Question

Solve the given inverse trigonometric function ${\tan ^{ - 1}}1 + {\cos ^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right) + {\sin ^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right)$

Answer 1

Hint:- We have to use properties of inverse trigonometric functions ${\cos ^{ - 1}}\left( {\cos x} \right) = x$, ${\tan ^{ - 1}}\left( {\tan x} \right) = x$, ${\sin ^{ - 1}}\left( {\sin x} \right) = x$ to solve the given problem.

Complete step by step answer:
We need to evaluate
${\tan ^{ - 1}}1 + {\cos ^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right) + {\sin ^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right)$
We can write it as
${\tan ^{ - 1}}\left( {\tan \dfrac{\pi }{4}} \right) + {\cos ^{ - 1}}\left( {\cos \dfrac{{2\pi }}{3}} \right) + {\sin ^{ - 1}}\left( {\sin \left( {\dfrac{{ - \pi }}{6}} \right)} \right)$$ \ldots \ldots \left( 1 \right)$
$
  \because \tan \dfrac{\pi }{4} = 1 \\
  \cos \dfrac{{2\pi }}{3} = \dfrac{{ - 1}}{2} \\
  \sin \left( {\dfrac{{ - \pi }}{6}} \right) = - \sin \left( {\dfrac{\pi }{6}} \right) = \dfrac{{ - 1}}{2} \\
$
${\tan ^{ - 1}}\left( {\tan \dfrac{\pi }{4}} \right) = \dfrac{\pi }{4}$ Because we know ${\tan ^{ - 1}}\left( {\tan x} \right) = x$
For $\dfrac{{ - \pi }}{2} < x < \dfrac{\pi }{2}$
${\cos ^{ - 1}}\left( {\cos \dfrac{{2\pi }}{3}} \right) = \dfrac{{2\pi }}{3}$ Because we know ${\cos ^{ - 1}}\left( {\cos x} \right) = x$
For $0 \leqslant x \leqslant \pi $
${\sin ^{ - 1}}\left( {\sin \dfrac{{ - \pi }}{6}} \right) = \dfrac{{ - \pi }}{6}$ Because we know ${\sin ^{ - 1}}\left( {\sin x} \right) = x$
For $\dfrac{{ - \pi }}{2} \leqslant x \leqslant \dfrac{\pi }{2}$
Now equation $\left( 1 \right)$becomes
$\dfrac{\pi }{4} + \dfrac{{2\pi }}{3} - \dfrac{\pi }{6}$=$\dfrac{{3\pi }}{4}$
Required answer is $\dfrac{{3\pi }}{4}$

Note: - Whenever we get these types of questions the key concept of solving these types of questions is we should have knowledge of changing inverse trigonometric functions according to requirement and remember the domain and range of these inverse trigonometric functions.