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Question

Mathematics Question on Inverse Trigonometric Functions

If sin−113+sin−123=sin−1x,sin^{-1} \frac{1}{3} + sin^{-1} \frac{2}{3} = sin^{-1} x , then the value of

A

0

B

(5−42)9\frac {(\sqrt{5} - 4 \sqrt{2})}{9}

C

(5+42)9\frac {(\sqrt{5} + 4 \sqrt{2})}{9}

D

π2\frac {\pi}{2}

Answer

(5+42)9\frac {(\sqrt{5} + 4 \sqrt{2})}{9}

Explanation

Solution

Given that, sin⁡−113+sin⁡−123=sin⁡1x\sin ^{-1} \frac{1}{3}+\sin ^{-1} \frac{2}{3}=\sin ^{1} x Taking sine on both sides ⇒(131−49+231−19)=x\Rightarrow\left(\frac{1}{3} \sqrt{1-\frac{4}{9}}+\frac{2}{3} \sqrt{1-\frac{1}{9}}\right)=x ⇒(13⋅53+23⋅83)=x\Rightarrow\left(\frac{1}{3} \cdot \frac{\sqrt{5}}{3}+\frac{2}{3} \cdot \frac{\sqrt{8}}{3}\right)=x ⇒(5+429)=x\Rightarrow\left(\frac{\sqrt{5}+4 \sqrt{2}}{9}\right)=x ∴x=(5+429)\therefore x=\left(\frac{\sqrt{5}+4 \sqrt{2}}{9}\right)