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Mathematics Question on Trigonometric Identities

If sin⁡−1(2a1+a2)−cos⁡−1(1−b21+b2)=tan⁡−1(2x1−x2),\sin^{-1} \left(\frac{2a}{1+a^{2}}\right) -\cos^{-1} \left(\frac{1-b^{2}}{1+b^{2}}\right) = \tan^{-1} \left(\frac{2x}{1-x^{2}}\right) , then what is the value of x?

A

a/b

B

ab

C

b/a

D

a−b1+ab\frac{a-b}{1+ab}

Answer

a−b1+ab\frac{a-b}{1+ab}

Explanation

Solution

Given ,
sin⁡−1(2a1+a2)−cos⁡−1(1−b21+b2)=tan⁡−1(2x1−x2)\sin^{-1} \left(\frac{2a}{1+a^{2}}\right) -\cos^{-1} \left(\frac{1-b^{2}}{1+b^{2}}\right) = \tan^{-1} \left(\frac{2x}{1-x^{2}}\right)
∴2tan⁡−1a−2tan⁡−1b=2tan⁡−1x\therefore 2 \tan^{-1} a - 2\tan^{-1} b =2 \tan^{-1} x
⇒tan⁡−1a−tan⁡−1b=tan⁡−1x\Rightarrow \tan^{-1}a -\tan^{-1} b =\tan^{-1}x
⇒tan⁡−1(a−b1+ab)=tan⁡−1x\Rightarrow \tan^{-1} \left(\frac{a-b}{1+ab}\right) = \tan^{-1}x
⇒x=a−b1+ab\Rightarrow x = \frac{a-b}{1+ ab}