MHT CET · Maths · Inverse Trigonometric Functions
If \(\cos ^{-1}\left(\frac{12}{13}\right)+\sin ^{-1}\left(\frac{3}{5}\right)=\sin ^{-1} P\), then the value of \(P\) is
- A \(\frac{63}{65}\)
- B \(\frac{56}{65}\)
- C \(\frac{48}{65}\)
- D \(\frac{36}{65}\)
Answer & Solution
Correct Answer
(B) \(\frac{56}{65}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \sin ^{-1}\left(\frac{3}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right) \\ & =\sin ^{-1}\left(\frac{3}{5}\right)+\sin ^{-1} \sqrt{1-\left(\frac{12}{13}\right)^2} \\ & =\sin ^{-1}\left(\frac{3}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right) \\ & =\sin ^{-1}\left[\frac{3}{5} \sqrt{1-\left(\frac{5}{13}\right)^2}+\frac{5}{13} \sqrt{1-\left(\frac{3}{5}\right)^2}\right] \\ & =\sin ^{-1}\left(\frac{3}{5} \times \frac{12}{13}+\frac{5}{13} \times \frac{4}{5}\right) \\ & =\sin ^{-1}\left(\frac{56}{65}\right)\end{aligned}\)
\(\therefore \quad P=\frac{56}{65}\)
\(\therefore \quad P=\frac{56}{65}\)
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