Question

Solve $\tan \left[ {\left( {\dfrac{1}{2}} \right){{\sin }^{ - 1}}\left( {\dfrac{{2a}}{{\left( {1 + {a^2}} \right)}}} \right) + \left( {\dfrac{1}{2}} \right){{\cos }^{ - 1}}\left( {\dfrac{{\left( {1 - {a^2}} \right)}}{{\left( {1 + {a^2}} \right)}}} \right)} \right] =$
A.$\dfrac{{2a}}{{\left( {1 + {a^2}} \right)}}$
B.$\dfrac{{\left( {1 - {a^2}} \right)}}{{\left( {1 + {a^2}} \right)}}$
C.$\dfrac{{2a}}{{\left( {1 - {a^2}} \right)}}$
D.None of these

Answer 1

Hint : In the given question the expression is related to the inverse trigonometric function with variables as angles . So , whenever the variables are present , try to substitute the value of the given variable with another trigonometric function to form an identity or formula and then solve it accordingly .

Complete step-by-step answer :
Given : $\tan \left[ {\left( {\dfrac{1}{2}} \right){{\sin }^{ - 1}}\left( {\dfrac{{2a}}{{\left( {1 + {a^2}} \right)}}} \right) + \left( {\dfrac{1}{2}} \right){{\cos }^{ - 1}}\left( {\dfrac{{\left( {1 - {a^2}} \right)}}{{\left( {1 + {a^2}} \right)}}} \right)} \right]$ ………..equation (A)
In this question we will solve the ${\sin ^{ - 1}}\left( {\dfrac{{2a}}{{\left( {1 + {a^2}} \right)}}} \right)$ and ${\cos ^{ - 1}}\left( {\dfrac{{\left( {1 - {a^2}} \right)}}{{\left( {1 + {a^2}} \right)}}} \right)$ differently for ease .
So , for ${\sin ^{ - 1}}\left( {\dfrac{{2a}}{{\left( {1 + {a^2}} \right)}}} \right)$ , we will put $a = \tan x$ , we get
$= {\sin ^{ - 1}}\left( {\dfrac{{2\tan x}}{{\left( {1 + {{\tan }^2}x} \right)}}} \right)$
The above expression represents the identity for $\sin 2x = \dfrac{{2\tan x}}{{1 + {{\tan }^2}x}}$ ,
Therefore on solving we get ,
$= {\sin ^{ - 1}}\left( {\sin 2x} \right)$
On simplifying we get ,
$= 2x$
Now solving the term ${\cos ^{ - 1}}\left( {\dfrac{{\left( {1 - {a^2}} \right)}}{{\left( {1 + {a^2}} \right)}}} \right)$ we get
Putting the value of $a = \tan y$, we get
$= {\cos ^{ - 1}}\left( {\dfrac{{\left( {1 - \tan {y^2}} \right)}}{{\left( {1 + \tan {y^2}} \right)}}} \right)$
The above expression represents the identity for $\cos 2x = \dfrac{{1 - {{\tan }^2}y}}{{1 + {{\tan }^2}y}}$
Therefore , we get
$= {\cos ^{ - 1}}\left( {\cos 2y} \right)$
On simplifying we get
$= 2y$
Now , putting the value of both the terms in the equation (A) , we get
$= \tan \left[ {\left( {\dfrac{1}{2}} \right) \times 2x + \left( {\dfrac{1}{2}} \right) \times 2y} \right]$
On solving we get
$= \tan \left[ {x + y} \right]$
Now using the identity of $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$ we get ,
$= \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}$
Now putting the values of $x = {\tan ^{ - 1}}a$ and $y = {\tan ^{ - 1}}a$ we get ,
$= \dfrac{{\tan \left( {{{\tan }^{ - 1}}a} \right) + \tan \left( {{{\tan }^{ - 1}}a} \right)}}{{1 - \tan \left( {{{\tan }^{ - 1}}a} \right)\tan \left( {{{\tan }^{ - 1}}a} \right)}}$ on simplifying we get ,
$= \dfrac{{a + a}}{{1 - a \times a}}$
On solving further we get ,
$= \dfrac{{2a}}{{1 - {a^2}}}$
Therefore , the correct answer is option (C) for the given question .
So, the correct answer is “Option C”.

Note : Inverse Trigonometric Functions plays an important role in calculus to find out the various integrals . Inverse trigonometric functions are also used in other areas such as science and engineering . The inverse of function is not cancelled out with same trigonometric function like ${\sin ^{ - 1}}(\sin x)$ , here trigonometric function is not cancelled out with its inverse it's just like equating the angles as the angles in range of $\sin x$ .