AP EAMCET · Maths · Inverse Trigonometric Functions
\(\sin ^{-1} \frac{4}{5}+2 \tan ^{-1} \frac{1}{3}\) is equal to
- A \(\frac{\pi}{3}\)
- B \(\frac{\pi}{4}\)
- C \(\frac{\pi}{2}\)
- D 0
Answer & Solution
Correct Answer
(C) \(\frac{\pi}{2}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \sin ^{-1} \frac{4}{5}+2 \tan ^{-1} \frac{1}{3} \\ & =\sin ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2\left(\frac{1}{3}\right)}{1-\left(\frac{1}{3}\right)^2} \\ & =\sin ^{-1} \frac{4}{5}+\tan ^{-1} \frac{\frac{2}{3}}{\frac{8}{9}} \\ & =\sin ^{-1} \frac{4}{5}+\tan…
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