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Question: If \[{{\sin }^{-1}}\left( \dfrac{2a}{1+{{a}^{2}}} \right)-{{\cos }^{-1}}\left( \dfrac{1-{{b}^{2}}}{1...

If sin⁡−1(2a1+a2)−cos⁡−1(1−b21+b2)=tan⁡−1(2x1+x2){{\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 xx is
A. a
B. b
C. a+b1−ab\dfrac{a+b}{1-ab}
D. a−b1+ab\dfrac{a-b}{1+ab}

Explanation

Solution

We first change the ratio of the given equations to one single form. We use the conversion forms of 2tan⁡−1(x)=sin⁡−1(2x1+x2)2{{\tan }^{-1}}\left( x \right)={{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right), 2tan⁡−1(x)=cos⁡−1(1−x21+x2)2{{\tan }^{-1}}\left( x \right)={{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right) and 2tan⁡−1(x)=tan⁡−1(2x1+x2)2{{\tan }^{-1}}\left( x \right)={{\tan }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right). We compress the ratio tan using tan⁡−1x−tan⁡−1y=tan⁡−1x−y1+xy{{\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 2tan⁡−1(x)=sin⁡−1(2x1+x2)2{{\tan }^{-1}}\left( x \right)={{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right), 2tan⁡−1(x)=cos⁡−1(1−x21+x2)2{{\tan }^{-1}}\left( x \right)={{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right) and 2tan⁡−1(x)=tan⁡−1(2x1+x2)2{{\tan }^{-1}}\left( x \right)={{\tan }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right).
We exchange the values and get

& {{\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{aligned}$$ We divide both sides with 2 and get $${{\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 $-\dfrac{\pi }{2}\le a\le \dfrac{\pi }{2}$.