Sin-1 (53)+Sin−1(135)=sinX−1 thenx=

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Sin-1 ( {3}{5})+{S}{i}{n}^{-1}({5}{13})=_{X}^{-1} then{x}=

Question

Sin-1 $(\frac{3}{5}) + Si n^{- 1} (\frac{5}{13}) = sin_{X}^{- 1}$ then $x =$

(^{\frac{63}{65})}

\frac{56}{65}

\frac{56}{63}

\frac{16}{63}

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Ava Discussed

Sin-1

(\frac{3}{5}) + Si n^{- 1} (\frac{5}{13}) = sin_{X}^{- 1}

then

x =

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Question Text

Sin-1

(\frac{3}{5}) + Si n^{- 1} (\frac{5}{13}) = sin_{X}^{- 1}

then

x =