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

Q: How to solve trigonometric equations in Class 11 Maths Chapter 3?

To solve trigonometric equations in Class 11 Maths Chapter 3, express the given terms in basic trigonometric functions like sin, cos, and tan, then try to apply proper trigonometric identities and use simplification.

Q: What are the key formulas of Trigonometric Functions in NCERT Class 11?

The key formulas of Trigonometric Functions in NCERT Class 11 are: $1) \cos^2x+\sin^2x=1$ $2) 1+\tan^2x =\sec^2x$ $3)1+\cot^2x =\operatorname{cosec}^2x$ $4)\cos (2n\pi + x)= \cos x$ $5)\sin (2n\pi + x) = \sin x$ $6) \sin (-x) = -\sin x$ $7) \cos (-x)= \cos x$ $8)\cos (x + y) = \cos x \cos y - \sin x \sin y$ $9)\cos (x - y) = \cos x \cos y + \sin x \sin y$ $10)\sin (x + y) = \sin x \cos y + \cos x \sin y$ $11)\sin (x - y) = \sin x \cos y - \cos x \sin y$ $12) \tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}$ $13) \tan(x-y)=\frac{\tan x-\tan y}{1+\tan x\tan y}$ $14) \cot(x+y)=\frac{\cot x\cot y-1}{\cot x+\cot y}$ $15) \cot(x-y)=\frac{1+\cot x\cot y}{\cot y-\cot x}$

Q: How to find the general solution of a trigonometric equation?

To find the general solution of a trigonometric equation, use the following standard equations: $\sin \theta=\sin \alpha \Rightarrow \theta=n \pi+(-1)^n \alpha, n \in \mathbb{Z}$ $\cos \theta=\cos \alpha \Rightarrow \theta=2 n \pi \pm \alpha, n \in \mathbb{Z}$ $\tan \theta=\tan \alpha \Rightarrow \theta=n \pi+\alpha, n \in \mathbb{Z}$

Q: What are the domain and range of trigonometric functions in Class 11?

In class 11 chapter 3 trigonometric functions, a domain basically denotes the input values and range denotes the output values. For example: The domain of $\sin\theta$ and $\cos\theta$ is $(-\infty,\infty)$ and range is (–1, 1).

Q: How to derive the sine and cosine function graphs?

To derive the sine and cosine function graphs, plot $y=\sin\theta$ and $y=\cos\theta$ for the values of $\theta$ like 0°, 30°, 60°, 90°, 180°, etc.. The sine graph starts at (0, 0) and goes up to 1 at 90°, drops to –1 at 270°, and comes to 0 again at 360° then repeats.

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

Edited By Komal Miglani | Updated on Mar 29, 2025 03:42 PM IST

Have you ever wondered how engineers design huge buildings, how sailors and pilots navigate through their journey, or how shadows change their length throughout the day? All of these answers can be found in Trigonometry, a fascinating branch of mathematics. From the latest NCERT syllabus for class 11, the chapter trigonometric functions contains the advanced concepts of trigonometry like radian measure, relation between degree and radian, sign of trigonometric functions, graphs of trigonometric functions, domain and range of trigonometric functions, and trigonometric functions of the sum and difference of two angles. Understanding these concepts will make students more efficient in solving problems involving height, distance, and angles. This chapter will also build a strong foundation for more advanced trigonometric concepts, which have many practical real-life applications like construction, navigation, architecture, etc.

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This Story also Contains

Trigonometric Functions Class 11 Questions And Answers PDF Free Download
Trigonometric Functions Class 11 Solutions - Important Formulae
Trigonometric Functions Class 11 NCERT Solutions (Exercises)
Trigonometry Class 11 Solutions - Exercise Wise
NCERT Solutions for Class 11 Mathematics - Chapter Wise
Importance of Solving NCERT Questions of Class 11 Maths Chapter 8
NCERT Solutions for Class 11 - Subject Wise
NCERT Books and NCERT Syllabus

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

This article on NCERT solutions for class 11 Maths Chapter 3 Trigonometric Functions offers clear and step-by-step solutions for the exercise problems in the NCERT Class 11 Maths Book. Students who are in need of trigonometric functions class 11 solutions will find this article very useful. It covers all the important class 11 maths chapter 3 question answers of trigonometric functions. These trigonometric functions class 11 ncert solutions are made by the Subject Matter Experts according to the latest CBSE syllabus, ensuring that students can grasp the basic concepts effectively. NCERT solutions for class 11 maths and NCERT solutions for other subjects and classes can be downloaded from the NCERT Solutions.

Trigonometric Functions Class 11 Questions And Answers PDF Free Download

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Trigonometric Functions Class 11 Solutions - Important Formulae

Angle Conversion:

Radian Measure = $\frac{π}{180}$ × Degree Measure

Degree Measure = $\frac{180}{π}$ × 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

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

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Trigonometric Ratios of Complementary Angles:

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

Periodic Trigonometric Ratios:

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

Trigonometric Identities:

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

Product to Sum Formulas:

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

Sum to Product Formulas:

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

General Trigonometric Formulas:

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

Sum and Difference Formulas for tan:

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

Double Angle Formulas for tan:

tan 2θ = (2 tan θ) / (1 - tan²θ)

Triple Angle Formulas for sin, cos, and tan:

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

Trigonometric Functions Class 11 NCERT Solutions (Exercises)

NCERT Trigonometric Functions Class 11 Questions and Answers
Exercise: 3.1, Page Number: 48-49, Total Questions: 7

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

(i) $25^{\circ}$
(ii) $- 47^{\circ}$ $30^{'}$
(iii) $240^{\circ}$
(iv) $520^{\circ}$

Answer:

It is solved using the relation between degree and radian
(i) $25^{\circ}$
We know that $180^{\circ}$ = $π$ radian
So, $1^{\circ} = \frac{π}{180}$ radian
$25^{\circ} = \frac{π}{180} \times 25$ radian $= \frac{5 π}{36}$ radian

(ii) $- 47^{\circ} 30^{'}$
We know that
$- 47^{\circ} 30^{'} = - 47 {\frac{1}{2}}^{\circ} = - {\frac{95}{2}}^{\circ}$
Now, we know that $180^{\circ} = π \Rightarrow 1^{\circ} = \frac{π}{180}$ radian
So, $- {\frac{95}{2}}^{\circ} = \frac{π}{180} \times (- \frac{95}{2})$ radian $= \frac{- 19 π}{72}$ radian

(iii) $240^{\circ}$
We know that
$180^{\circ} = π \Rightarrow 1^{\circ} = \frac{π}{180}$ radian
So, $240^{\circ} = \frac{π}{180} \times 240 = \frac{4 π}{3}$ radian

(iv) $520^{\circ}$
We know that
$180^{\circ} = π \Rightarrow 1^{\circ} = \frac{π}{180}$ radian
So, $520^{\circ} = \frac{π}{180} \times 520$ radian $= \frac{26 π}{9}$ radian

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

$(i) \frac{11}{16}$
$(i i) - 4$
$(i i i) \frac{5 π}{3}$
$(i v) \frac{7 π}{6}$

Answer:

(i) $\frac{11}{16}$
We know that
$π$ radian $= 180^{\circ} \Rightarrow 1 r a d i a n = {\frac{180}{π}}^{\circ}$
So, $\frac{11}{16} r a d i a n = \frac{180}{π} \times {\frac{11}{16}}^{\circ}$ (we need to take $π = \frac{22}{7}$ )
$\frac{11}{16} r a d i a n = \frac{180 \times 7}{22} \times {\frac{11}{16}}^{\circ} = {\frac{315}{8}}^{\circ}$
(we use $1^{\circ} = 60^{'}$ and 1' = 60'')
Here 1' represents 1 minute and 60" represents 60 seconds
Now,
${\frac{315}{8}}^{\circ} = 39 {\frac{3}{8}}^{\circ} = 39^{\circ} + {\frac{3 \times 60}{8}}^{'} = 39^{\circ} + 22^{'} + {\frac{1}{2}}^{'} = 39^{\circ} + 22^{'} + 30^{″} = {\frac{315}{8}}^{\circ} = 39^{\circ} 22^{'} 30^{″}$

(ii) $- 4$
We know that
$π$ radian $= 180^{\circ} \Rightarrow 1 r a d i a n = {\frac{180}{π}}^{\circ}$ (we need to take $π = \frac{22}{7}$ )
So, $- 4 r a d i a n = \frac{- 4 \times 180}{π} = \frac{- 4 \times 180 \times 7}{22} = - {\frac{2520}{11}}^{\circ}$
(we use $1^{\circ} = 60^{'}$ and 1' = 60'')
$\Rightarrow {\frac{- 2520}{11}}^{\circ} = - 229 {\frac{1}{11}}^{\circ} = - 229^{\circ} + {\frac{1 \times 60}{11}}^{'} = - 229^{\circ} + 5^{'} + {\frac{5}{11}}^{'} = - 229^{\circ} + 5^{'} + 27^{″} = - 229^{\circ} 5^{'} 27^{″}$

(iii) $\frac{5 π}{3}$
We know that
$π$ radian $= 180^{\circ} \Rightarrow 1 r a d i a n = {\frac{180}{π}}^{\circ}$ (we need to take $π = \frac{22}{7}$ )
So, $\frac{5 π}{3} r a d i a n = \frac{180}{π} \times {\frac{5 π}{3}}^{\circ} = 300^{\circ}$

(iv) $\frac{7 π}{6}$
We know that
$π$ radian $= 180^{\circ} \Rightarrow 1 r a d i a n = {\frac{180}{π}}^{\circ}$ (we need to take $π = \frac{22}{7}$ )
So, $\frac{7 π}{6} r a d i a n = \frac{180}{π} \times \frac{7 π}{6} = 210^{\circ}$

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
$∴$ Number of revolutions made by the wheel in 1 second = $\frac{360}{60} = 6$
( $∵$ 1 minute = 60 seconds)
In one revolution, the wheel will cover $2 π$ radian
So, in 6 revolutions it will cover = $6 \times 2 π = 12 π$ radian
$∴$ In 1 the second wheel will turn $12 π$ 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 $π = \frac{22}{7}$ )

Answer:
We know that
$l = r θ$ ( where $l$ is the length of the arc, $r$ is the radius of the circle and $θ$ is the angle subtended)
here $r$ = 100 cm
and $l$ = 22 cm
Now,
$θ = \frac{l}{r} = \frac{22}{100}$ radian
We know that
$π r a d i a n = 180^{\circ}$
So, 1 radian = ${\frac{180}{π}}^{\circ}$
$∴ \frac{22}{100} r a d i a n = \frac{180}{π} \times {\frac{22}{100}}^{\circ} = \frac{180 \times 7}{22} \times \frac{22}{100} = {\frac{63}{5}}^{\circ}$
So, ${\frac{63}{5}}^{\circ} = 12 {\frac{3}{5}}^{\circ} = 12^{\circ} + {\frac{3 \times 60}{5}}^{'} = 12^{\circ} + 36^{'} = 12^{\circ} 36^{'}$
Thus, the angle subtended at the centre of a circle $θ = 12^{\circ} 36^{'}$ .

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 = $\frac{D i a m e t e r}{2} = \frac{40}{2} = 20$ cm
length of chord = 20 cm
We know that
$θ = \frac{l}{r}$ ( $r$ = 20 cm , $l$ = ? , $θ$ = ?)
Now,
1654678632225
AB is the chord of length 20 cm and OA and OB are radii of circle i.e. 20 cm each
The angle subtended by OA and OB at centre = $θ$
$∵$ OA = OB = AB
$∴$ $Δ$ OAB is equilateral triangle
So, each angle is $60^{\circ}$
$∴$ $θ = 60^{\circ}$ $= \frac{π}{3}$ radian
Now, we have $θ$ and $r$
So, $l = r θ = 20 \times \frac{π}{3} = \frac{20 π}{3}$
$∴$ the length of the minor arc of the chord ( $l$ ) = $\frac{20 π}{3}$ 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: $θ_{1} = 60^{\circ}, θ_{2} = 75^{\circ}$ and $l_{1} = l_{2}$
We need to find the ratio of their radii $\frac{r_{1}}{r_{2}}$
We know that arc length $l = r θ$
So, $l_{1} = r_{1} θ_{1}$ and $l_{2} = r_{2} θ_{2}$
Now, $l_{1} = l_{2}$
So, $\frac{r_{1}}{r_{2}} = \frac{θ_{2}}{θ_{1}} = \frac{75}{60} = \frac{5}{4}$ , which 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
$l = r θ$
Now,
$r$ = 75 cm
$l$ = 10 cm
So,
$θ = \frac{l}{r} = \frac{10}{75} = \frac{2}{15}$ radian

(ii) We know that
$l = r θ$
Now,
$r$ = 75 cm
$l$ = 15 cm
So,
$θ = \frac{l}{r} = \frac{15}{75} = \frac{1}{5}$ radian

(iii) We know that
$l = r θ$
Now,
$r$ = 75 cm
$l$ = 21 cm
So,
$θ = \frac{l}{r} = \frac{21}{75} = \frac{7}{25}$ radian

NCERT Trigonometric Functions Class 11 Questions and Answers
Exercise: 3.2, Page Number: 57, Total Questions: 10

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

Answer:
$\cos x = - \frac{1}{2}$
$∵ \sec x = \frac{1}{\cos x} = \frac{1}{- \frac{1}{2}} = - 2$
$x$ lies in III quadrants. Therefore sec x is negative
$\sin^{2} x + \cos^{2} x = 1 \Rightarrow \sin^{2} x = 1 - \cos^{2} x$
$\Rightarrow \sin^{2} x = 1 - {(- \frac{1}{2})}^{2} \Rightarrow \sin^{2} x = 1 - \frac{1}{4} = \frac{3}{4} \Rightarrow \sin x = \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2}$
$x$ lies in III quadrants. Therefore sin x is negative
$∴ \sin x = - \frac{\sqrt{3}}{2}$
$∵ c o s e c x = \frac{1}{\sin x} = \frac{1}{- \frac{\sqrt{3}}{2}} = - \frac{2}{\sqrt{3}}$
$x$ 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 as $\sin x = \frac{3}{5}$ , $x$ lies in second quadrant.

Answer:
$\sin x = \frac{3}{5}$
$c o s e c x = \frac{1}{\sin x} = \frac{1}{\frac{3}{5}} = \frac{5}{3}$
$x$ lies in the second quadrant. Therefore cosec x is positive
$\sin^{2} x + \cos^{2} x = 1 \Rightarrow \cos^{2} x = 1 - \sin^{2} x$
$\Rightarrow \cos^{2} x = 1 - {(\frac{3}{5})}^{2} \Rightarrow \cos^{2} x = 1 - \frac{9}{25} = \frac{16}{25} \Rightarrow \cos x = \sqrt{\frac{16}{25}} = \pm \frac{4}{5}$
$x$ lies in the second quadrant. Therefore cos x is negative
$∴ \cos x = - \frac{4}{5}$
$\sec x = \frac{1}{\cos x} = \frac{1}{- \frac{4}{5}} = - \frac{5}{4}$
$x$ 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 as $\cot x = \frac{3}{4}$ , $x$ lies in third quadrant.

Answer:
$\cot x = \frac{3}{4}$
$\tan x = \frac{1}{\cot x} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$
$1 + \tan^{2} x = \sec^{2} x \Rightarrow 1 + \frac{4^{2}}{3^{2}} = \sec^{2} x \Rightarrow 1 + \frac{16}{9} = \sec^{2} x \Rightarrow \frac{25}{9} = \sec^{2} x$
$\Rightarrow \sec x = \sqrt{\frac{25}{9}} = \pm \frac{5}{3}$
$x$ lies in x lies in third quadrant. therefore sec x is negative
$\sec x = - \frac{5}{3}$
$\cos x = \frac{1}{\sec x} = \frac{1}{- \frac{5}{3}} = - \frac{3}{5}$
$\sin^{2} x + \cos^{2} x = 1 \Rightarrow \sin^{2} x = 1 - \cos^{2} x$
$\Rightarrow \sin^{2} x = 1 - {(- \frac{3}{5})}^{2} \Rightarrow \sin^{2} x = 1 - \frac{9}{25} \Rightarrow \sin^{2} x = \frac{16}{25} \Rightarrow \sin x = \sqrt{\frac{16}{25}} = \pm \frac{4}{5}$
$x$ lies in third quadrant. Therefore sin x is negative
$\sin x = - \frac{4}{5}$
$c o s e c x = \frac{1}{\csc} = \frac{1}{- \frac{4}{5}} = - \frac{5}{4}$ .

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

Answer:
$\sec x = \frac{13}{5}$
$\cos x = \frac{1}{\sec x} = \frac{1}{\frac{13}{5}} = \frac{5}{13}$
$\sin^{2} x + \cos^{2} x = 1 \Rightarrow \sin^{2} x = 1 - \cos^{2} x$
$\Rightarrow \sin^{2} x = 1 - (\frac{5}{13})^{2} \Rightarrow \sin^{2} x = 1 - \frac{25}{169} = \frac{144}{169} \Rightarrow \sin x = \sqrt{\frac{144}{169}} = \pm \frac{12}{13}$
$x$ 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 as $\tan x = - \frac{5}{12}$ , $x$ lies in second quadrant.

Answer:
$\tan x = - \frac{5}{12}$
$\cot x = \frac{1}{\tan x} = \frac{1}{- \frac{5}{12}} = - \frac{12}{5}$
$1 + \tan^{2} x = \sec^{2} x \Rightarrow 1 + {(- \frac{5}{12})}^{2} = \sec^{2} x \Rightarrow 1 + \frac{25}{144} = \sec^{2} x \Rightarrow \frac{169}{144} = \sec^{2} x$
$\Rightarrow \sec x = \sqrt{\frac{169}{144}} = \pm \frac{13}{12}$
$x$ lies in second quadrant. Therefore the value of $\sec x$ is negative
$\sec x = - \frac{13}{12}$
$\cos x = \frac{1}{\sec x} = \frac{1}{- \frac{13}{12}} = - \frac{12}{13}$
$\sin^{2} x + \cos^{2} x = 1 \Rightarrow \sin^{2} x = 1 - \cos^{2} x$
$\Rightarrow \sin^{2} x = 1 - {(- \frac{12}{13})}^{2} \Rightarrow \sin^{2} x = 1 - \frac{144}{169} \Rightarrow \sin^{2} x = \frac{25}{169} \Rightarrow \sin x = \sqrt{\frac{25}{169}} = \pm \frac{5}{13}$
$x$ 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 $\sin 765^{\circ}$

Answer:
We know that values of $\sin x$ repeat after an interval of $2 π$ or $360^{\circ}$
$\sin 765^{\circ} = \sin (2 \times 360^{\circ} + 45^{\circ}) = \sin 45^{\circ} = \frac{1}{\sqrt{2}}$ .

Question:7 Find the values of the trigonometric functions $c o s e c (- 1410^{\circ})$

Answer:
We know that value of $cosec x$ repeats after an interval of $2 π$ or $360^{\circ}$ .
$cosec (- 1410^{\circ}) = cosec (360^{\circ} \times 4 + (- 1410^{\circ})) = cosec 30^{\circ} = 2$

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

Answer:
We know that $\tan x$ repeats after an interval of $π$ or 180 $^{\circ}$ .
$\tan (\frac{19 π}{3}) = \tan (6 π + \frac{π}{3}) = \tan \frac{π}{3} = \tan 60^{\circ} = \sqrt{3}$

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

Answer:
We know that $\sin x$ repeats after an interval of $2 π$ or $360^{\circ}$
$\sin (- \frac{11 π}{3}) = \sin (4 π + (- \frac{11 π}{3})) = \sin \frac{π}{3} = \frac{\sqrt{3}}{2}$

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

Answer:
We know that $\cot x$ repeats after an interval of $π$ or 180 $^{\circ}$ .
$\cot (- \frac{15 π}{4}) = \cot (4 π + (- \frac{15 π}{4})) = \cot (\frac{π}{4}) = 1$

NCERT Trigonometric Functions Class 11 Questions and Answers
Exercise: 3.3, Page Number: 67-68, Total Questions: 25

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

Answer:
We know the values of sin 30°, cos 60°, and tan 45°.
$\sin (\frac{π}{6}) = (\frac{1}{2}), \cos (\frac{π}{3}) = (\frac{1}{2}), \tan (\frac{π}{4}) = 1$
L.H.S. $= \sin^{2} \frac{π}{6} + \cos^{2} \frac{π}{3} - \tan^{2} \frac{π}{4} =$ ${(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{2} - 1^{2}$
$= \frac{1}{4} + \frac{1}{4} - 1 = - \frac{1}{2} =$ R.H.S.

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

Answer:
$\sin \frac{π}{6} = \frac{1}{2}, cosec \frac{7 π}{6} = cosec (π + \frac{π}{6}) = - cosec \frac{π}{6} = - 2, \cos \frac{π}{3} = \frac{1}{2}$
L.H.S. = $2 \sin^{2} \frac{π}{6} + {cosec}^{2} \frac{7 π}{6} \cos^{2} \frac{π}{3} = 2 {(\frac{1}{2})}^{2} + {(- 2)}^{2} {(\frac{1}{2})}^{2} = 2 \times \frac{1}{4} + 4 \times \frac{1}{4} = \frac{1}{2} + 1 = \frac{3}{2}$
= R.H.S.

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

Answer:
We know the values of cot 30°, tan 30°, and cosec 30°.
$\cot \frac{π}{6} = \sqrt{3}, cosec \frac{5 π}{6} = cosec (π - \frac{π}{6}) = cosec \frac{π}{6} = 2, \tan \frac{π}{6} = \frac{1}{\sqrt{3}}$
L.H.S. $= \cot^{2} \frac{π}{6} + cosec \frac{5 π}{6} + 3 \tan^{2} \frac{π}{6} = {(\sqrt{3})}^{2} + 2 + 3 \times {(\frac{1}{\sqrt{3}})}^{2} = 3 + 2 + 1 = 6 =$ R.H.S.

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

Answer:
$\sin \frac{3 π}{4} = \sin (π - \frac{π}{4}) = \sin \frac{π}{4} = \frac{1}{\sqrt{2}}, \cos \frac{π}{4} = \frac{1}{\sqrt{2}}, \sec \frac{π}{3} = 2$
Using the above values
L.H.S. $= 2 \sin^{2} \frac{3 π}{4} + 2 \cos^{2} \frac{π}{4} + 2 \sec^{2} \frac{π}{3} = 2 \times {(\frac{1}{\sqrt{2}})}^{2} + 2 \times {(\frac{1}{\sqrt{2}})}^{2} + 2 {(2)}^{2} \Rightarrow 1 + 1 + 8 = 10 =$ R.H.S.

Question:5(i) Find the value of $(i) \sin 75^{\circ}$

Answer:
$\sin 75^{\circ} = \sin (45^{\circ} + 30^{\circ})$
We know that
$\sin (x + y) = \sin x \cos y + \cos x \sin y$
Using this idendity
$\sin 75^{\circ} = \sin (45^{\circ} + 30^{\circ}) = \sin 45^{\circ} \cos 30^{\circ} + \cos 45^{\circ} \sin 30^{\circ}$
$= \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2} \Rightarrow \frac{\sqrt{3}}{2 \sqrt{2}} + \frac{1}{2 \sqrt{2}} = \frac{\sqrt{3} + 1}{2 \sqrt{2}}$

Question:5(ii) Find the value of
$(i i) \tan 15^{\circ}$

Answer:
$\tan 15^{\circ} = \tan (45^{\circ} - 30^{\circ})$
We know that,
$[\tan (x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}]$
By using this we can write,
$\tan (45^{\circ} - 30^{\circ}) = \frac{\tan 45^{\circ} - t a n 30^{\circ}}{1 + \tan 45^{\circ} \tan 30^{\circ}} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 (\frac{1}{\sqrt{3}})} = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$
$= \frac{{(\sqrt{3} - 1)}^{2}}{(\sqrt{3} + 1) (\sqrt{3} - 1)} = \frac{3 + 1 - 2 \sqrt{3}}{{(\sqrt{3})}^{2} - {(1)}^{2}} \Rightarrow \frac{4 - 2 \sqrt{3}}{3 - 1} = \frac{2 (2 - \sqrt{3})}{2} = 2 - \sqrt{3}$

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

Answer:
$\cos (\frac{π}{4} - x) \cos (\frac{π}{4} - y) - \sin (\frac{π}{4} - x) \sin (\frac{π}{4} - y)$
Multiply and divide by 2 both cos and sin functions
We get,
$\frac{1}{2} [2 \cos (\frac{π}{4} - x) \cos (\frac{π}{4} - y)] + \frac{1}{2} [- 2 \sin (\frac{π}{4} - x) \sin (\frac{π}{4} - y)]$
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 = $(\frac{π}{4} - x)$
B = $(\frac{π}{4} - y)$
So,
$\frac{1}{2} [\cos {(\frac{π}{4} - x) + (\frac{π}{4} - y)} + \cos {(\frac{π}{4} - x) - (\frac{π}{4} - y)}] + \frac{1}{2} [\cos {(\frac{π}{4} - x) + (\frac{π}{4} - y)} - \cos {(\frac{π}{4} - x) + (\frac{π}{4} - y)}]$
$= 2 \times \frac{1}{2} [\cos {(\frac{π}{4} - x) + (\frac{π}{4} - y)}]$
$= \cos [\frac{π}{2} - (x + y)]$
As we know that
$(\cos (\frac{π}{2} - A) = \sin A)$
By using this
$= \cos [\frac{π}{2} - (x + y)]$ $= \sin (x + y) =$ R.H.S.

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

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

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

Answer:
As we know,
$\cos (π + x) = - \cos x$ , $\sin (π - x) = \sin x$ , $\cos (\frac{π}{2} + x) = - \sin x$ and $\cos (- x) = \cos x$
By using these our equation simplify to
$\frac{\cos x \times - \cos x}{\sin x \times - \sin x} = \frac{- \cos^{2} x}{- \sin^{2} x} = \cot^{2} x$ $(∵ \cot x = \frac{\cos x}{\sin x}) =$ R.H.S.

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

Answer:
We know that
$\cos (\frac{3 π}{2} + x) = \sin x, \cos (2 π + x) = \cos x, \cot (\frac{3 π}{2} - x) = \tan x, \cot (2 π + x) = \cot x$
So, by using these our equation simplifies to
$\cos (\frac{3 π}{2} + x) \cos (2 π + x) [\cot (\frac{3 π}{2} - x) + \cot (2 π + x)]$
$= \sin x \cos x [\tan x + \cot x] = \sin x \cos x [\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}]$
$= \sin x \cos x [\frac{\sin^{2} x + \cos^{2} x}{\sin x \cos x}] = \sin^{2} x + \cos^{2} x = 1 =$ R.H.S.

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

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)
Now, $\frac{{- (\cos (2 n + 3) x - \cos (- x)) + (\cos (2 n + 3) + \cos (- x))}}{2} (∵ \cos (- x) = \cos x)$
$= \frac{2 \cos x}{2} = \cos x =$ R.H.S.

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

Answer:
We know that
[ cos(A+B) - cos (A-B) = -2sinAsinB ]
By using this identity
$\cos (\frac{3 π}{4} + x) - \cos (\frac{3 π}{4} - x) = - 2 \sin \frac{3 π}{4} \sin x = - 2 \times \frac{1}{\sqrt{2}} \sin x = - \sqrt{2} \sin x =$ R.H.S.

Question:12 Prove the following $\sin^{2} 6 x - \sin^{2} 4 x = \sin 2 x \sin 10 x$

Answer:
We know that
$a^{2} - b^{2} = (a + b) (a - b)$
So, $\sin^{2} 6 x - \sin^{2} 4 x = (\sin 6 x + \sin 4 x) (\sin 6 x - \sin 4 x)$
Now, we know that
$\sin A + \sin B = 2 \sin (\frac{A + B}{2}) \cos (\frac{A - B}{2}), \sin A - \sin B = 2 \cos (\frac{A + B}{2}) \sin (\frac{A - B}{2})$
By using these identities
sin6x + sin4x = 2sin5x cosx
sin6x - sin4x = 2cos5x sinx
$\Rightarrow \sin^{2} 6 x - \sin^{2} 4 x = (2 \cos 5 x \sin 5 x) (2 \sin x \cos x)$
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
So, $\sin^{2} 6 x - \sin^{2} 4 x = \sin 2 x \sin 10 x$

Question:13 Prove the following $\cos^{2} 2 x - \cos^{2} 6 x = \sin 4 x \sin 8 x$

Answer:
As we know that
$a^{2} - b^{2} = (a - b) (a + b)$
$∴ \cos^{2} 2 x - \cos^{2} 6 x = (\cos 2 x - \cos 6 x) (\cos 2 x + \cos 6 x)$
Now, $\cos A - \cos B = - 2 \sin (\frac{A + B}{2}) \sin (\frac{A - B}{2}) \cos A + \cos B = 2 \cos (\frac{A + B}{2}) \cos (\frac{A - B}{2})$
By using these identities
cos2x - cos6x = -2sin(4x)sin(-2x) = 2sin4xsin2x ( $∵$ sin(-x) = -sin x and cos(-x) = cosx)
cos2x + cos 6x = 2cos4xcos(-2x) = 2cos4xcos2x
So our equation becomes
$(\cos 2 x - \cos 6 x) (\cos 2 x + \cos 6 x) = (2 \sin 4 x \sin 2 x) (2 \cos 4 x \cos 2 x) = (2 \sin 2 x \cos 2 x) (2 \sin 4 x \cos 4 x)$
$= \sin 4 x \sin 8 x =$ R.H.S.

Question:14 Prove the following $\sin 2 x + 2 \sin 4 x + \sin 6 x = 4 \cos^{2} x \sin 4 x$

Answer:
We know that
$\sin A + \sin B = 2 \sin (\frac{A + B}{2}) \cos (\frac{A - B}{2})$
We are using this identity
sin2x + 2sin4x + sin6x = (sin2x + sin6x) + 2sin4x
sin2x + sin6x = 2sin4xcos(-2x) = 2sin4xcos(2x) ( $∵$ 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) ( $∵ \cos 2 x = 2 \cos^{2} x - 1$ )
= $2 \sin 4 x (2 \cos^{2} x - 1$ +1 )
= $2 \sin 4 x (2 \cos^{2} x$ )
= $4 \sin 4 x \cos^{2} x =$ R.H.S.

Question:15 Prove the following $\cot 4 x (\sin 5 x + \sin 3 x) = \cot x (\sin 5 x - \sin 3 x)$

Answer:
We know that
$\sin x + \sin y = 2 \sin (\frac{x + y}{2}) \cos (\frac{x - y}{2})$
By using this , we get
sin5x + sin3x = 2sin4xcosx
$\frac{\cos 4 x}{\sin 4 x} (2 \sin 4 x \cos x) = 2 \cos 4 x \cos x$
Now nultiply and divide by sin x
$\frac{2 \cos 4 x \cos x \sin x}{\sin x} = \cot x (2 \cos 4 x \sin x) (∵ \frac{\cos x}{s i n x} = \cot x)$
Now we know that
$2 \cos x \sin y = \sin (x + y) - \sin (x - y)$
By using this our equation becomes
$= \cot x (\sin 5 x - \sin 3 x) =$ R.H.S.

Question:16 Prove the following $\frac{\cos 9 x - \cos 5 x}{\sin 17 x - \sin 3 x} = - \frac{\sin 2 x}{\cos 10 x}$

Answer:
As we know that
$\cos x - \cos y = - 2 \sin \frac{x + y}{2} \sin \frac{x - y}{2}, \cos 9 x - \cos 5 x = - 2 \sin 7 x \sin 2 x$
$\sin x - \sin y = 2 \cos \frac{x + y}{2} \sin \frac{x - y}{2}, \sin 17 x - \sin 3 x = 2 \cos 10 x \sin 7 x$
Now, $\frac{\cos 9 x - \cos 5 x}{\sin 17 x - \sin 3 x} = \frac{- 2 \sin 7 x \sin 2 x}{2 \cos 10 x \sin 7 x} = - \frac{\sin 2 x}{\cos 10 x} =$ R.H.S.

Question:17 Prove the following $\frac{\sin 5 x + \sin 3 x}{\cos 5 x + \cos 3 x} = \tan 4 x$

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
$\sin 5 x + \sin 3 x = 2 \sin 4 x \cos x, \cos 5 x + \cos 3 x = 2 \cos 4 x \cos x$
Now, $\frac{\sin 5 x + \sin 3 x}{\cos 5 x + \cos 3 x} = \frac{2 \sin 4 x \cos x}{2 \cos 4 x \cos x} = \frac{\sin 4 x}{\cos 4 x} = \tan 4 x =$ R.H.S.

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

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
$\frac{\sin x - \sin y}{\cos x + \cos y} = \frac{2 \cos \frac{x + y}{2} \sin \frac{x - y}{2}}{2 \cos \frac{x + y}{2} \cos \frac{x - y}{2}} = \frac{\sin \frac{x - y}{2}}{\cos \frac{x - y}{2}} = \tan \frac{x - y}{2} =$ R.H.S.

Question:19 Prove the following $\frac{\sin x + \sin 3 x}{\cos x + \cos 3 x} = \tan 2 x$

Answer:
We know that
$\sin x + \sin y = 2 \sin \frac{x + y}{2} \cos \frac{x - y}{2}$
$\cos x + \cos y = 2 \cos \frac{x + y}{2} \cos \frac{x - y}{2}$
We use these equations,
$\sin x + \sin 3 x = 2 \sin 2 x \cos (- x) = 2 \sin 2 x \cos x (∵ \cos (- x) = \cos x)$
$\cos x + \cos 3 x = 2 \cos 2 x \cos (- x) = 2 \cos 2 x \cos x (∵ \cos (- x) = \cos x)$
Now, $\frac{\sin x + \sin 3 x}{\cos x + \cos 3 x} = \frac{2 \sin 2 x \cos x}{2 \cos 2 x \cos x} = \frac{\sin 2 x}{\cos 2 x} = \tan 2 x =$ R.H.S.

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

Answer:
We know that
$\sin 3 x = 3 \sin x - 4 \sin^{3} x, \cos^{2} x - \sin^{2} x = \cos 2 x$ and $\cos 2 x = 1 - 2 \sin^{2} x$
We use these identities
$\sin x - \sin 3 x = \sin x - (3 \sin x - 4 \sin^{3} x) = 4 \sin^{3} x - 2 \sin x = 2 \sin x (2 \sin^{2} x - 1)$
$\sin^{2} x - \cos^{2} x = \sin^{2} x - (1 - \sin^{2} x) = 2 \sin^{2} x - 1$
Now, $\frac{\sin x - \sin 3 x}{\sin^{2} - \cos^{2} x} = \frac{2 \sin x (2 \sin^{2} x - 1)}{2 \sin^{2} x - 1} = 2 \sin x =$ R.H.S.

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

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
$\frac{(\cos 4 x + \cos 2 x) + \cos 3 x}{(\sin 4 x + \sin 2 x) + \sin 3 x} = \frac{2 \cos 3 x \cos x + \cos 3 x}{2 \sin 3 x \cos x + \sin 3 x} = \frac{2 \cos 3 x (\cos x + 1)}{2 \sin 3 x (\cos x + 1)} = \cot 3 x =$ R.H.S.

Question:22 prove the following $\cot x \cot 2 x - \cot 2 x \cot 3 x - \cot 3 x \cot x = 1$

Answer:
L.H.S.
= $\cot x \cot 2 x - \cot 3 x (\cot 2 x - \cot x)$
Now we can write $\cot 3 x = \cot (2 x + x)$
⇒ $\cot (a + b) = \frac{\cot a \cot b - 1}{\cot a + \cot b}$
So, $\cot x \cot 2 x - \frac{\cot 2 x \cot x - 1}{\cot 2 x + \cot x} (\cot 2 x + \cot x)$
= $\cot x \cot 2 x - (\cot 2 x \cot x - 1)$
= $\cot x \cot 2 x - \cot 2 x \cot x + 1 = 1 =$ R.H.S.

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

Answer:
We know that
$\tan 2 A = \frac{2 \tan A}{1 - \tan^{2} A}$
and we can write tan 4x = tan 2(2x)
So, $\tan 4 x = \frac{2 \tan 2 x}{1 - \tan^{2} 2 x}$ = $\frac{2 (\frac{2 \tan x}{1 - \tan^{2} x})}{1 - (\frac{2 \tan x}{1 - \tan^{2} x})^{2}}$
= $\frac{2 (2 \tan x) (1 - \tan^{2} x)}{(1 - \tan x)^{2} - (4 \tan^{2} x)}$
= $\frac{(4 \tan x) (1 - \tan^{2} x)}{(1)^{2} + (\tan^{2} x)^{2} - 2 \tan^{2} x - (4 \tan^{2} x)}$
= $\frac{(4 \tan x) (1 - \tan^{2} x)}{1^{2} + \tan^{4} x - 6 \tan^{2} x}$ = R.H.S.

Question:24 Prove the following $\cos 4 x = 1 - 8 \sin^{2} x \cos^{2} x$

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

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

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

NCERT Trigonometric Functions Class 11 Questions and Answers -
Exercise: Miscellaneous, Page Number: 71-72, Total Questions: 10

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

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
$2 \cos \frac{π}{13} \cos \frac{9 π}{13} + 2 \cos \frac{(\frac{3 π}{13} + \frac{5 π}{13})}{2} \cos \frac{(\frac{3 π}{13} - \frac{5 π}{13})}{2}$
= $2 \cos \frac{π}{13} \cos \frac{9 π}{13} + 2 \cos \frac{4 π}{13} \cos \frac{- π}{13}$ ( we know that cos(-x) = cos x )
= $2 \cos \frac{π}{13} \cos \frac{9 π}{13} + 2 \cos \frac{4 π}{13} \cos \frac{π}{13}$
= $2 \cos \frac{π}{13} (\cos \frac{9 π}{13} + \cos \frac{4 π}{13})$
again use the above identity
= $2 \cos \frac{π}{13} (2 \cos (\frac{\frac{9 π}{13} + \frac{4 π}{13}}{2}) \cos (\frac{\frac{9 π}{13} - \frac{4 π}{13}}{2})$
= $2 \cos \frac{π}{13} 2 \cos \frac{π}{2} \cos \frac{5 π}{26}$
we know that $\cos \frac{π}{2}$ = 0
So, $2 \cos \frac{π}{13} 2 \cos \frac{π}{2} \cos \frac{5 π}{26}$ = 0 = R.H.S.

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

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

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

Answer:
We know that $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ and $(a - b)^{2} = a^{2} - 2 a b + b^{2}$
We use these two in our problem
$(\sin x - \sin y)^{2} = \sin^{2} x - 2 \sin x \sin y + \sin^{2} y$ and $(\cos x + \cos y)^{2} = \cos^{2} x + 2 \cos x \cos y + \cos^{2} y$
$(\cos x + \cos y)^{2} + (\sin x - \sin y)^{2}$ = $\cos^{2} x + 2 \cos x \cos y + \cos^{2} y$ + $\sin^{2} x - 2 \sin x \sin y + \sin^{2} y$
= 1 + 2 cos x cos y + 1 - 2 sin x sin y $(∵ \sin^{2} x + \cos^{2} x = 1 a n d \sin^{2} y + \cos^{2} y = 1)$
= 2 + 2(cos x cos y - sin x sin y)
= 2 + 2cos(x + y)
= 2(1 + cos(x + y) )
Now we can write
$\cos (x + y) = 2 \cos^{2} \frac{(x + y)}{2} - 1$ $(∵ \cos 2 x = 2 \cos^{2} x - 1 \Rightarrow \cos x = 2 \cos^{2} \frac{x}{2} - 1)$
= $2 (1 + 2 \cos^{2} \frac{(x + y)}{2} - 1)$
= $4 \cos^{2} \frac{(x + y)}{2}$
= R.H.S.

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

Answer:
We know that $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ and $(a - b)^{2} = a^{2} - 2 a b + b^{2}$
We use these two in our problem
$(\sin x - \sin y)^{2} = \sin^{2} x - 2 \sin x \sin y + \sin^{2} y$ and $(\cos x - \cos y)^{2} = \cos^{2} x - 2 \cos x \cos y + \cos^{2} y$
$(\cos x - \cos y)^{2} + (\sin x - \sin y)^{2}$ = $\cos^{2} x - 2 \cos x \cos y + \cos^{2} y$ + $\sin^{2} x - 2 \sin x \sin y + \sin^{2} y$
= 1 - 2cos x cos y + 1 - 2sin x sin y $(∵ \sin^{2} x + \cos^{2} x = 1 a n d \sin^{2} y + \cos^{2} y = 1)$
= 2 - 2(cos x cos y + sin x sin y)
= 2 - 2cos(x - y) $(∵ \cos (x - y) = \cos x \cos y + \sin x \sin y)$
= 2(1 - cos(x - y) )
Now we can write
$\cos (x - y) = 1 - 2 \sin^{2} \frac{(x - y)}{2}$ $(∵ \cos 2 x = 1 - 2 \sin^{2} x \Rightarrow \cos x = 1 - 2 \sin^{2} \frac{x}{2})$
So, $2 (1 - \cos (x - y)) = 2 (1 - (1 - 2 \sin^{2} \frac{(x - y)}{2}))$
= $4 \sin^{2} \frac{(x - y)}{2}$ = R.H.S.

Question:5 Prove that $\sin x + \sin 3 x + \sin 5 x + \sin 7 x = 4 \cos x \cos 2 x \sin 4 x$

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
If we notice we need sin4x in our final result so it is better if we need a combination of sin7x and sin x , sin3x and sin5x to get sin4x.
$(\sin 7 x + \sin x) + (\sin 5 x + \sin 3 x) = 2 \sin \frac{7 x + x}{2} \cos \frac{7 x - x}{2}$ $+ 2 \sin \frac{5 x + 3 x}{2} \cos \frac{5 x - 3 x}{2}$
$=$ $2 \sin 4 x \cos 3 x + 2 \sin 4 x \cos x$
take 2sin4x common
= 2sin4x(cos3x + cosx)
We know that
$\cos A + \cos B = 2 \cos \frac{A + B}{2} \cos \frac{A - B}{2}$
We use this
$\cos 3 x + \cos x = 2 \cos \frac{3 x + x}{2} \cos \frac{3 x - x}{2}$
= $2 \cos 2 x \cos x$
Now 2sin4x(cos3x + cosx) = 2sin4x( $2 \cos 2 x \cos x$ )
= $4 \cos x \cos 2 x \sin 4 x$ = R.H.S.

Question:6 Prove that $\frac{(\sin 7 x + \sin 5 x) + (\sin 9 x + \sin 3 x)}{(\cos 7 x + \cos 5 x) + (\cos 9 x + \cos 3 x)} = \tan 6 x$

Answer:
We know that
$\sin A + \sin B = 2 \sin \frac{A + B}{2} \cos \frac{A - B}{2}$
$\cos A + \cos B = 2 \cos \frac{A + B}{2} \cos \frac{A - B}{2}$
We use these two identities in our problem
sin7x + sin5x = $2 \sin \frac{7 x + 5 x}{2} \cos \frac{7 x - 5 x}{2}$ = $2 \sin 6 x \cos x$
sin 9x + sin 3x = $2 \sin \frac{9 x + 3 x}{2} \cos \frac{9 x - 3 x}{2}$ = $2 \sin 6 x \cos 3 x$
cos 7x + cos5x = $2 \cos \frac{7 x + 5 x}{2} \cos \frac{7 x - 5 x}{2}$ = $2 \cos 6 x \cos x$
cos 9x + cos3x = $2 \cos \frac{9 x + 3 x}{2} \cos \frac{9 x - 3 x}{2}$ = $2 \cos 6 x \cos 3 x$
$\frac{(\sin 7 x + \sin 5 x) + (\sin 9 x + \sin 3 x)}{(\cos 7 x + \cos 5 x) + (\cos 9 x + \cos 3 x)}$ = $\frac{(2 \sin 6 x \cos x) + (2 \sin 6 x \cos 3 x)}{(2 \cos 6 x c o s x) + (2 \cos 6 x \cos 3 x)}$
= $\frac{2 \sin 6 x (\cos x + \cos 3 x)}{2 \cos 6 x (\cos x + \cos 3 x)} = \tan 6 x$ = R.H.S. $(∵ \frac{\sin x}{\cos x} = \tan x)$

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

Answer:
We know that
$\cos A + \cos B = 2 \cos \frac{A + B}{2} \cos \frac{A - B}{2}$
$\sin A - \sin B = 2 \cos \frac{A + B}{2} \sin \frac{A - B}{2}$
We use these identities
$\sin 3 x - \sin x = 2 \cos \frac{3 x + x}{2} \sin \frac{3 x - x}{2}$
$= 2 \cos 2 x \sin x$
sin2x + $2 \cos 2 x \sin x$ = 2sinx cosx + $2 \cos 2 x \sin x$
take 2 sinx common
$2 \sin x (\cos x + \cos 2 x) = 2 \sin x (2 \cos \frac{2 x + x}{2} \cos \frac{2 x - x}{2})$
$= 2 \sin x (2 \cos \frac{3 x}{2} \cos \frac{x}{2})$
$= 4 \sin x \cos \frac{3 x}{2} \cos \frac{x}{2} =$ R.H.S.

Question:8 Find $\sin \frac{x}{2}, \cos \frac{x}{2}, a n d \tan \frac{x}{2}$ in $\tan x = - \frac{4}{3}$ , x in quadrant II.

Answer:
tan x = $- \frac{4}{3}$
We know that ,
$\sec^{2} x = 1 + \tan^{2} x$
$= 1 + {(- \frac{4}{3})}^{2}$
$= 1 + \frac{16}{9}$ = $\frac{25}{9}$
$\sec x = \sqrt{\frac{25}{9}}$ = $\pm \frac{5}{3}$
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,
$\cos x = 2 \cos^{2} \frac{x}{2} - 1$ ( $∵ \cos 2 x = 2 \cos^{2} x - 1 \Rightarrow \cos x = 2 \cos^{2} \frac{x}{2} - 1$ )
⇒ $- \frac{3}{5} + 1 = 2$ $\cos^{2} \frac{x}{2}$
⇒ $\frac{- 3 + 5}{5}$ = $2 \cos^{2} \frac{x}{2}$
⇒ $\frac{2}{5}$ = $2 \cos^{2} \frac{x}{2}$
⇒ $\cos^{2} \frac{x}{2}$ = $\frac{1}{5}$
⇒ $\cos \frac{x}{2}$ = $\sqrt{\frac{1}{5}}$ = $\pm \frac{1}{\sqrt{5}}$
x lies in II quadrant so value of $\cos \frac{x}{2}$ is +ve
$\cos \frac{x}{2}$ = $\frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}$
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
$\sin \frac{x}{2} = \frac{2}{\sqrt{} 5} = \frac{2 \sqrt{5}}{5}$
$\tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} = \frac{\frac{2 \sqrt{5}}{5}}{(\frac{\sqrt{5}}{5})} = 2$

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

Answer:
$π < x < \frac{3 π}{2} \Rightarrow \frac{π}{2} < \frac{x}{2} < \frac{3 π}{4}$
We know that
cos x = $2 \cos^{2} \frac{x}{2} - 1$
$2 \cos^{2} \frac{x}{2} =$ cos x + 1 = $(- \frac{1}{3})$ + 1 = $(\frac{- 1 + 3}{3})$ = $\frac{2}{3}$
⇒ $\cos \frac{x}{2} = \sqrt{\frac{1}{3}} = \pm \frac{1}{\sqrt{3}}$
⇒ $\cos \frac{x}{2} = - \frac{1}{\sqrt{3}} = - \frac{\sqrt{3}}{3}$ (As x lies in 3rd quadrant)
we know that
cos x = $1 - 2 \sin^{2} \frac{x}{2}$
⇒ $2 \sin^{2} \frac{x}{2} = 1 - \cos x$ = 1 - $(- \frac{1}{3})$ = $\frac{3 + 1}{3}$ = $\frac{4}{3}$
⇒ $2 \sin^{2} \frac{x}{2} = \frac{4}{3} \Rightarrow \sin^{2} \frac{x}{2} = \frac{2}{3} \Rightarrow \sin \frac{x}{2} = \pm \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3}$
Because $\sin \frac{x}{2}$ is +ve in given quadrant
$\tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} = \frac{\frac{\sqrt{6}}{3}}{\frac{- \sqrt{3}}{3}} = - \sqrt{2}$

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

Answer:
$\frac{π}{2} < x < π \Rightarrow \frac{π}{4} < \frac{x}{2} < \frac{π}{2}$ all functions are positive in this range
We know that
$\cos^{2} x = 1 - \sin^{2} x$ = 1 - ${(\frac{1}{4})}^{2}$ = $1 - \frac{1}{16}$ = $\frac{15}{16}$
cos x = $\sqrt{\frac{15}{16}} = \pm \frac{\sqrt{1} 5}{4} = - \frac{\sqrt{1} 5}{4}$ (cos x is -ve in II quadrant)
We know that
cosx = $2 \cos^{2} \frac{x}{2} - 1$
$2 \cos^{2} \frac{x}{2} = \cos x + 1 = - \frac{\sqrt{1} 5}{4} + 1 = \frac{- \sqrt{1} 5 + 4}{4}$
$\cos^{2} \frac{x}{2} = \frac{- \sqrt{1} 5 + 4}{8}$
$\cos \frac{x}{2} = \pm \sqrt{\frac{- \sqrt{1} 5 + 4}{8}} = \frac{\sqrt{- \sqrt{1} 5 + 4}}{2 \sqrt{2}} = \frac{\sqrt{8 - 2 \sqrt{1} 5}}{4}$ (because all functions are posititve in given range)
Similarly,
cos x = $1 - 2 \sin^{2} \frac{x}{2}$
$2 \sin^{2} \frac{x}{2} = 1 - \cos x \Rightarrow 2 \sin^{2} \frac{x}{2} = 1 - (\frac{- \sqrt{1} 5}{4}) = \frac{4 + \sqrt{1} 5}{4}$
$\sin \frac{x}{2} = \pm \sqrt{\frac{\sqrt{1} 5 + 4}{8}} = \frac{\sqrt{\sqrt{1} 5 + 4}}{2 \sqrt{2}} = \frac{\sqrt{8 + 2 \sqrt{1} 5}}{4}$ (because all functions are posititve in given range)
$\tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} = \frac{\frac{\sqrt{8 + 2 \sqrt{1} 5}}{4}}{\frac{\sqrt{8 - 2 \sqrt{1} 5}}{4}} = \frac{8 + 2 \sqrt{1} 5}{\sqrt{64 - 15 \times 4}} = \frac{8 + 2 \sqrt{1} 5}{\sqrt{4}} = 4 + \sqrt{1} 5$

Trigonometry Class 11 Solutions - Exercise Wise

Interested students can study Trigonometric Functions Exercise using following links-

NCERT Solutions for Class 11 Mathematics - Chapter Wise

NCERT Solutions for Class 11 Mathematics Chapters

NCERT Solutions for Class 11 Mathematics Chapter 1: Sets

NCERT Solutions for Class 11 Mathematics Chapter 2: Relations and Functions

NCERT Solutions for Class 11 Mathematics Chapter 3: Trigonometric Functions

NCERT Solutions for Class 11 Mathematics Chapter 4: Principle of Mathematical Induction

NCERT Solutions for Class 11 Mathematics Chapter 5: Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Mathematics Chapter 6: Linear Inequalities

NCERT Solutions for Class 11 Mathematics Chapter 7: Permutations and Combinations

NCERT Solutions for Class 11 Mathematics Chapter 8: Binomial Theorem

NCERT Solutions for Class 11 Mathematics Chapter 9: Sequences and Series

NCERT Solutions for Class 11 Mathematics Chapter 10: Straight lines

NCERT Solutions for Class 11 Mathematics Chapter 11: Conic Sections

NCERT Solutions for Class 11 Mathematics Chapter 12: Introduction to Three Dimensional Geometry

NCERT Solutions for Class 11 Mathematics Chapter 13: Limits and Derivatives

NCERT Solutions for Class 11 Mathematics Chapter 14: Mathematical Reasoning

NCERT Solutions for Class 11 Mathematics Chapter 15: Statistics

NCERT Solutions for Class 11 Mathematics Chapter 16: Probability

Importance of Solving NCERT Questions of Class 11 Maths Chapter 8

Solving these NCERT questions will help students understand the basic concepts of trigonometry easily.
Students can practice various types of questions which will improve their problem-solving skills.
These NCERT exercises cover all the important topics and concepts so that students can be well-prepared for various exams.
By solving these NCERT trigonometry problems students will get to know about all the real-life applications of trigonometry.

NCERT Solutions for Class 11 - Subject Wise

Here are the subject-wise links for the NCERT solutions of class 11

NCERT Books and NCERT Syllabus

Here are some useful links for NCERT books and NCERT syllabus for class 11

Frequently Asked Questions (FAQs)

1. How to solve trigonometric equations in Class 11 Maths Chapter 3?

To solve trigonometric equations in Class 11 Maths Chapter 3, express the given terms in basic trigonometric functions like sin, cos, and tan, then try to apply proper trigonometric identities and use simplification.

2. What are the key formulas of Trigonometric Functions in NCERT Class 11?

The key formulas of Trigonometric Functions in NCERT Class 11 are:

$1) \cos^{2} x + \sin^{2} x = 1$

$2) 1 + \tan^{2} x = \sec^{2} x$

$3) 1 + \cot^{2} x = {cosec}^{2} x$

$4) \cos (2 n π + x) = \cos x$

$5) \sin (2 n π + x) = \sin x$

$6) \sin (- x) = - \sin x$

$7) \cos (- x) = \cos x$

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

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

$10) \sin (x + y) = \sin x \cos y + \cos x \sin y$

$11) \sin (x - y) = \sin x \cos y - \cos x \sin y$

$12) \tan (x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$

$13) \tan (x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}$

$14) \cot (x + y) = \frac{\cot x \cot y - 1}{\cot x + \cot y}$

$15) \cot (x - y) = \frac{1 + \cot x \cot y}{\cot y - \cot x}$

3. How to find the general solution of a trigonometric equation?

To find the general solution of a trigonometric equation, use the following standard equations:

$\sin θ = \sin α \Rightarrow θ = n π + (- 1)^{n} α, n \in Z$

$\cos θ = \cos α \Rightarrow θ = 2 n π \pm α, n \in Z$

$\tan θ = \tan α \Rightarrow θ = n π + α, n \in Z$

4. What are the domain and range of trigonometric functions in Class 11?

In class 11 chapter 3 trigonometric functions, a domain basically denotes the input values and range denotes the output values.

For example: The domain of $\sin θ$ and $\cos θ$ is $(- \infty, \infty)$ and range is (–1, 1).

5. How to derive the sine and cosine function graphs?

To derive the sine and cosine function graphs, plot $y = \sin θ$ and $y = \cos θ$ for the values of $θ$ like 0°, 30°, 60°, 90°, 180°, etc..

The sine graph starts at (0, 0) and goes up to 1 at 90°, drops to –1 at 270°, and comes to 0 again at 360° then repeats.

Also Read

NCERT Class 11 Biology Chapter 7 Notes Structural Organisation In Animals- Download PDF Notes

NCERT Class 11 Biology Chapter 13 Notes Photosynthesis In Higher Plants - Download PDF Notes

NCERT Class 11 Biology Chapter 8 Notes Cell The Unit Of Life- Download PDF Notes

NCERT Class 11 Biology Chapter 21 Notes Neural Control And Coordination- Download PDF Notes

NCERT Class 11 Chemistry Chapter 4 Notes Chemical Bonding and Molecular Structure - Download PDF

Hydrogen Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 9 Notes - Download PDF

NCERT Exemplar Class 11 Maths Solutions Chapter 7 Permutations and Combinations

NCERT Exemplar Class 11 Maths Solutions Chapter 11 Conic Sections

NCERT Exemplar Class 11 Biology Solutions Chapter 10 Cell Cycle and Cell Division

NCERT Exemplar Class 11 Biology Solutions Chapter 8 Cell The Unit Of Life

NCERT Exemplar Class 11 Biology Solutions Chapter 14 Respiration in Plants

NCERT Exemplar Class 11 Biology Solutions Chapter 1 The Living World

NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations

NCERT Exemplar Class 11 Maths Solutions Chapter 15 Statistics

NCERT Solutions for Class 11 Chemistry Chapter 13 Hydrocarbons

NCERT Solutions for Class 11 Chemistry Chapter 12 Organic Chemistry Some Basic Principles and Techniques

NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reactions

NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
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NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

Trigonometric Functions Class 11 Questions And Answers PDF Free Download

Trigonometric Functions Class 11 Solutions - Important Formulae

Trigonometric Functions Class 11 NCERT Solutions (Exercises)

Trigonometry Class 11 Solutions - Exercise Wise

NCERT Solutions for Class 11 Mathematics - Chapter Wise

Importance of Solving NCERT Questions of Class 11 Maths Chapter 8

NCERT Solutions for Class 11 - Subject Wise

NCERT Books and NCERT Syllabus

Frequently Asked Questions (FAQs)

Also Read

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