x}{2} = \frac{-\sqrt15+4}{8}$
$\cos\frac{x}{2} = \pm \sqrt\frac{-\sqrt15+4}{8} = \frac{\sqrt{-\sqrt15+4}}{2\sqrt2} = \frac{\sqrt{8-2\sqrt15}}{4}$ (because all functions are posititve in given range)

similarly,
cos x = $1-2\sin^{2}\frac{x}{2}$
$2\sin^{2}\frac{x}{2} = 1 - \cos x\\ \\ 2\sin^{2}\frac{x}{2} = 1 -\left (\frac{-\sqrt15}{4} \right ) = \frac{4+\sqrt15}{4}$
$\sin\frac{x}{2} = \pm \sqrt\frac{\sqrt15+4}{8} = \frac{\sqrt{\sqrt15+4}}{2\sqrt2} = \frac{\sqrt{8+2\sqrt15}}{4}$ (because all functions are posititve in given range)
$\tan\frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \frac{\frac{\sqrt{8+2\sqrt15}}{4}}{\frac{\sqrt{8-2\sqrt15}}{4}} = \frac{{8+2\sqrt15}}{\sqrt{64 - 15\times4}} = \frac{{8+2\sqrt15}}{\sqrt{4}} = 4 + \sqrt15$

Trigonometric Functions Class 11 Solutions - Topics

Interested students can study Class 11 NCERT Solutions at a single place. Here are some highlights of Maths NCERT Solutions for Class 11 Chapter 3 encompasses the following topics and sub-topics:

Introduction (3.1): This section of class 11 chapter 3 covers the basic trigonometric ratios and identities, along with their applications in solving word problems related to heights and distances.
Angles (3.2): This section of class 11 chapter 3 maths discusses different terminologies used in trigonometry, such as terminal side, initial sides, measuring an angle in degrees and radians, etc. It also covers degree and radian measures and their relation to real numbers.
Trigonometric Functions (3.3): Students can understand the generalised trigonometric functions with signs and learn about the domain and range of trigonometric functions with examples in this section.
Trigonometric Functions of Sum and Difference of Two Angles (3.4): This section provides formulas related to the sum and difference of two angles in trigonometric functions.

Interested students can study Trigonometric Functions Exercise using following links-

NCERT Solutions for Class 11 Mathematics - Chapter Wise

chapter-1	Sets
chapter-2	Relations and Functions
chapter-3	Trigonometric Functions
chapter-4	Principle of Mathematical Induction
chapter-5	Complex Numbers and Quadratic equations
chapter-6	Linear Inequalities
chapter-7	Permutation and Combinations
chapter-8	Binomial Theorem
chapter-9	Sequences and Series
chapter-10	Straight Lines
chapter-11	Conic Section
chapter-12	Three Dimensional Geometry
chapter-13	Limits and Derivatives
chapter-14	Mathematical Reasoning
chapter-15	Statistics
chapter-16	Probability

Key Features of NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

Comprehensive Coverage: The solutions provide a comprehensive explanation of all the topics covered in Chapter 3, ensuring a strong understanding of trigonometric functions.

Step-by-Step Solutions: Each problem and example in the NCERT textbook is solved step by step, making it easier for students to follow the logical progression of trigonometric concepts.

Clear and Concise Language: The solutions are presented in clear and concise language, making complex trigonometric concepts more accessible to students.

NCERT Solutions for Class 11 - Subject Wise

Trigonometric Functions Basic Identities

$\\1) cos^2x+sin^2x=1\\2)\ 1+tan^2x\ \ \ \ =sec^2x\\3)1+cot^2x\ \ \ \ \ \ =cosec^2x\\4)cos (2n\pi + x) \ = cos x \\5)sin (2n\pi + x) \ = sin x \\6) sin (-x) \ \ \ \ \ \ \ = -sinx \\7) cos (-x) \ \ \ \ \ \ \ = cos x$

The above identities you may have studied in your high school classes also. Here are a few more identities that you have to remember and understand from the NCERT class 11 maths chapter 3 trigonometric functions

$\\8)cos (x + y) = cos x cos y - sin x sin y \\9)cos (x - y) = cos x cos y + sin x sin y\\10)sin (x + y) = sin x cos y + cos x sin y \\11)sin (x - y) = sin x cos y - cos x sin y$

Some conditional identities from the NCERT solutions for class 11 maths chapter 3 Trigonometric Functions

If angles x, y and (x ± y) is not an odd multiple of π 2, then

$\\a) tan(x+y)=\frac{tanx+tany}{1-tanxtany}\\b)tan(x-y)=\frac{tanx-tany}{1+tanxtany}$

If angles x, y and (x ± y) is not a multiple of π, then

$\\a) cot(x+y)=\frac{cotxcoty-1}{cotx+coty}\\b)cot(x-y)=\frac{1+cotxcoty}{coty-cotx}$

There are a few more identities used in the NCERT solutions for class 11 maths chapter 3 trigonometric functions which can be derived using the above identities. Try to derive it by your self.

NCERT Books and NCERT Syllabus

Happy Reading !!!

Frequently Asked Questions (FAQs)

1. What are important topics of the chapter Trigonometric Functions ?

In the chapter 3 class 11 maths Trigonometric functions, identities of trigonometric functions, trigonometric functions of some standard angles, trigonometric functions of sum and difference of two angles and trigonometric equations are the important topics of this chapter.

2. How does the NCERT solutions are helpful ?

NCERT solutions will help the students, if the stuck while solving the NCERT problems. Trigonometry class 11 solutions provides practice problems for students that help them to command and get indepth understanding. finally class 11 maths trigonometry help to build strong foundation for higher classes. For ease Interested students can study trigonometric functions class 11 pdf both online and offline. therefore students can practice class 11 maths chapter 3 solutions.

3. Which are the most difficult chapters in the NCERT class 11 maths ?

Complex Numbers and Quadratic Equations : This chapter introduces students to complex numbers, which can be a new and abstract concept for many. Understanding operations with complex numbers and solving quadratic equations involving complex roots can be challenging.
Limits and Derivatives : Calculus, in general, tends to be difficult for many students, especially when they encounter the concept of limits for the first time. Understanding the concept of limits and their applications, as well as learning differentiation rules and techniques, can be challenging.
Permutations and Combinations : This chapter involves various counting principles and problem-solving techniques, which can be confusing for some students. Understanding when to use permutations versus combinations and solving problems involving both can be tricky.
Binomial Theorem : The binomial theorem involves expanding expressions raised to positive integer powers, which can require a good understanding of algebraic manipulation and pattern recognition.
Statistics : This chapter involves understanding various measures of central tendency and dispersion, as well as probability distributions. It can be challenging for some students to grasp the concepts and apply them to solve problems.
Trigonometric Functions : Trigonometry, particularly when it involves proving identities and solving trigonometric equations, can be challenging due to the abstract nature of the concepts and the numerous identities to remember.

Ultimately, the toughest chapter can vary based on individual preferences and strengths. Some students may find certain chapters easier than others based on their prior knowledge or interests in specific topics. Regular practice, seeking clarification when needed, and utilizing resources such as textbooks, online tutorials, and practice problems can help students overcome challenges in any chapter.

4. Which is the official website of NCERT ?

NCERT official is the official website of the NCERT where you can get NCERT textbooks and syllabus from class 1 to 12.

5. Does CBSE provides the solutions of NCERT class 11 ?

No, CBSE doesn’t provided NCERT solutions for any class or subject.

6. Where can I find the complete solutions of NCERT class 11 maths ?

Here you will get the detailed NCERT solutions for class 11 maths by clicking on the link.

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NCERT Solutions for Exercise 10.1 Class 11 Maths Chapter 10 - Straight Lines

NCERT Solutions for Miscellaneous Exercise Chapter 16 Class 11 - Probability

NCERT Exemplar Class 11 Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry

NCERT Exemplar Class 11 Maths Solutions Chapter 10 Straight Lines

NCERT Exemplar Class 11 Maths Solutions Chapter 6 Linear Inequalities

NCERT Exemplar Class 11 Maths Solutions Chapter 16 Probability

NCERT Exemplar Class 11 Maths Solutions Chapter 14 Mathematical Reasoning

NCERT Exemplar Class 11 Maths Solutions Chapter 7 Permutations and Combinations

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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

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