If s i n − 1 3 5 + c o s − 1 ( 12 13 ) = s i n − 1 C , t h e n C = Options: A . &nbsp 65 56 B . &nbsp 24 65 C . &nbsp 16 65 D . &nbsp 56 65

Answer: Option D : D G i v e n s i n − 1 C = s i n − 1 3 5 + c o s − 1 12 13 ⇒ C = s i n ( s i n − 1 3 5 + c o s − 1 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 ⇒ C = 3 5 × 12 13 + √ 1 − 9 25 √ 1 − 144 169 ⇒ C = 56 65 .

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

Discussion Forum : Inverse Trigonometric Functions

Question -

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

Options:

A .

\frac{65}{56}

B .

\frac{24}{65}

C .

\frac{16}{65}

D .

\frac{56}{65}

Answer: Option D
:
D

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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Discussion Forum : Inverse Trigonometric Functions

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