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Question: If \(\sin^{- 1}\frac{2a}{1 + a^{2}} - \cos^{- 1}\frac{1 - b^{2}}{1 + b^{2}} = \tan^{- 1}\frac{2x}{1 ...

If sin⁡−12a1+a2−cos⁡−11−b21+b2=tan⁡−12x1−x2,\sin^{- 1}\frac{2a}{1 + a^{2}} - \cos^{- 1}\frac{1 - b^{2}}{1 + b^{2}} = \tan^{- 1}\frac{2x}{1 - x^{2}}, then x=x =

A

a

B

b

C

a+b1−ab\frac{a + b}{1 - ab}

D

a−b1+ab\frac{a - b}{1 + ab}

Answer

a−b1+ab\frac{a - b}{1 + ab}

Explanation

Solution

Put a=tan⁡θ,b=tan⁡φa = \tan\theta,b = \tan\varphi and x=tan⁡ψ,x = \tan\psi, then reduced form is

sin⁡−1(sin⁡2θ)−cos⁡−1(cos⁡2φ)=tan⁡−1(tan⁡2ψ)\sin^{- 1}(\sin 2\theta) - \cos^{- 1}(\cos 2\varphi) = \tan^{- 1}(\tan 2\psi)

⇒ 2θ−2φ=2ψ⇒θ−φ=ψ2\theta - 2\varphi = 2\psi \Rightarrow \theta - \varphi = \psi

Taking tan on both sides, we get tan⁡(θ−φ)\tan(\theta - \varphi) = tan⁡ψ\tan\psi

⇒ tan⁡θ−tan⁡φ1+tan⁡θ.tan⁡φ=tan⁡ψ\frac{\tan\theta - \tan\varphi}{1 + \tan\theta.\tan\varphi} = \tan\psi

Substituting these values, we get a−b1+ab=x\frac{a - b}{1 + ab} = x