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NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

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

Trigonometric Functions Class 11 Questions And Answers

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

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

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Trigonometric Functions Class 11 Solutions - Important Formulae

Angle Conversion:

Radian Measure = π/180 × Degree Measure

Degree Measure = 180/π × Radian Measure

Trigonometric Ratios:

  • sin θ = P / H

  • cos θ = B / H

  • tan θ = P / B

  • cosec θ = H / P

  • sec θ = H / B

  • cot θ = B / P

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Reciprocal Trigonometric Ratios:

  • sin θ = 1 / cosec θ

  • cosec θ = 1 / sin θ

  • cos θ = 1 / sec θ

  • sec θ = 1 / cos θ

  • tan θ = 1 / cot θ

  • cot θ = 1 / tan θ

Trigonometric Ratios of Complementary Angles:

  • sin (90° – θ) = cos θ

  • cos (90° – θ) = sin θ

  • tan (90° – θ) = cot θ

  • cot (90° – θ) = tan θ

  • sec (90° – θ) = cosec θ

  • cosec (90° – θ) = sec θ

Periodic Trigonometric Ratios:

  • sin(π/2-θ) = cos θ

  • cos(π/2-θ) = sin θ

  • sin(π-θ) = sin θ

  • cos(π-θ) = -cos θ

  • sin(π+θ) = -sin θ

  • cos(π+θ) = -cos θ

  • sin(2π-θ) = -sin θ

  • cos(2π-θ) = cos θ

Trigonometric Identities:

  • sin² θ + cos² θ = 1

  • cosec² θ – cot² θ = 1

  • sec² θ – tan² θ = 1

Product to Sum Formulas:

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

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

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

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

Sum to Product Formulas:

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

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

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

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

General Trigonometric Formulas:

  • sin (x+y) = sin x × cos y + cos x × sin y

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

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

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

Sum and Difference Formulas for tan:

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

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

Double Angle Formulas for tan:

  • tan 2θ = (2 tan θ) / (1 - tan²θ)

Triple Angle Formulas for sin, cos, and tan:

  • sin 3θ = 3sin θ – 4sin³θ

  • cos 3θ = 4cos³θ – 3cos θ

  • tan 3θ = [3tan θ – tan³θ] / [1 − 3tan²θ]

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

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

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

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

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

Answer:

It is solved using relation between degree and radian

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

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


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

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

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

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

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

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

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

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

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

Answer:

(1) \frac{11}{16}

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

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

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

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

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

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

(ii) -4
We know that

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


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


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

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

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

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


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

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


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

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

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

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

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

Answer:

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

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

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

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

Answer:

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

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

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

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

Answer:

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

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

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

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

Answer:

(i) We know that

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

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

(ii) We know that

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

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

(iii) We know that

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

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


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

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

Answer:

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

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

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

x lies in III quadrants. Therefore cosec x is negative

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

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

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

Answer:

Solution

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

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

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

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

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

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

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

Answer:

Solution

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

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

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

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

Answer:

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

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

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

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

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

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

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

Answer:

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

or

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

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

Answer:

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

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

Answer:

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

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

Answer:

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

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

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

Answer:

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


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

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

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

Answer:

The solutions for the given problem is done as follows.

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

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

Answer:

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

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

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

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

Answer:

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

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

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

Answer:

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

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

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

Answer:

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

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

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

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

Answer:

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

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

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

Now, we know that

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

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

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

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

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

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

As we know that

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

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

R.H.S

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

Answer:

As we know that

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

So, by using these identities

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

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

Answer:

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

By using these our equation simplify to

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

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

Answer:

We know that

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

So, by using these our equation simplifies to

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

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

Answer:

Multiply and divide by 2

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

Now by using identities


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

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

R.H.S.

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

Answer:

We know that

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

By using this identity

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

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

Answer:

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

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

Now, we know that


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

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

Now,

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

by using these identities

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

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

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

Answer:

As we know that

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

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

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

So our equation becomes


R.H.S.

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

Answer:

We know that

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

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

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

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

Answer:

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

sin5x + sin3x = 2sin4xcosx

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

now nultiply and divide by sin x

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

Now we know that

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

By using this our equation becomes

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

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

Answer:

As we know that

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

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

Answer:

We know that

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

We use these identities

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

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

Answer:

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

We use these identities

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

R.H.S.

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

Answer:

We know that

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

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

Answer:
We know that

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

We use these identities

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

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

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

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

Answer:

We know that

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

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

=RHS

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

Answer:

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

and we know that

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

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

Answer:

We know that

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

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


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

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

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

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

Answer:

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

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

Answer:

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

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

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

Answer:

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

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

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

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

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

Answer:

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

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

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

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

Answer:

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

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


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

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

Answer:

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

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

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


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

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

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

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

where n \ \epsilon \ Z

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

Answer:

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

Therefore, the general solution is

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

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

Answer:

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

\therefore the general solution is

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

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

Answer:

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

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

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

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

Answer:

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

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

Answer:

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

Class 11 maths chapter 3 NCERT solutions - Miscellaneous Exercise

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

Answer:

We know that

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

we use this in our problem

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

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

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

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

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

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

Answer:

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

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

Answer:

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

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

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

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

= R.H.S.

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

Answer:

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

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

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

so

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


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

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

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

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

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

Answer:

We know that

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

We use these two identities in our problem

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

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

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

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


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

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

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

Answer:

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

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

= 2\cos2x\sin x


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

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

= R.H.S.

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

Answer:

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

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

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

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

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

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

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

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

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

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

Answer:

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

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

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

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

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

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

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

Answer:

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

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

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

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

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

Trigonometric Functions Class 11 Solutions - Topics

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

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

Interested students can study Trigonometric Functions Exercise using following links-

NCERT Solutions for Class 11 Mathematics - Chapter Wise

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

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

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

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

NCERT Solutions for Class 11 - Subject Wise

Trigonometric Functions Basic Identities

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

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

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

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

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

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

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

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

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

NCERT Books and NCERT Syllabus

Happy Reading !!!

Frequently Asked Question (FAQs)

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

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

2. How does the NCERT solutions are helpful ?

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

3. Which are the most difficult chapters in the NCERT class 11 maths ?
  1. Complex Numbers and Quadratic Equations : This chapter introduces students to complex numbers, which can be a new and abstract concept for many. Understanding operations with complex numbers and solving quadratic equations involving complex roots can be challenging.

  2. Limits and Derivatives : Calculus, in general, tends to be difficult for many students, especially when they encounter the concept of limits for the first time. Understanding the concept of limits and their applications, as well as learning differentiation rules and techniques, can be challenging.

  3. Permutations and Combinations : This chapter involves various counting principles and problem-solving techniques, which can be confusing for some students. Understanding when to use permutations versus combinations and solving problems involving both can be tricky.

  4. Binomial Theorem : The binomial theorem involves expanding expressions raised to positive integer powers, which can require a good understanding of algebraic manipulation and pattern recognition.

  5. Statistics : This chapter involves understanding various measures of central tendency and dispersion, as well as probability distributions. It can be challenging for some students to grasp the concepts and apply them to solve problems.

  6. Trigonometric Functions : Trigonometry, particularly when it involves proving identities and solving trigonometric equations, can be challenging due to the abstract nature of the concepts and the numerous identities to remember.

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

4. Which is the official website of NCERT ?

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

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

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

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

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

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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