NCERT Class 12 Maths Inverse Trigonometric Functions

NCERT Class 12 Maths Inverse Trigonometric Functions

Chapter 2 Ex.2.1 Question 1

Find the principal value of \({\sin ^{ - 1}}\left( { - \frac{1}{2}} \right)\).

Solution

Let, \({\sin ^{ - 1}}\left( { - \frac{1}{2}} \right) = y\)

Hence,

\[\begin{align}\sin y &= \left( { - \frac{1}{2}} \right)\\&= - \sin \left( {\frac{\pi }{6}} \right)\\&= \sin \left( { - \frac{\pi }{6}} \right)\end{align}\]

Range of the principal value of \({\sin ^{ - 1}}\left( x \right)\) is \(\left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)\)

Thus, principal value of \({\sin ^{ - 1}}\left( { - \frac{1}{2}} \right) = \left( { - \frac{\pi }{6}} \right)\).

Chapter 2 Ex.2.1 Question 2

Find the principal value of \({\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right)\).

Solution

Let, \({\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) = y\)

Hence,

\[\begin{align}\cos y &= \left( {\frac{{\sqrt 3 }}{2}} \right)\\&= \cos \frac{\pi }{6}\end{align}\]

Range of the principal value of \({\cos ^{ - 1}}\left( x \right)\) is \(\left( {0,\pi } \right)\).

Thus, principal value of \({\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) = \left( {\frac{\pi }{6}} \right)\)

Chapter 2 Ex.2.1 Question 3

Find the principal value of \(\ {\rm{cosec}}{^{ - 1}}\left( 2 \right)\).

Solution

Let, \(\ {\rm{cosec}}^{ - 1}\left( 2 \right) = y\)

Hence,

\[\begin{align}\ {\rm{cosec}}\,y &= 2\\ &= {\rm{cosec}}\left( {\frac{\pi }{6}} \right)\end{align}\]

Range of the principal value of \(\ {\rm{cosec}}{^{ - 1}}\left( x \right) = \left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right] - \left\{ 0 \right\}\)

Thus, principal value of \(\ {\rm{cosec}}{^{ - 1}}\left( 2 \right) = \left( {\frac{\pi }{6}} \right)\).

Chapter 2 Ex.2.1 Question 4

Find the principal value of \({\tan ^{ - 1}}\left( { - \sqrt 3 } \right)\)

Solution

Let, \({\tan ^{ - 1}}\left( { - \sqrt 3 } \right) = y\)

Hence,

\[\begin{align}\tan y &= - \sqrt 3 \\&= - \tan \left( {\frac{\pi }{3}} \right)\\&= \tan \left( { - \frac{\pi }{3}} \right)\end{align}\]

Range of the principal value of \({\tan ^{ - 1}}\left( x \right) = \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)\)

Thus, principal value of \({\tan ^{ - 1}}\left( { - \sqrt 3 } \right) = \left( { - \frac{\pi }{3}} \right)\).

Chapter 2 Ex.2.1 Question 5

Find the principal value of \({\cos ^{ - 1}}\left( { - \frac{1}{2}} \right)\)

Solution

Let, \({\cos ^{ - 1}}\left( { - \frac{1}{2}} \right) = y\)

Hence,

\[\begin{align}\cos y &= - \frac{1}{2}\\&= - \cos \left( {\frac{\pi }{3}} \right)\\&= \cos \left( {\pi - \frac{\pi }{3}} \right)\\&= \cos \left( {\frac{{2\pi }}{3}} \right)\end{align}\]

Range of the principal value of \({\cos ^{ - 1}}\left( x \right) = \left[ {0,\pi } \right]\)

Thus, principal value of \({\cos ^{ - 1}}\left( { - \frac{1}{2}} \right) = \left( {\frac{{2\pi }}{3}} \right)\).

Chapter 2 Ex.2.1 Question 6

Find the principal value of \({\tan ^{ - 1}}\left( { - 1} \right)\)

Solution

Let, \({\tan ^{ - 1}}\left( { - 1} \right) = y\)

Hence,

\[\begin{align}\tan {\rm{ }}y &= - 1\\&= - \tan \left( {\frac{\pi }{4}} \right)\\&= \tan \left( { - \frac{\pi }{4}} \right)\end{align}\]

Range of the principal value of \({\tan ^{ - 1}}\left( x \right) = \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)\)

Thus, principal value of \({\tan ^{ - 1}}\left( { - 1} \right) = \left( { - \frac{\pi }{4}} \right)\).

Chapter 2 Ex.2.1 Question 7

Find the principal value of \({\sec ^{ - 1}}\left( {\frac{2}{{\sqrt 3 }}} \right)\)

Solution

Let, \({\sec ^{ - 1}}\left( {\frac{2}{{\sqrt 3 }}} \right)=y \)

Hence,

\[\begin{align} \sec \;y &= \frac{2}{{\sqrt 3 }} \\ &= \sec \left( {\frac{\pi }{6}} \right) \end{align} \]

Range of the principal value of \({\sec ^{ - 1}}\left( x \right) = \left[ {0,\pi } \right] - \left\{ {\frac{\pi }{2}} \right\}\)

Thus, principal value of \({\sec ^{ - 1}}\left( {\frac{2}{{\sqrt 3 }}} \right) = \left( {\frac{\pi }{6}} \right)\).

Chapter 2 Ex.2.1 Question 8

Find the principal value of \({\cot ^{ - 1}}\left( {\sqrt 3 } \right)\)

Solution

Let, \({\cot ^{ - 1}}\left( {\sqrt 3 } \right) = y\)

Hence,

\[\begin{align}\cot \;y &= \sqrt 3 \\&= \cot \left( {\frac{\pi }{6}} \right)\end{align}\]

Range of the principal value of \({\cot ^{ - 1}}\left( x \right) = \left( {0,\pi } \right)\)

Thus, principal value of \({\cot ^{ - 1}}\left( {\sqrt 3 } \right) = \left( {\frac{\pi }{6}} \right)\).

Chapter 2 Ex.2.1 Question 9

Find the principal value of \({\cos ^{ - 1}}\left( { - \frac{1}{{\sqrt 2 }}} \right)\)

Solution

Let, \({\cos ^{ - 1}}\left( { - \frac{1}{{\sqrt 2 }}} \right) = y\)

Hence,

\[\begin{align}\cos \;y &= - \frac{1}{{\sqrt 2 }}\\&= - \cos \left( {\frac{\pi }{4}} \right)\\&= \cos \left( { - \frac{\pi }{4}} \right)\\&= \cos \left( {\pi - \frac{\pi }{4}} \right)\\&= \cos \left( {\frac{{3\pi }}{4}} \right)\end{align}\]

Range of the principal value of \({\cos ^{ - 1}}\left( x \right) = \left[ {0,\pi } \right]\)

Thus, principal value of \({\cos ^{ - 1}}\left( { - \frac{1}{{\sqrt 2 }}} \right) = \left( {\frac{{3\pi }}{4}} \right)\).

Chapter 2 Ex.2.1 Question 10

Find the principal value of \({\rm{cose}}{{\rm{c}}^{ - 1}}\left( { - \sqrt 2 } \right)\)

Solution

Let, \({\rm{cose}}{{\rm{c}}^{ - 1}}\left( { - \sqrt 2 } \right) = y\)

Hence,

\[\begin{align}{\rm{cosec }}\;y &= - \sqrt 2 \\&= - {\rm{cosec}}\left( {\frac{\pi }{4}} \right)\\&= {\rm{cosec}}\left( { - \frac{\pi }{4}} \right)\end{align}\]

Range of the principal value of \({\rm{cose}}{{\rm{c}}^{ - 1}}\left( x \right) = \left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right] - \left\{ 0 \right\}\)

Thus, principal value of \({\rm{cose}}{{\rm{c}}^{ - 1}}\left( { - \sqrt 2 } \right) = \left( { - \frac{\pi }{4}} \right)\).

Chapter 2 Ex.2.1 Question 11

Find the value of \({\tan ^{ - 1}}\left( 1 \right) + {\cos ^{ - 1}}\left( { - \frac{1}{2}} \right) + {\sin ^{ - 1}}\left( { - \frac{1}{2}} \right)\).

Solution

Let, \({\tan ^{ - 1}}\left( 1 \right) = x\)

Hence,

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

Therefore,

\({\tan ^{ - 1}}\left( 1 \right) = \left( {\frac{\pi }{4}} \right)\)

Now, let \({\cos ^{ - 1}}\left( { - \frac{1}{2}} \right) = y\)

Hence,

\[\begin{align}\cos y &= - \frac{1}{2}\\&= - \cos \left( {\frac{\pi }{3}} \right)\\&= \cos \left( {\pi - \frac{\pi }{3}} \right)\\&= \cos \left( {\frac{{2\pi }}{3}} \right)\end{align}\]

Therefore,

\({\cos ^{ - 1}}\left( { - \frac{1}{2}} \right) = \frac{{2\pi }}{3}\)

Again, let \({\sin ^{ - 1}}\left( { - \frac{1}{2}} \right) = z\)

Hence,

\[\begin{align}\sin z &= - \frac{1}{2}\\&= - \sin \left( {\frac{\pi }{6}} \right)\\&= \sin \left( { - \frac{\pi }{6}} \right)\end{align}\]

Therefore,

\({\sin ^{ - 1}}\left( { - \frac{1}{2}} \right) = - \frac{\pi }{6}\)

Thus,

\[\begin{align}{\tan ^{ - 1}}\left( 1 \right) + {\cos ^{ - 1}}\left( { - \frac{1}{2}} \right) + {\sin ^{ - 1}}\left( { - \frac{1}{2}} \right) &= \frac{\pi }{4} + \frac{{2\pi }}{3} - \frac{\pi }{6}\\&= \frac{{3\pi + 8\pi - 2\pi }}{{12}}\\&= \frac{{9\pi }}{{12}}\\&= \frac{{3\pi }}{4}\end{align}\]

Chapter 2 Ex.2.1 Question 12

Find the value of \({\cos ^{ - 1}}\left( {\frac{1}{2}} \right) + 2{\sin ^{ - 1}}\left( {\frac{1}{2}} \right)\)

Solution

Let, \({\tan ^{ - 1}}\left( 1 \right) = x\)

Hence,

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

Therefore,

\({\cos ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{3}\)

Let, \({\sin ^{ - 1}}\left( {\frac{1}{2}} \right) = y\)

Hence,

\[\begin{align}\sin y &= \frac{1}{2}\\&= \sin \left( {\frac{\pi }{6}} \right)\end{align}\]

Therefore,

\({\sin ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{6}\)

Thus

\[\begin{align}{\cos ^{ - 1}}\left( {\frac{1}{2}} \right) + 2{\sin ^{ - 1}}\left( {\frac{1}{2}} \right) &= \frac{\pi }{3} + 2\left( {\frac{\pi }{6}} \right)\\&= \frac{{2\pi }}{3}\end{align}\]

Chapter 2 Ex.2.1 Question 13

Find the value of \({\sin ^{ - 1}}x = y\), then

(A) \(0 \le y \le \pi \)

(B) \( - \frac{\pi }{2} \le y \le \frac{\pi }{2}\)

(C) \(0 \le y \le \pi \)

(D) \( - \frac{\pi }{2} < y < \frac{\pi }{2}\)

Solution

It is given that \({\sin ^{ - 1}}x = y\)

Range of the principal value of \({\sin ^{ - 1}}x = \left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]\)

Thus, \( - \frac{\pi }{2} \le y \le \frac{\pi }{2}\)

The answer is B.

Chapter 2 Ex.2.1 Question 14

Find the value of \({\tan ^{ - 1}}\sqrt 3 - {\sec ^{ - 1}}\left( { - 2} \right)\)is equal to

(A) \(0\)

(B) \( - \frac{\pi }{3}\)

(C) \(\frac{\pi }{3}\)

(D) \(\frac{{2\pi }}{3}\)

Solution

Let

Hence,

\[\begin{align}\tan x &= \sqrt 3 \\&= \tan \left( {\frac{\pi }{3}} \right)\end{align}\]

Range of the principal value of \({\tan ^{ - 1}}x = \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)\)

Therefore,\({\tan ^{ - 1}}\left( {\sqrt 3 } \right) = \left( {\frac{\pi }{3}} \right)\)

Let \({\sec ^{ - 1}}\left( { - 2} \right) = y\)

Hence,

\[\begin{align}\sec y &= \left( { - 2} \right)\\&= - \sec \left( {\frac{\pi }{3}} \right)\\&= \sec \left( { - \frac{\pi }{3}} \right)\\&= \sec \left( {\pi - \frac{\pi }{3}} \right)\\&= \sec \left( {\frac{{2\pi }}{3}} \right)\end{align}\]

Range of the principal value of \({\rm{se}}{{\rm{c}}^{ - 1}}{\rm{x = }}\left[ {0,\pi } \right] - \left\{ {\frac{\pi }{2}} \right\}\)

Therefore, \({\sec ^{ - 1}}\left( { - 2} \right) = \frac{{2\pi }}{3}\)

Thus,

\[\begin{align}{\tan ^{ - 1}}\sqrt 3 - {\sec ^{ - 1}}\left( { - 2} \right)& = \frac{\pi }{3} - \frac{{2\pi }}{3}\\&= - \frac{\pi }{3}\end{align}\]

The answer is \(B\).

Inverse Trigonometric Functions | NCERT Solutions
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