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Mathematics Question on Inverse Trigonometric Functions

If a>b>0,Sec−1(a+ba−b)=2  Sin−1xa > b > 0, Sec^{-1} \left(\frac{a+b}{a-b} \right)= 2 \; Sin^{-1}x then x=x = _____

A

−aa+b-\sqrt{\frac{a}{a+b}}

B

aa+b\sqrt{\frac{a}{a+b}}

C

−ba+b-\sqrt{\frac{b}{a+b}}

D

ba+b\sqrt{\frac{b}{a+b}}

Answer

ba+b\sqrt{\frac{b}{a+b}}

Explanation

Solution

If a>b>0,sec⁡−1(a+ba−b)=2sin⁡−1xa>b>0, \sec ^{-1}\left(\frac{a+b}{a-b}\right)=2 \sin ^{-1} x
⇒cos⁡−1(a−ba+b)=2sin⁡−1x\Rightarrow \cos ^{-1}\left(\frac{a-b}{a+b}\right)=2 \sin ^{-1} x
⇒cos⁡−1(1−ba1+ba)=2sin⁡−1x\Rightarrow \cos ^{-1}\left(\frac{1-\frac{b}{a}}{1+\frac{b}{a}}\right)=2 \sin ^{-1} x
\Rightarrow \cos ^{-1}\left\\{\frac{1-(\sqrt{b / a})^{2}}{1+(\sqrt{b / a})^{2}}\right\\}=2 \sin ^{-1} x
⇒2tan⁡−1(b/a)=2sin⁡−1x\Rightarrow 2 \tan ^{-1}(\sqrt{b / a})=2 \sin ^{-1} x
[∵2tan⁡−1x=cos⁡−1(1−x21+x2)]\left[\because 2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right]
⇒sin⁡−1(ba+b)=sin⁡−1x\Rightarrow \sin ^{-1}\left(\frac{\sqrt{b}}{\sqrt{a+b}}\right)=\sin ^{-1} x
[∵tan⁡−1x=sin⁡−1x1+x2]\left[\because \tan ^{-1} x=\sin ^{-1} \frac{x}{\sqrt{1+x^{2}}}\right]
⇒x=ba+b\Rightarrow x=\sqrt{\frac{b}{a+b}}