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TS EAMCET · Maths · Inverse Trigonometric Functions

\(\sin ^{-1} \frac{4}{5}+2 \tan ^{-1} \frac{1}{3}\) is equal to

  1. A \(\frac{\pi}{3}\)
  2. B \(\frac{\pi}{4}\)
  3. C \(\frac{\pi}{2}\)
  4. D 0
Verified Solution

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…