NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

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

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. Also, it will give them the new ways to solve the NCERT problems.

Q: Which are the most difficult chapters in the NCERT class 11 maths ?

Permutation and combination, trigonometry are considered to be difficult chapters by the most of students but with the regular practice students can get command on them also.

Q: 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.

Q: Does CBSE provides the solutions of NCERT class 11 ?

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

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

Updated on Aug 4, 2020 - 10:09 p.m. IST by Ravindra Pindel

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions: Trigonometry 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 NCERT solutions for class 11 maths chapter 3 trigonometric functions. You will use the trigonometric identities in other chapters of mathematics and throughout the NCERT physics for both class 11 and 12 also. So it is very important to memorise and understand the basic identities and properties of trigonometric functions. NCERT solutions for class 11 maths chapter 3 trigonometric functions will help you for the same. Command on this chapter 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 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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Let's understand this chapter with one example.

If we want to measure the height of a building such that we are standing at 50m away from building at point P and the angle of elevation made with the ground at P is 45 degree (as shown in the below figure). Then what will be the height of the building?

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Let the height of the building be QR and distance from the base of the building to point P is 50 meter

Using the trigonometric function

$tan(45)=\frac{QR}{QP}\\\Rightarrow 1=\frac{QR}{50}\\\Rightarrow QR=50\ m$

There are many other examples such as trigonometry is used in electric circuit analysis, predicting the heights of tides in the ocean, analysing a musical tone and in seismology, etc.

The main topics of this chapter are listed below:

3.1 Introduction

3.2 Angles

3.3 Trigonometric Functions

3.4 Trigonometric Functions of Sum and Difference of Two Angles

3.5 Trigonometric Equations

NCERT solutions for class 11 maths chapter 3 trigonometric functions-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,

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 solutions for class 11 maths chapter 3 trigonometric functions-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 solutions for class 11 maths chapter 3 trigonometric functions-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 solutions for class 11 maths chapter 3 trigonometric function-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 $n \ \epsilon \ Z$

NCERT solutions for class 11 maths chapter 3 trigonometric functions-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)
$\tan\frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \frac{\frac{\sqrt{8+2\sqrt15}}{4}}{\frac{\sqrt{8-2\sqrt15}}{4}} = \frac{{8+2\sqrt15}}{\sqrt{64 - 15\times4}} = \frac{{8+2\sqrt15}}{\sqrt{4}} = 4 + \sqrt15$

NCERT solutions for class 11 mathematics

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

NCERT solutions for class 11- Subject wise

NCERT solutions for class 11 biology

NCERT solutions for class 11 maths

NCERT solutions for class 11 chemistry

NCERT solutions for Class 11 physics

The basic identities used in NCERT solutions for class 11 maths chapter 3 Trigonometric Functions are listed below

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

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 solutions for 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.

Happy Reading !!!

Solutions

Frequently Asked Question (FAQs) - NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

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

Answer:

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.

Question: How does the NCERT solutions are helpful ?

Answer:

NCERT solutions will help the students, if the stuck while solving the NCERT problems. Also, it will give them the new ways to solve the NCERT problems.

Question: Which are the most difficult chapters in the NCERT class 11 maths ?

Answer:

Permutation and combination, trigonometry are considered to be difficult chapters by the most of students but with the regular practice students can get command on them also.

Question: Which is the official website of NCERT ?

Answer:

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

Question: Does CBSE provides the solutions of NCERT class 11 ?

Answer:

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

Question: Where can I find the complete solutions of NCERT class 11 maths ?

Answer:

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

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NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

NCERT solutions for class 11 maths chapter 3 trigonometric functions-Exercise: 3.1

NCERT solutions for class 11 maths chapter 3 trigonometric functions-Exercise: 3.2

NCERT solutions for class 11 maths chapter 3 trigonometric functions-Exercise: 3.3

NCERT solutions for class 11 maths chapter 3 trigonometric function-Exercise: 3.4

NCERT solutions for class 11 maths chapter 3 trigonometric functions-Miscellaneous Exercise

Frequently Asked Question (FAQs) - NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

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

Question: How does the NCERT solutions are helpful ?

Question: Which are the most difficult chapters in the NCERT class 11 maths ?

Question: Which is the official website of NCERT ?

Question: Does CBSE provides the solutions of NCERT class 11 ?

Question: Where can I find the complete solutions of NCERT class 11 maths ?

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