# NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

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 cahpter 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 memorize and understand the basic identities and properties of trigonometric functions. CBSE 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. There are four exercises and a miscellaneous exercise in this chapter which are explained below.

Exercise:3.1

Exercise:3.2

Exercise:3.3

Exercise:3.4

Miscellaneous Exercise

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?

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, analyzing 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

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

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

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

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

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       $\Theta = 12\degree36'$

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

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

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

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

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

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

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
$\sin x = -\frac {4}{5}$
$cosec x = \frac {1}{\csc} = \frac {1}{-\frac{4}{5}} = - \frac{5}{4}$

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
$\sin x =- \frac {12}{13}$
$\csc x = \frac {1}{\sin x} = \frac {1}{-\frac {12}{13}} = -\frac {13}{12}$
$\tan x = \frac {\sin x}{\cos x} = \frac {-\frac{12}{13}}{\frac{5}{13}} = -\frac {12}{5}$
$\cot x = \frac {1}{\tan x} = \frac {1}{-\frac{12}{5}} = -\frac{5}{12}$

$\tan x = -\frac {5}{12}$
$\cot x = \frac {1}{\tan x} = \frac {1}{-\frac{5}{12}} = -\frac {12}{5}$
$1 + \tan^{2}x = \sec^{2}x\\ 1 + \left ( -\frac{5}{12} \right )^{2} = \sec^{2}x\\ 1 + \frac {25}{144} = \sec^{2}x\\ \\ \frac {169}{144} = \sec^{2}x\\ \sec x = \sqrt {\frac {169}{144}} = \pm \frac {13}{12}$
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
$\sin x = \frac {5}{13}$
$\csc = \frac {1}{\sin x} = \frac {1}{\frac {5}{13}} = \frac {13}{5}$

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

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

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

or

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

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

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

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

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

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.

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.

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.

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

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

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

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

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

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

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

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

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

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.

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.

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.

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.

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

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.

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.

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.

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.

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.

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.

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.

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.

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

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.

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.

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.

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

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

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

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  = $n\pi \pm \frac{5\pi}{6} \ where \ n \ \epsilon \ Z$

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 $n \ \epsilon \ Z$

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

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

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

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

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

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

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

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 $n \ \epsilon \ Z$

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

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.

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.

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.

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
$cos(x + y) = 1 -2sin^{2}\frac{(x + y)}{2}$                                             $\left ( \because \cos2x = 1 - 2\sin^{2}x \ \Rightarrow \cos x = 1 - 2\sin^{2}\frac{x}{2} \right )$

so

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

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

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.

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.                       $\small \left ( \because \frac{\sin x}{\cos x} = \tan x\right )$

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.

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

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

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

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

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

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

$\frac{\pi}{2} < x < \pi\\ \\ \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$ 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 Solutions of NCERT for class 11 chapter 2 Relations and Functions chapter-3 CBSE NCERT solutions for class 11 maths chapter 3 Trigonometric Functions chapter-4 NCERT solutions for class 11 chapter 4 Principle of Mathematical Induction chapter-5 Solutions of NCERT for class 11 chapter 5 Complex Numbers and Quadratic equations chapter-6 CBSE 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 Solutions of NCERT for class 11 maths chapter 8 Binomial Theorem chapter-9 CBSE 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 Solutions of NCERT for class 11 maths chapter 11 Conic Section chapter-12 CBSE 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 Solutions of NCERT for class 11 maths chapter 14 Mathematical Reasoning chapter-15 CBSE 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

 Solutions of NCERT for class 11 biology CBSE NCERT solutions for class 11 maths NCERT solutions for class 11 chemistry Solutions of NCERT 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.