The correct option is C 56/65Given sin^ 1C = sin^ 13/5 + cos^ 112/13 ⇒ C = sin ( sin^ 13/5 + cos^ 112/13 ) Using sin^(A + B) = sin A cos B+ cos A sin B ⇒ C = ...

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Relation between Inverses of Trigonometric Functions and Their Reciprocal Functions

then C =1

Question

If $s i n^{- 1} \frac{3}{5} + c o s^{- 1} (\frac{12}{13}) = s i n^{- 1} C, t h e n C =$

\frac{65}{56}

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\frac{24}{65}

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\frac{16}{65}

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\frac{56}{65}

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Solution

The correct option is D $\frac{56}{65}$
$G i v e n s i n^{- 1} C = s i n^{- 1} \frac{3}{5} + c o s^{- 1} \frac{12}{13} \Rightarrow C = s i n (s i n^{- 1} \frac{3}{5} + c o s^{- 1} \frac{12}{13}) U s i n g s i n^{(} A + B) = s i n A c o s B + c o s A s i n B \Rightarrow C = \frac{3}{5} \times \frac{12}{13} + \sqrt{1 - \frac{9}{25}} \sqrt{1 - \frac{144}{169}} \Rightarrow C = \frac{56}{65} .$

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If sin−135+cos−1(1213)=sin−1 C,then C=

The correct option is D 5665Given sin−1C=sin−135+cos−11213⇒C=sin(sin−135+cos−11213)Using sin(A+B)=sin A cos B+cos A sin B⇒C=35×1213+√1−925√1−144169⇒C=5665.

If $s i n^{- 1} \frac{3}{5} + c o s^{- 1} (\frac{12}{13}) = s i n^{- 1} C, t h e n C =$