Question

If \[{{\sin }^{-1}}\left( \dfrac{2a}{1+{{a}^{2}}} \right)-{{\cos }^{-1}}\left( \dfrac{1-{{b}^{2}}}{1+{{b}^{2}}} \right)={{\tan }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)\], then the value of $x$ is
A. a
B. b
C. \[\dfrac{a+b}{1-ab}\]
D. \[\dfrac{a-b}{1+ab}\]

Answer 1

Hint: We first change the ratio of the given equations to one single form. We use the conversion forms of \[2{{\tan }^{-1}}\left( x \right)={{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)\], \[2{{\tan }^{-1}}\left( x \right)={{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)\] and \[2{{\tan }^{-1}}\left( x \right)={{\tan }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)\]. We compress the ratio tan using \[{{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\dfrac{x-y}{1+xy}\]. We equate the values to find the solution.

Complete step by step answer:
We first convert all the trigonometric inverses into the same form of tan.
We use the theorems of \[2{{\tan }^{-1}}\left( x \right)={{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)\], \[2{{\tan }^{-1}}\left( x \right)={{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)\] and \[2{{\tan }^{-1}}\left( x \right)={{\tan }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)\].
We exchange the values and get
\[\begin{align}
  & {{\sin }^{-1}}\left( \dfrac{2a}{1+{{a}^{2}}} \right)-{{\cos }^{-1}}\left( \dfrac{1-{{b}^{2}}}{1+{{b}^{2}}} \right)={{\tan }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right) \\
 & \Rightarrow 2{{\tan }^{-1}}\left( a \right)-2{{\tan }^{-1}}\left( b \right)=2{{\tan }^{-1}}\left( x \right) \\
\end{align}\]
We divide both sides with 2 and get \[{{\tan }^{-1}}x={{\tan }^{-1}}a-{{\tan }^{-1}}b\].
We also use the inverse sum rule of \[{{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\dfrac{x-y}{1+xy}\].
Therefore, \[{{\tan }^{-1}}x={{\tan }^{-1}}a-{{\tan }^{-1}}b={{\tan }^{-1}}\dfrac{a-b}{1+ab}\].
Equating both sides we get \[x=\dfrac{a-b}{1+ab}\]. The correct option is option (D).

Note:
Although for elementary knowledge the principal domain is enough to solve the problem. But if mentioned to find the general solution then the domain changes to $-\infty \le x\le \infty $. In that case we have to use the formula $x=n\pi +{{\left( -1 \right)}^{n}}a$ for $\sin \left( x \right)=\sin a$ where $-\dfrac{\pi }{2}\le a\le \dfrac{\pi }{2}$.