Question

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

Answer 1

Hint: Generally, in Mathematics, the trigonometric Identities are useful whenever trigonometric functions are involved in an expression or an equation and these identities are useful whenever expressions involving trigonometric functions need to be simplified.
Since from given we are asked to find the value of $sin\left[ {{{\tan }^{ - 1}}\dfrac{{\left[ {1 - {x^2}} \right]}}{{2x}} + {{\cos }^{ - 1}}\dfrac{{\left[ {1 - {x^2}} \right]}}{{\left[ {1 + {x^2}} \right]}}} \right]$.
We need to apply the trigonometric identities to obtain the required answer.
Formula to be used:
The trigonometric identities that are used to solve the given problem are as follows.
a) \[cot2\theta = \dfrac{{co{t^2}\theta - 1}}{{2cot\theta }}\]
b)\[cos2\theta = \dfrac{{1 - ta{n^2}\theta }}{{1 + ta{n^2}\theta }}\;\]
c) $\tan \theta = \dfrac{1}{{\cot \theta }}$
d) $\cot \theta = \tan \left( {\dfrac{\pi }{2} - \theta } \right)$
e) ${\cos ^{ - 1}}\cos \theta = \theta $
f)${\tan ^{ - 1}}\tan \theta = \theta $

Complete step by step answer:
Since from the given that, we are asked to calculate the expression on the trigonometric functions $sin\left[ {{{\tan }^{ - 1}}\dfrac{{\left[ {1 - {x^2}} \right]}}{{2x}} + {{\cos }^{ - 1}}\dfrac{{\left[ {1 - {x^2}} \right]}}{{\left[ {1 + {x^2}} \right]}}} \right]$
Let us put $x = \tan \theta $ in the given expression.
Thus, we get
$sin\left[ {{{\tan }^{ - 1}}\dfrac{{\left[ {1 - {x^2}} \right]}}{{2x}} + {{\cos }^{ - 1}}\dfrac{{\left[ {1 - {x^2}} \right]}}{{\left[ {1 + {x^2}} \right]}}} \right] = sin\left[ {{{\tan }^{ - 1}}\dfrac{{\left[ {1 - {{\tan }^2}\theta } \right]}}{{2\tan \theta }} + {{\cos }^{ - 1}}\dfrac{{\left[ {1 - {{\tan }^2}\theta } \right]}}{{\left[ {1 + {{\tan }^2}\theta } \right]}}} \right]$
Now, we shall apply$\tan \theta = \dfrac{1}{{\cot \theta }}$ in the above equation.
 \[ \Rightarrow sin\left[ {{{\tan }^{ - 1}}\dfrac{{\left[ {1 - {x^2}} \right]}}{{2x}} + {{\cos }^{ - 1}}\dfrac{{\left[ {1 - {x^2}} \right]}}{{\left[ {1 + {x^2}} \right]}}} \right] = sin\left[ {{{\tan }^{ - 1}}\dfrac{{\left[ {1 - \dfrac{1}{{{{\cot }^2}\theta }}} \right]}}{{2\dfrac{1}{{\cot \theta }}}} + {{\cos }^{ - 1}}\dfrac{{\left[ {1 - {{\tan }^2}\theta } \right]}}{{\left[ {1 + {{\tan }^2}\theta } \right]}}} \right]\]
\[ = sin\left[ {{{\tan }^{ - 1}}\left[ {\dfrac{{{{\cot }^2}\theta - 1}}{{{{\cot }^2}\theta }}} \right] \times \dfrac{{\cot \theta }}{2} + {{\cos }^{ - 1}}\dfrac{{\left[ {1 - {{\tan }^2}\theta } \right]}}{{\left[ {1 + {{\tan }^2}\theta } \right]}}} \right]\]
\[ = sin\left[ {{{\tan }^{ - 1}}\left[ {\dfrac{{{{\cot }^2}\theta - 1}}{{2\cot \theta }}} \right] + {{\cos }^{ - 1}}\dfrac{{\left[ {1 - {{\tan }^2}\theta } \right]}}{{\left[ {1 + {{\tan }^2}\theta } \right]}}} \right]\]
Now, we need to apply the formulae\[cot2\theta = \dfrac{{co{t^2}\theta - 1}}{{2cot\theta }}\]and \[cos2\theta = \dfrac{{1 - ta{n^2}\theta }}{{1 + ta{n^2}\theta }}\;\]in the above equation.
\[ \Rightarrow sin\left[ {{{\tan }^{ - 1}}\dfrac{{\left[ {1 - {x^2}} \right]}}{{2x}} + {{\cos }^{ - 1}}\dfrac{{\left[ {1 - {x^2}} \right]}}{{\left[ {1 + {x^2}} \right]}}} \right] = sin\left[ {{{\tan }^{ - 1}}\cot 2\theta + {{\cos }^{ - 1}}\cos 2\theta } \right]\]
\[ = sin\left[ {{{\tan }^{ - 1}}\tan \left( {\dfrac{\pi }{2} - 2\theta } \right) + {{\cos }^{ - 1}}\cos 2\theta } \right]\] (Here we applied the formula$\cot \theta = \tan \left( {\dfrac{\pi }{2} - \theta } \right)$ )
Now, we need to apply the formulas${\cos ^{ - 1}}\cos \theta = \theta $and${\tan ^{ - 1}}\tan \theta = \theta $in the above equation.
Thus, we get
\[ \Rightarrow sin\left[ {{{\tan }^{ - 1}}\dfrac{{\left[ {1 - {x^2}} \right]}}{{2x}} + {{\cos }^{ - 1}}\dfrac{{\left[ {1 - {x^2}} \right]}}{{\left[ {1 + {x^2}} \right]}}} \right] = sin\left[ {\dfrac{\pi }{2} - 2\theta + 2\theta } \right]\]
$ = \sin \dfrac{\pi }{2}$
$ = 1$
Hence, $sin\left[ {{{\tan }^{ - 1}}\dfrac{{\left[ {1 - {x^2}} \right]}}{{2x}} + {{\cos }^{ - 1}}\dfrac{{\left[ {1 - {x^2}} \right]}}{{\left[ {1 + {x^2}} \right]}}} \right] = 1$

So, the correct answer is “Option A”.

Note:
 If we are asked to calculate the value of a trigonometric expression, we need to first analyze the given problem where we are able to apply the trigonometric identities.
Here, some trigonometric identities/formulae that we applied are needed to know to obtain the desired answer. Hence, we got $sin\left[ {{{\tan }^{ - 1}}\dfrac{{\left[ {1 - {x^2}} \right]}}{{2x}} + {{\cos }^{ - 1}}\dfrac{{\left[ {1 - {x^2}} \right]}}{{\left[ {1 + {x^2}} \right]}}} \right] = 1$.