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# NCERT Solutions of class 11 maths chapter 3-Trigonometric functions

Here NCERT Solutions of class 11 maths of chapter 3 – Trigonometric functions are solved by an expert teacher of maths. All questions of chapter 3-Trigonometric functions of class 11 are solved by a step-by-step method to help the students in their preparation for the forthcoming exams.

The chapter Trigonometric functions of NCERT textbook of class 11 contains the questions based on the relation between radian and degree measures of angles, the length of the arc of the circle, verification of different kinds of Trigonometric identities

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## NCERT Solutions of class 11 maths chapter 3-Trigonometric functions

Exercise 3.1 and 3.2- Trigonometric Functions

Exercise 3.3 – Trigonometric Functions

Miscellaneous Exercise Chapter 3-Trigonometric Functions

### CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

Study notes of Maths and Science NCERT and CBSE from class 9 to 12

Class 11 NCERT solutions of Physics and Chemistry

Chapter 1 – Some concepts of the chemistry

Chapter 1-Physical World

## NCERT Solutions of class 11 maths chapter 3 -Trigonometric Functions

### The Relation Between Radian and Degree →

${\color{DarkBlue} \mathbf{\Theta\left (in\; radian \right ) =\frac{length \; of \; arc}{radius\; of\; circle}=\frac{l}{r}}}$

A circle makes the angle(θ) of 360° around its center and an arc of the length(l)= 2πr(circumference)

${\color{DarkBlue} \mathbf{Therefore,\; 360^{\circ} =\frac{2\Pi r}{r}= 2\Pi }}$

Which shows that 1° = 2Π/360

$\dpi{100} \mathbf{{\color{DarkBlue} {\color{DarkBlue} \mathbf{So,1^{\circ}= \frac{2\Pi }{360}=\frac{\Pi }{180}}}\; Radian}}$

${\color{DarkBlue} \mathbf{1\; rad.\; =\frac{180}{\Pi }\; deg.}}$

### Exercise 3.1

Q1. Find the radian measure corresponding to the following degree measures.

(i) 25°  (ii) –47°30′  (iii) 240°   (iv) 520°

Answer.(i) Using the following relationship between rad. and deg. in all questions.

${\color{DarkBlue} \mathbf{1^{\circ}= \frac{\Pi }{180}\; rad.}}$

$\mathbf{{\color{DarkBlue} \mathbf{25^{\circ}= \frac{\Pi }{180}\; \times 25}=\frac{5\Pi }{36}}}$

(ii) –47°30′ = –47°+ 30/60(1°= 60′)

${\color{DarkBlue} \mathbf{-47\frac{1}{2}=-\frac{95}{2}}}$

${\color{DarkBlue} \mathbf{-\frac{95}{2}=-\frac{95}{2}\times \frac{\Pi }{180}}}$

$\mathbf{={}\color{DarkBlue} \mathbf{-\frac{19\Pi }{36}}}$

${\color{DarkBlue}\mathbf{(iii)} \; \mathbf{240^{\circ}=\frac{\Pi }{180}\times 240 = \frac{20\Pi }{15}\; rad.}}$

${\color{DarkBlue}\mathbf{(iv)\; }\mathbf{520^{\circ}= \frac{\Pi }{180}\times 520 = \frac{26\Pi }{9}\; rad.}}$

Q2.Find the degree measures of the following radian measures.

${\color{DarkGreen} \mathbf{(i) \frac{11}{16}\; \; (ii) -4\; \; (iii)\frac{5\Pi }{3}\; \; (iv)\frac{7\Pi }{6}}}$

Answers. Using the following relationship between radian and degree in all questions.

${\color{DarkBlue} \mathbf{1 rad.= \frac{180}{\Pi }\; deg.}}$

${\color{DarkBlue} \mathbf{(i)\;} \mathbf{\frac{11}{16}\; rad.=\frac{11}{16}\times \frac{180}{\Pi }\; degree}}$

Substituting π = 22/7

${\color{DarkBlue} \mathbf{=\frac{11\times 7\times 180}{16\times 22}}}$

${\color{DarkBlue}= \mathbf{\frac{315}{8} =39\frac{3}{8}=39 + \frac{3}{8}\times 60\; (1^{\circ}=60')}}$

${\color{DarkBlue} \mathbf{39+\frac{45}{2}= 39+22\frac{1}{2}=39^{\circ}22'+\frac{1}{2}'}}$

${\color{DarkBlue} \mathbf{\frac{1}{2}\times 60=30'' (1'=60'')}}$

${\color{DarkBlue} T\mathbf{herefore \; \frac{11}{16}rad.= 39^{\circ}22'30''}}$

${\color{DarkBlue} \mathbf{=\frac{-4\times 7\times 180}{22}=-\frac{2520}{11}deg.}}$

${\color{DarkBlue}\mathbf{= }\mathbf{-229\frac{1}{11}= -\left ( 229+\frac{1}{11} \times 60\right )\; \because 1^{\circ}=60'}}$

${\color{DarkBlue} \mathbf{=-\left ( 229+\frac{60}{11} \right )=-\left ( 229^{\circ}5'+\frac{5}{11}' \right )}}$

${\color{DarkBlue} \mathbf{\frac{5}{11}'=\frac{5}{11}\times 60=\frac{300}{11}}}\mathbf{\simeq 27''}\; \mathbf{\because 1'=60''}$

${\color{DarkBlue} \mathbf{(iii)\; \frac{5\Pi }{3}=\frac{5\Pi }{3}\times \frac{180}{\Pi }=150^{\circ}\left ( 1rad=\frac{180}{\Pi }^{\circ} \right )}}$

${\color{DarkBlue} \mathbf{(iv)\; \frac{7\Pi }{6}=\frac{7\Pi }{6}\times \frac{180}{\Pi }=210^{\circ}}}$

See the video for Q1,Q2,Q3 and Q4

Q3.A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second.

In 1 minute  the wheel makes = 360 revolution

360 revolution is equivalent to = 360 ×2π = 720π rad.

1 minute = 60”

So, in 1second the wheel will turn=720π/60=12π rad.

Therefore in 1 second, the wheel will turn 12π rad.

Q4.Find the degree measure of the angle subtended at the center of the circle of radius 100 cm by an arc of length 22 cm.

Anwer. Let the given arc subtend an angle of θ.

${\color{DarkBlue} \mathbf{\Theta = \frac{length of arc}{radius}=\frac{l}{r}}}$

${\color{DarkBlue} \mathbf{\Theta =\frac{22}{100}=\frac{11}{50}\; rad.}}$

1 rad. =180/π, where π =22/7

${\color{DarkBlue} \mathbf{\Theta =\frac{11}{50}\times \frac{180}{\Pi }=\frac{11\times 7\times 180}{22\times 50}}}$

${\color{DarkBlue} \mathbf{\Theta =\frac{63}{5}=12\frac{3}{5}}}$

${\color{DarkBlue} \mathbf{12^{\circ}+\frac{3}{5}^{\circ}\times 60,\; 1^{\circ}=60'}}$

θ= 12°36”

Q5. In a circle diameter 40 cm, the length of a chord is 20 cm, find the length of the minor arc of the chord.

Answer. The diameter of circle = 40 cm, the length of  the chord =20 cm,the radius =40/2=20 cm

Drawing a chord AB in a circle of center O and joining its both ends to the center,

in ΔOAB, OA =OB(radii of same circle), OA =OB =OC=20 cm

therefore θ = 60°(ΔOAB is an equilateral Δ)

${\color{DarkBlue} \mathbf{\Theta = \frac{the \; length\; of \; minor\; arc}{radius}}}$

${\color{DarkBlue} \mathbf{60^{\circ}=\frac{\Pi }{180}\times 60\; rad.=\frac{22}{21}\; rad.}}$

${\color{DarkBlue} \mathbf{length \; of \; minor \; arc= 20 \times \frac{22}{21}=\frac{440}{21}cm}}$

Q6.In two circles,  arcs of the same length subtend  angles 60° and 75 ° at the center,find  the ratio of thier radii.

Answer. Let r₁ and  r₂   are the radii of the circles having the arcs subtending the angles 60°and 75° respectively.

Applying the formula θ = l/r, l=length of the arc and r=radius of the circle

${\color{DarkBlue} \mathbf{60\times \frac{\Pi }{180}=\frac{l}{r_{1}}}}$

${\color{DarkBlue} \mathbf{\frac{\Pi }{3}=\frac{l}{r_{1}}}}$

${\color{DarkBlue} \mathbf{r_{1}=\frac{3}{\Pi }}}$

${\color{DarkBlue} \mathbf{75\times \frac{\Pi }{180}=\frac{l}{r_{2}}}}$

${\color{DarkBlue} \mathbf{=\frac{5\Pi }{12}= \frac{l}{r_{2}}}}$

${\color{DarkBlue} \mathbf{r_{2}= \frac{12}{5\Pi }}}$

${\color{DarkBlue} \mathbf{r_{1}:r_{2} = \frac{3}{\Pi }:\frac{12}{5\Pi }}}$

${\color{DarkBlue} \mathbf{r_{1}:r_{2}= 5:4}}$

Therefore the ratio between radii of the circles will be 5:4.

Q7.Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of length.

(a)10 cm  (b) 15 cm  (c) 21 cm

(a) Answer. Length of pendulum makes a radius of  arcs formed by it

So,r =10 cm, l = 10 cm

${\color{DarkBlue} \mathbf{\Theta =\frac{l}{r}=\frac{10}{75}=\frac{2}{15}}}$

So, the angle in radian will be 2/15  rad. by which pendulum swings.

All questions (b) and (c) you can solve in a similar way.

### Exercise-3.2

Q1.Find the value of 5 other trigonometric functions if cos x = -1/2 if x lies in third quadrant.

${\color{DarkBlue} \mathbf{cosx = -\frac{1}{2}}}$

${\color{DarkBlue} \mathbf{\therefore sinx= \pm \sqrt{1-cos^{2}x}= \pm \sqrt{1-\left ( -\frac{1}{2} \right )^{2}}}}$

${\color{DarkBlue} \mathbf{sinx =\pm \frac{\sqrt{3}}{2}}}$

The value of sinx is negative at third quadrant, so

${\color{DarkBlue} \mathbf{sinx =-\frac{\sqrt{3}}{2}}}$

${\color{DarkBlue} \mathbf{tanx =\frac{sinx}{cosx}=\frac{\left ( -\frac{\sqrt{3}}{2} \right )}{\left ( -\frac{1}{2} \right )}=\sqrt{3}}}$

${\color{DarkBlue} \mathbf{cotx=\frac{1}{tanx}=\frac{1}{\sqrt{3}}}}$

${\color{DarkBlue} \mathbf{secx=\frac{1}{cosx}=\frac{1}{\left ( -\frac{1}{2} \right )}=-2}}$

${\color{DarkBlue} \mathbf{cosecx=\frac{1}{sinx}=\frac{1}{\left ( -\frac{\sqrt{3}}{2} \right )}=-\frac{2}{\sqrt{3}}}}$

Q2. Find the value of other 5 trigonometric functions if ${\color{DarkBlue} \mathbf{sinx=\frac{3}{5}}}$ ,x lies in the second quadrant.

${\color{DarkBlue} \mathbf{cosx=\pm \sqrt{1-sin^{2}x}=\pm \sqrt{1-\left ( \frac{3}{5} \right )^{2}}}}$

${\color{DarkBlue} \mathbf{cosx=\pm \frac{4}{5}}}$

x lies in second quadrant and value of cosx is negative at second quadrant,so

${\color{DarkBlue} \mathbf{cosx= - \frac{4}{5}}}$

${\color{DarkBlue} \mathbf{tanx =\frac{sinx}{cosx}= \frac{\frac{3}{5}}{-\frac{4}{5}}=-\frac{3}{4}}}$

${\color{DarkBlue} \mathbf{cotx =\frac{1}{tanx}=\frac{1}{-\frac{3}{4}}=-\frac{4}{3}}}$

${\color{DarkBlue} \mathbf{secx =\frac{1}{cosx}=\frac{1}{-\frac{4}{5}}=-\frac{5}{4}}}$

${\color{DarkBlue} \mathbf{cosecx=\frac{1}{sinx}=\frac{1}{\frac{3}{5}}=\frac{5}{3}}}$

Q3. Find the value of other 5 trigonometric functions,if cotx =${\color{DarkBlue} \mathbf{\frac{3}{4}}}$ ,x lies in third quadrant.

${\color{DarkBlue} \mathbf{tanx=\frac{1}{cotx}=\frac{1}{\frac{3}{4}}=\frac{4}{3}}}$

$\mathbf{secx = \pm \sqrt{1+tan^{2}x}= \pm \sqrt{1+\left ( \frac{4}{3} \right )^{2}}}$

${\color{DarkBlue} \mathbf{secx=\pm \frac{5}{3}}}$

Since,x lies in third quadrant and value of secx is negative in third quadrant.

${\color{DarkBlue}\therefore \mathbf{secx= -\frac{5}{3}}}$

${\color{DarkBlue} \mathbf{cosx =\frac{1}{secx}=\frac{1}{-\frac{5}{3}}=-\frac{3}{5}}}$

${\color{DarkBlue} \mathbf{sinx =cosx.tanx = -\frac{3}{5}\times \frac{4}{3}=-\frac{4}{5}}}$

${\color{DarkBlue} \mathbf{cosecx=\frac{1}{sinx}=\frac{1}{-\frac{4}{5}}= -\frac{5}{4}}}$

Q4.Find the value of other 5 trigonometric functions,if secx= ${\color{DarkBlue} \mathbf{\frac{13}{5}}}$, x lies in fourth quadrant.

${\color{DarkBlue} \mathbf{cosx=\frac{1}{secx}=\frac{1}{\frac{13}{5}}=\frac{5}{13}}}$

${\color{DarkBlue} \mathbf{sinx =\pm \sqrt{1-cos^{2}x}=\pm \sqrt{1-\left ( \frac{5}{13} \right )^{2}}}}$

${\color{DarkBlue} \mathbf{sinx =\pm \frac{12}{13}}}$

x lies in fourth quadrant and value of sinx is negative at 4 th quadrant.

Therefore

${\color{DarkBlue} \mathbf{sinx = - \frac{12}{13}}}$

${\color{DarkBlue} \mathbf{tanx =\frac{sinx}{cosx}=\frac{-\frac{12}{13}}{\frac{5}{13}}}}$

${\color{DarkBlue} \mathbf{tanx = -\frac{12}{5}}}$

${\color{DarkBlue} \mathbf{cotx =-\frac{5}{12}}}$

${\color{DarkBlue} \mathbf{cosecx=\frac{1}{sinx}=\frac{1}{-\frac{12}{13}}}}$

${\color{DarkBlue} \mathbf{cosecx = -\frac{13}{12}}}$

Q5. Find the value of other  5 trigonometric function,if tanx =${\color{DarkBlue} \mathbf{ -\frac{5}{12}}}$, x lies in the second quadrant.

${\color{DarkBlue} \mathbf{cotx =\frac{1}{tanx}=-\frac{12}{5}}}$

${\color{DarkBlue} \mathbf{cosecx=\pm \sqrt{1+cot^{2}x}= \pm \sqrt{1+\left ( -\frac{12}{5} \right )^{2}}}}$

${\color{DarkBlue} \mathbf{cosecx= \pm \frac{13}{5}}}$

x lies in 2nd quadrant and value of cosecx is + ve in 2nd quadrant.

therefore

${\color{DarkBlue} \mathbf{cosecx= \frac{13}{5}}}$

${\color{DarkBlue} \mathbf{sinx= \frac{5}{13}}}$

${\color{DarkBlue} \mathbf{cosx= cotx.sinx=-\frac{12}{5} \times \frac{5}{13}}}$

${\color{DarkBlue} \mathbf{cosx=-\frac{12}{13}}}$

${\color{DarkBlue} \mathbf{secx =\frac{1}{cosx}=-\frac{13}{12}}}$

${\color{DarkBlue} \mathbf{cosecx =\frac{1}{sinx}=\frac{13}{5}}}$

Q6. Find the value of the trigonometric function sin765°.

As we know sin(2nπ + θ) = sinθ, where n∈ N, means the value of sin repeats after every interval of 2π.

2π = 360°

So dividing 765 by 360 ,we get

765 = 2 ×360 + 45= 2×2π+ 45

sin765 =sin(4π + 45) =sin45

${\color{DarkBlue} \mathbf{sin45^{\circ}=\frac{1}{\sqrt{2}}}}$

Therefore

${\color{DarkBlue} \mathbf{sin765^{\circ}=\frac{1}{\sqrt{2}}}}$

Q7.Find the value of the trigometric function cosec(-1410°) .

Since ,sin(–θ) = –sinθ

∴cosec(–θ) = –cosecθ

cosec(–1410) = –cosec(1410)

Moreover ,sin(2nπ +θ) = sinθ

On the same way cosec(2nπ +θ) = cosecθ(n∈N) after every interval of 2π, the value of cosecθ also repeats like sinθ.

Therefore dividing 1410 by 360(i.e 2π) ,we get

1410 = 4 ×360  –30 =8π –π/6

${\color{DarkBlue} \mathbf{-cosec(8\Pi -\frac{\Pi }{6}) =-cosec\left ( -\frac{\Pi }{6} \right )}}$

${\color{DarkBlue} \mathbf{-cosec\left ( -\frac{\Pi }{6} \right )=-\left ( -cosec\frac{\Pi }{6} \right )=cosec\frac{\Pi }{6}}}$

${\color{DarkBlue} \mathbf{Therefore ,\; cosec\left ( -1410 \right )= cosec\frac{\Pi }{6}=2}}$

Q8. Find the value of the trigonometric function $\dpi{100} {\color{DarkBlue} \mathbf{tan\frac{19\Pi }{3}}}$.

As we know the value of tan repeats after an interval of π i.e tan so dividing 19π by 3 we get

$\dpi{100} {\color{DarkBlue} \mathbf{\frac{19\Pi }{3} = 6\Pi +\frac{\Pi }{3}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{tan\frac{19\Pi }{3}= tan(6\Pi +\frac{\Pi }{3})= tan\frac{\Pi }{3}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{tan\frac{19\Pi }{3}=\sqrt{3}}}$

Q9.Find the value of the trigonometric function $\dpi{100} {\color{DarkGreen} \mathbf{sin\left ( -\frac{11\Pi }{3} \right )}}$.

$\dpi{100} {\color{DarkBlue} \mathbf{sin\left ( -\frac{11\Pi }{3} \right )}}\mathbf{{\color{DarkBlue} =-sin\frac{11\Pi }{3}}}$

Since the value of sin repeats after an interval of 2π(360°) ,

11π/3 =11×180/3=660° ,dividing it by 360,we get

660=2×360–60=2×2π –π/3= 4π –π/3

$\dpi{100} {\color{DarkBlue} \mathbf{-sin\frac{11\Pi }{3}=-sin\left ( 4\Pi -\frac{\Pi }{3} \right )}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=-sin\left ( -\frac{\Pi }{3} \right )}}$

As we know sin(–θ)= –sinθ, so

$\dpi{100} {\color{DarkBlue} \mathbf{-\left ( -sin\frac{\Pi }{3} \right )=sin\frac{\Pi }{3}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{sin\left ( -\frac{11\Pi }{3} \right )=\frac{\sqrt{3}}{2}}}$

Q10.Find the value of the trigonometric function .$\dpi{100} {\color{DarkGreen} \mathbf{cot\left ( -\frac{15\Pi }{4} \right )}}$.

As we know cot(–θ) = –cotθ,therefore

$\dpi{100} {\color{DarkBlue} \mathbf{cot(-\frac{15\Pi }{4})}}\mathbf{\color{DarkBlue} {=-cot\frac{15\Pi }{4}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=-cot\left ( 3\frac{3}{4}\Pi \right )= -cot\left ( 3\Pi +\frac{3\Pi }{4} \right )}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=-cot\left ( 3\Pi +\Pi +\frac{3\Pi }{4}-\Pi \right )=cot\left ( 4\Pi -\frac{\Pi }{4} \right )}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=-cot\left ( -\frac{\Pi }{4} \right )=-\left ( -cot\frac{\Pi }{4} \right )}}$

$\dpi{100} {=\color{DarkBlue} \mathbf{cot\frac{\Pi }{4}=1}}$

$\dpi{100} {\color{DarkBlue} \mathbf{Therefore \; cot\left ( -\frac{15\Pi }{4} \right )=1}}$

## NCERT Solutions of Science and Maths for Class 9,10,11 and 12

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