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 \[\sin x\] .