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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main · Mathematics · Inverse Trigonometric Functions

^ -1 (4)/(5)+2 ^ -1 (1)/(3) is equal to

sin⁡−145+2tan⁡−113\sin ^{-1} \frac{4}{5}+2 \tan ^{-1} \frac{1}{3} is equal to

Options

  1. Aπ3\frac{\pi}{3}
  2. Bπ4\frac{\pi}{4}
  3. Cπ2\frac{\pi}{2}
  4. D0

Correct answer

C. π2\frac{\pi}{2}

Step-by-step solution

sin⁡−145+2tan⁡−113=sin⁡−145+tan⁡−12(13)1−(13)2=sin⁡−145+tan⁡−12389=sin⁡−145+tan⁡−134=sin⁡−145+cos⁡−145[∵tan⁡−134=cos⁡−145]=π2[∵sin⁡−1x+cos⁡−1x=π2]\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 ^{-1} \frac{3}{4} \\ & =\sin ^{-1} \frac{4}{5}+\cos ^{-1} \frac{4}{5} \quad\left[\because \tan ^{-1} \frac{3}{4}=\cos ^{-1} \frac{4}{5}\right] \\ & =\frac{\pi}{2} \\ & {\left[\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right]} \\ & \end{aligned}
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