Question

The value of $\sin \left[ {{{\tan }^{ - 1}}\dfrac{{1 - {x^2}}}{{2x}} + {{\cos }^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right]$ is
1. $1$
2. $0$
3. $ - 1$
4. $\dfrac{\pi }{2}$

Answer 1

Hint: Here, we are given the function $\sin \left[ {{{\tan }^{ - 1}}\dfrac{{1 - {x^2}}}{{2x}} + {{\cos }^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right]$ and we have to find the value of it. First step is to let $x = \tan \theta $ because we already have the formulas same as the given angle in the function. So, to make them similar we take the value of $x$. Now we’ll put the values and solve the inverse trigonometric function and convert $\cot $ in terms of $\tan $ using trigonometric sign convention.

Formula Used:
Trigonometric formula –
$\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}$
$\cos 2\theta = \dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}$

Complete step by step Solution:
Given that,
$\sin \left[ {{{\tan }^{ - 1}}\dfrac{{1 - {x^2}}}{{2x}} + {{\cos }^{ - 1}}\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right] - - - - - \left( 1 \right)$
Let, $x = \tan \theta - - - - - \left( 2 \right)$
\[ \Rightarrow \theta = {\tan ^{ - 1}}x\]
From equation (1) and (2)
$ = \sin \left[ {{{\tan }^{ - 1}}\dfrac{{1 - {{\left( {\tan \theta } \right)}^2}}}{{2\left( {\tan \theta } \right)}} + {{\cos }^{ - 1}}\dfrac{{1 - {{\left( {\tan \theta } \right)}^2}}}{{1 + {{\left( {\tan \theta } \right)}^2}}}} \right]$
Now, using trigonometric formulas as the angle of inverse functions are similar to the formula $\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}$ and $\cos 2\theta = \dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}$
$ = \sin \left[ {{{\tan }^{ - 1}}\left( {\cot 2\theta } \right) + {{\cos }^{ - 1}}\left( {\cos 2\theta } \right)} \right]$
Covert the function using sign convention formulas,
$ = \sin \left[ {{{\tan }^{ - 1}}\left( {\tan \left( {\dfrac{\pi }{2} - 2\theta } \right)} \right) + 2\theta } \right]$
$ = \sin \left[ {\dfrac{\pi }{2} - 2\theta + 2\theta } \right]$
Using trigonometry table, write the value of $\sin \left[ {\dfrac{\pi }{2}} \right]$
$ = \sin \left[ {\dfrac{\pi }{2}} \right]$
$ = 1$

Hence, the correct option is 1.

Note: The key concept involved in solving this problem is a good knowledge of Trigonometric formulas and identities. Students must know that to solve such questions always take the variable as the trigonometric function to make it similar to the formulas. In such a question, where the inverse is there. When you will shift the inverse trigonometric function to the other side its angle will be equal to the trigonometric function with the angle of the other side term and vice versa. For example, Let the function be $\sin \theta = x$ . Now, if you’ll shift the trigonometric function equation will be $x = {\sin ^{ - 1}}\theta $ and again you will repeat the procedure. It will be the same. Also, when the same inverse and trigonometric functions are there they get canceled.