NCERT Solutions For Class 11 Maths Chapter 3 Exercise 3.2

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Chapter 3 Ex.3.2 Question 1

Find the values of trigonometric function \(\cos x = - \frac{1}{2}\), \(x \) lies in third quadrant.

Solution

As we know that

\[\begin{align}\sec x &= \frac{1}{{\cos x}}\\ &= \frac{1}{{\left( { - \frac{1}{2}} \right)}}\\ &= - 2\end{align}\]

Now, \({\sin ^2}x + {\cos ^2}x = 1\)

\[\begin{align}{\sin ^2}x &= 1 - {\cos ^2}x\\\sin x &= \pm \sqrt {1 - {{\cos }^2}x} \\ &= \pm \sqrt {1 - {{\left( { - \frac{1}{2}} \right)}^2}} \\& = \pm \sqrt {1 - \frac{1}{4}} \\ &= \pm \sqrt {\frac{3}{4}} \\ &= \pm \frac{{\sqrt 3 }}{2}\end{align}\]

Since, \(x\) lies in third quadrant, the value of \(\sin x\) will be negative.

Therefore,

\[\sin x = - \frac{\sqrt 3 }{2}\]

Now,

\[\begin{align}{\rm{cosec}}\;x &= \frac{1}{\sin x}\\ &= \frac{1}{\left( { - \frac{{\sqrt 3 }}{2}} \right)}\\ &= - \frac{2}{\sqrt 3 }\end{align}\]

Now,

\[\begin{align}\tan x &= \frac{\sin x}{\cos x}\\ &= \frac{{\left( { - \frac{1}{2}} \right)}}{{\left( { - \frac{{\sqrt 3 }}{2}} \right)}}\\& = \sqrt 3 \end{align}\]

Now,

\[\begin{align}\cot x &= \frac{1}{{\tan x}}\\& = \frac{1}{{\sqrt 3 }}\end{align}\]

Hence, \(\sin x = - \frac{{\sqrt 3 }}{2}\), \({\rm{cosec}}\;x = - \frac{2}{{\sqrt 3 }}\), \(\sec x = - 2\) , \(\tan x = \sqrt 3 \), and \(\cot x = \frac{1}{{\sqrt 3 }}\)

Chapter 3 Ex.3.2 Question 2

Find the values of trigonometric function \(\sin x = \frac{3}{5}\), \(x\) lies in second quadrant.

Solution

As we know that

\[\begin{align}{\rm{cosec}}\;x &= \frac{1}{\sin x}\\ &= \frac{1}{{\left( {\frac{3}{5}} \right)}}\\ &= \frac{5}{3}\end{align}\]

Now, \({\sin ^2}x + {\cos ^2}x = 1\)

\[\begin{align}{\cos ^2}x &= 1 - {\sin ^2}x\\\cos x &= \pm \sqrt {1 - {{\sin }^2}x} \\ &= \pm \sqrt {1 - {{\left( {\frac{3}{5}} \right)}^2}} \\ &= \pm \sqrt {1 - \frac{9}{{25}}} \\ &= \pm \sqrt {\frac{{16}}{{25}}} \\ &= \pm \frac{4}{5}\end{align}\]

Since, \(x\) lies in second quadrant, the value of \(\cos x\) will be negative.

Therefore,

\[\cos x = - \frac{4}{5}\]

Now,

\[\begin{align}\sec x &= \frac{1}{{\cos x}}\\ &= \frac{1}{{\left( { - \frac{4}{5}} \right)}}\\ &= - \frac{5}{4}\end{align}\]

Now,

\[\begin{align}\tan x &= \frac{{\sin x}}{{\cos x}}\\ &= \frac{{\left( {\frac{3}{5}} \right)}}{{\left( { - \frac{4}{5}} \right)}}\\ &= - \frac{3}{4}\end{align}\]

Now,

\[\begin{align}\cot x &= \frac{1}{\tan x}\\ &= \frac{1}{\left( { - \frac{3}{4}} \right)}\\ &= - \frac{4}{3}\end{align}\]

Hence, \(\rm{cosec}\;x = \frac{5}{3}\), \(\cos x = - \frac{4}{5}\), \(\sec x = - \frac{5}{4}\), \(\tan x = - \frac{3}{4}\), and \(\cot x = - \frac{4}{3}\)

Chapter 3 Ex.3.2 Question 3

Find the values of trigonometric function \(\cot x = \frac{3}{4}\), \(x\) lies in third quadrant.

Solution

As we know that

\[\begin{align}\tan x &= \frac{1}{{\cot x}}\\ &= \frac{1}{{\left( {\frac{3}{4}} \right)}}\\ &= \frac{4}{3}\end{align}\]

Now, \(1 + {\tan ^2}x = {\sec ^2}x\)

\[\begin{align}\sec x &= \pm \sqrt {1 + {{\tan }^2}x} \\ &= \pm \sqrt {1 + {{\left( {\frac{4}{3}} \right)}^2}} \\ &= \pm \sqrt {1 + \frac{{16}}{9}} \\ &= \pm \sqrt {\frac{{25}}{9}} \\ &= \pm \frac{5}{3}\end{align}\]

Since, \(x\) lies in third quadrant, the value of \(\sec x\) will be negative.

Therefore,

\[\sec x = - \frac{5}{3}\]

Now,

\[\begin{align}\cos x &= \frac{1}{\sec x}\\ &= \frac{1}{\left( { - \frac{5}{3}} \right)}\\& = - \frac{3}{5}\end{align}\]

Now,

\[\begin{align}\tan x &= \frac{{\sin x}}{{\cos x}}\\\sin x &= \tan x\cos x\\& = \left( {\frac{4}{3}} \right) \times \left( { - \frac{3}{5}} \right)\\ &= - \frac{4}{5}\end{align}\]

Now,

\[\begin{align}{\rm{cosec}}\;x &= \frac{1}{\sin x}\\ &= \frac{1}{\left( { - \frac{4}{5}} \right)}\\ &= - \frac{5}{4}\end{align}\]

Hence, \(\sin x = - \frac{4}{5}\), \({\rm{cosec}}\;x = - \frac{5}{4}\) , \(\cos x=-\frac{3}{5}\), \(\sec x=-\frac{5}{3}\), and \(\tan x=\frac{4}{3}\)

 

Chapter 3 Ex.3.2 Question 4

Find the values of trigonometric function \(\sec x = \frac{{13}}{5}\), \(x\) lies in fourth quadrant.

Solution

As we know that

\[\begin{align}\cos x &= \frac{1}{{\sec x}}\\ &= \frac{1}{{\left( {\frac{{13}}{5}} \right)}}\\ &= \frac{5}{{13}}\end{align}\]

Now, \({\sin ^2}x + {\cos ^2}x = 1\)

\[\begin{align}{\sin ^2}x &= 1 - {\cos ^2}x\\\sin x &= \pm \sqrt {1 - {{\cos }^2}x} \\ &= \pm \sqrt {1 - {{\left( {\frac{5}{13}} \right)}^2}} \\& = \pm \sqrt {1 - \frac{25}{169}} \\ &= \pm \sqrt {\frac{144}{169}} \\& = \pm \frac{12}{13}\end{align}\]

Since, \(x\) lies in fourth quadrant, the value of \(\sin x\) will be negative.

Therefore,

\[\sin x=-\frac{12}{13}\]

Now,

\[\begin{align}{\rm{cosec}}\;x &= \frac{1}{\sin x}\\ &= \frac{1}{\left( { - \frac{12}{13}} \right)}\\ &= - \frac{13}{12}\end{align}\]

Now,

\[\begin{align}\tan x &= \frac{\sin x}{\cos x}\\ &= \frac{\left( { - \frac{12}{13}} \right)}{\left( {\frac{5}{{13}}} \right)}\\ &= - \frac{12}{5}\end{align}\]

Now,

\[\begin{align}\cot x &= \frac{1}{\tan x}\\ &= \frac{1}{{\left( { - \frac{12}{5}} \right)}}\\ &= - \frac{5}{{12}} \end{align}\]

Hence, \(\sin x = - \frac{12}{13}\), \({\rm{cosec}}\;x = - \frac{13}{12}\), \(\cos x = \frac{5}{{13}}\), \(\tan x = - \frac{{12}}{5}\), and \(\cot x = - \frac{5}{12}\)

Chapter 3 Ex.3.2 Question 5

Find the values of trigonometric function \(\tan x = - \frac{5}{{12}}\), \(x\) lies in second quadrant.

Solution

As we know that

\[\begin{align}\cot x &= \frac{1}{{\tan x}}\\&= \frac{1}{{\left( { - \frac{5}{{12}}} \right)}}\\ &= - \frac{{12}}{5}\end{align}\]

Now, \(1 + {\tan ^2}x = {\sec ^2}x\)

\[\begin{align}\sec x &= \pm \sqrt {1 + {{\tan }^2}x} \\ &= \pm \sqrt {1 + {{\left( { - \frac{5}{{12}}} \right)}^2}} \\ &= \pm \sqrt {1 + \frac{25}{144}} \\ &= \pm \sqrt {\frac{{169}}{144}} \\ &= \pm \frac{13}{12}\end{align}\]

Since, \(x\) lies in second quadrant, the value of \(\sec x\) will be negative.

Therefore,

\[\sec x = - \frac{13}{12}\]

Now,

\[\begin{align}\cos x &= \frac{1}{\sec x}\\ &= \frac{1}{\left( { - \frac{13}{12}} \right)}\\ &= - \frac{12}{13}\end{align}\]

Now,

\[\begin{align}\tan x &= \frac{\sin x}{\cos x}\\\sin x &= \tan x\cos x\\ &= \left( { - \frac{5}{12}} \right) \times \left( { - \frac{12}{13}} \right)\\ &= \frac{5}{13}\end{align}\]

Now,

\[\begin{align}{\rm{cosec}}\;x &= \frac{1}{{\sin x}}\\ &= \frac{1}{{\left( {\frac{5}{{13}}} \right)}}\\& = \frac{{13}}{5}\end{align}\]

Hence, \(\sin x = \frac{5}{{13}}\), \({\rm{cosec}}\;x = \frac{{13}}{5}\), \(\cos x = - \frac{{12}}{{13}}\), \(\sec x = - \frac{{13}}{{12}}\), and \(\cot x = - \frac{{12}}{5}\)

Chapter 3 Ex.3.2 Question 6

Find the values of the trigonometric function \(\sin 765^\circ \)

Solution

It is known that the value of \(\sin x\) repeat after an interval of \(2n\) or \(360^\circ\).

Therefore,

\[\begin{align}\sin 765^\circ &= \sin \left( {2 \times 360^\circ + 45^\circ } \right)\\ &= \sin 45^\circ \\ &= \frac{1}{{\sqrt 2 }}\end{align}\]

Chapter 3 Ex.3.2 Question 7

Find the values of the trigonometric function \({\rm{cosec}}\left( { - 1410^\circ } \right)\)

Solution

It is known that the value of \(\sin x\) repeat after an interval of \(2n\) or \(360^\circ\).

Therefore,

\[\begin{align}{\rm{cosec}}\left( { - 1410^\circ } \right) &= {\rm{cosec}}\left( {4 \times 360^\circ - 1410^\circ } \right)\\& = {\rm{cosec}}\left( {1440^\circ - 1410^\circ } \right)\\& = {\rm{cosec}}30^\circ \\ &= 2\end{align}\]

Chapter 3 Ex.3.2 Question 8

Find the values of the trigonometric functions in \(\tan \frac{{19\pi }}{3}\)

Solution

It is known that the value of \(\tan x\) repeat after an interval of \(n\) or \(180^{\circ}\).

Therefore,

\[\begin{align}\tan \frac{{19\pi }}{3} &= \tan 6\frac{1}{3}\pi \\ &= \tan \left( {6\pi + \frac{1}{3}\pi } \right)\\ &= \tan \frac{\pi }{3}\\& = \tan 60^\circ \\& = \sqrt 3 \end{align}\]

Chapter 3 Ex.3.2 Question 9

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

Solution

It is known that the value of \(\sin x\) repeat after an interval of \(2n\) or 360°.

Therefore,

\[\begin{align}\sin \left( { - \frac{11\pi }{3}} \right) &= \sin \left( {2 \times 2\pi - \frac{{11\pi }}{3}} \right)\\ &= \sin \frac{\pi }{3}\\ &= \sin 60^\circ \\ &= \frac{{\sqrt 3 }}{2}\end{align}\]

Chapter 3 Ex.3.2 Question 10

Find the values of the trigonometric functions in \(\cot \left( { - \frac{15\pi }{4}} \right)\)

Solution

It is known that the value of \(\cos x\) repeat after an interval of \(n\) or \(180^{\circ}.\)

Therefore,

\[\begin{align}\cot \left( { - \frac{15\pi }{4}} \right) &= \cot \left( {4\pi - \frac{15\pi}{4}} \right)\\ &= \cot \frac{\pi }{4}\\ &= \cot 45^\circ \\ &= 1\end{align}\]