Chapter 2: Inverse Trigonometric Functions

Ans: Let’s assume when  ${\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = y$, Then $\sin y = \left( { - \dfrac{1}{2}} \right) =  - \sin \left( {\dfrac{\pi }{6}} \right) = \sin \left( { - \dfrac{\pi }{6}} \right)$.

As we know that the range of the principal value branch of ${\sin ^{ - 1}}$ is $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$ and $\sin \left( { - \dfrac{\pi }{6}} \right) = \dfrac{1}{2}$

Hence, the principal value of ${\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$ is $ - \dfrac{\pi }{6}$.

Ans: Let’s consider, ${\cos ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right) = y.$ Then $\cos y = \dfrac{{\sqrt 3 }}{2} = \cos \left( {\dfrac{\pi }{6}} \right)$

As we know that the range of the principal value branch of ${\cos ^{ - 1}}$ is $[0,\pi ]$ and $\cos \left( {\dfrac{\pi }{6}} \right) = \dfrac{{\sqrt 3 }}{2}$

Therefore, the principal value of ${\cos ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)$ is $\dfrac{\pi }{6}$.

Ans: Let’s consider,${\operatorname{cosec} ^{ - 1}}(2) = y$.Then, $\operatorname{cosec} {\text{y}} = 2 = \operatorname{cosec} \left( {\dfrac{\pi }{6}} \right)$.

As we know that the range of the principal value branch of ${\operatorname{cosec} ^{ - 1}}$ is$\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] - \{ 0\} $. 

Therefore, the principal value of ${\operatorname{cosec} ^{ - 1}}(2)$ is $\dfrac{\pi }{6}$.

Ans: Let’s consider ${\tan ^{ - 1}}( - \sqrt 3 ) = y$ Then, $\tan y =  - \sqrt 3  =  - \tan \dfrac{\pi }{3} = \tan \left( { - \dfrac{\pi }{3}} \right)$.

As we know that the range of the principal value branch of ${\tan ^{ - 1}}$ is $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$ and$\tan \left( { - \dfrac{\pi }{3}} \right)$ is $ - \sqrt 3 $

Hence, the principal value of ${\tan ^{ - 1}}( - \sqrt 3 )$ is $ - \dfrac{\pi }{3}$.

Ans: Let’s consider, ${\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = y.$

Then, $\cos y =  - \dfrac{1}{2} =  - \cos \left( {\dfrac{\pi }{3}} \right) = \cos \left( {\pi  - \dfrac{\pi }{3}} \right) = \cos \left( {\dfrac{{2\pi }}{3}} \right)$.

As we know that the range of the principal value branch of ${\cos ^{ - 1}}$ is $[0,\pi ]$ and $\cos \left( {\dfrac{{2\pi }}{3}} \right) =  - \dfrac{1}{2}$

Therefore, the principal value of ${\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$ is $\left( {\dfrac{{2\pi }}{3}} \right)$.

Ans: Let’s assume that ${\tan ^{ - 1}}( - 1) = {\text{y}}$.

Then, $\tan y =  - 1 =  - \tan \left( {\dfrac{\pi }{4}} \right) = \tan \left( { - \dfrac{\pi }{4}} \right)$.

As we know that the range of the principal value branch of ${\tan ^{ - 1}}$ is $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$ and$\tan \left( { - \dfrac{\pi }{4}} \right) =  - 1$

Therefore, the principal value of ${\tan ^{ - 1}}( - 1)$ is $ - \dfrac{\pi }{4}$.

Ans: Let’s consider ${\sec ^{ - 1}}\left( {\dfrac{2}{{\sqrt 3 }}} \right) = y$. 

Then, $\sec y = \dfrac{2}{{\sqrt 3 }} = \sec \left( {\dfrac{\pi }{6}} \right)$.

As we know that the range of the principal value branch of ${\sec ^{ - 1}}$ is $[0,\pi ] - \left\{ {\dfrac{\pi }{2}} \right\}$ and $\sec \left( {\dfrac{\pi }{6}} \right) = \dfrac{2}{{\sqrt 3 }}$

Ans: Let’s consider ${\cot ^{ - 1}}(\sqrt 3 ) = y$. Then $\cot y = \sqrt 3  = \cot \left( {\dfrac{\pi }{6}} \right)$.

As we know that the range of the principal value branch of ${\cot ^{ - 1}}$ is $(0,\pi )$ and $\cot \left( {\dfrac{\pi }{6}} \right) = \sqrt 3 $.

Hence, the principal value of ${\cot ^{ - 1}}(\sqrt 3 )$ is $\dfrac{\pi }{6}$.

Ans: Let’s ${\cos ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right) = y$.

Then $\cos y =  - \dfrac{1}{{\sqrt 2 }} =  - \cos \left( {\dfrac{\pi }{4}} \right) = \cos \left( {\pi  - \dfrac{\pi }{4}} \right) = \cos \left( {\dfrac{{3\pi }}{4}} \right)$.

As we know that the range of the principal value branch of ${\cos ^{ - 1}}$ is $[0,\pi ]$ and $\cos \left( {\dfrac{{3\pi }}{4}} \right) =  - \dfrac{1}{{\sqrt 2 }}$.

Therefore, the principal value of ${\cos ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right)$ is $\dfrac{{3\pi }}{4}$.

Ans: Let’s consider, ${\operatorname{cosec} ^{ - 1}}( - \sqrt 2 ) = y$. Then, $\cos ecy =  - \sqrt 2  =  - \cos ec\left( {\dfrac{\pi }{4}} \right) = \cos ec\left( { - \dfrac{\pi }{4}} \right)$

As we know that the range of the principal value branch of ${\operatorname{cosec} ^{ - 1}}$ is $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] - \{ 0\} $ and $\operatorname{cosec} \left( { - \dfrac{\pi }{4}} \right) =  - \sqrt 2 $

Therefore, the principal value of ${\operatorname{cosec} ^{ - 1}}( - \sqrt 2 )$ is $ - \dfrac{\pi }{4}$.