If $\sin ^{-1} \frac{1}{3}+\sin ^{-1} \frac{2}{3}=\sin ^{-1} x$, then $x$ is equal to:

JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Application +4 | -1

\sin ^{-1} \frac{1}{3}+\sin ^{-1} \frac{2}{3}=\sin ^{-1} x

, then

x

is equal to:

View answer and explanation

Correct option: C

\frac{\sqrt{5}+4 \sqrt{2}}{9}

Option C (

\frac{{\sqrt{{5}}+4 \sqrt{{2}}}}{{9}}

) is correct because applying the sine addition formula and simplifying results in this value for x.

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\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1

x

{x^2} + {y^2} + {z^2} = {r^2},

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2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right)