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NDA 2014
Exam
NDA
Year
2014
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

NDA 2014 · Mathematics · Inverse Trigonometric Functions

If x and y are positive and x y 1 , then what is ^ -1 x + ^ -1 y...

If xx and yy are positive and xy>1x y \gt 1, then what is tan⁡−1x\tan ^{-1} x +tan⁡−1y+\tan ^{-1} y equal to?

Options

  1. Atan⁡−1(x+y1−xy)\tan ^{-1}\left(\frac{x+y}{1-x y}\right)
  2. Bπ+tan⁡−1(x+y1−xy)\pi+\tan ^{-1}\left(\frac{x+y}{1-x y}\right)
  3. Cπ−tan⁡−1(x+y1−xy)\pi-\tan ^{-1}\left(\frac{x+y}{1-x y}\right)
  4. Dtan⁡−1(x−y1+xy)\tan ^{-1}\left(\frac{x-y}{1+x y}\right)

Correct answer

B. π+tan⁡−1(x+y1−xy)\pi+\tan ^{-1}\left(\frac{x+y}{1-x y}\right)

Step-by-step solution

tan⁡−1x+tan⁡−1y=tan⁡−1(x+y)1−xy\tan ^{-1} x+\tan ^{-1} y=\frac{\tan ^{-1}(x+y)}{1-x y} But xy>1x y \gt 1 ∴tan⁡−1x+tan⁡−1y=π+tan⁡−1(x+y)1−xy.\therefore \tan ^{-1} x+\tan ^{-1} y=\pi+\tan ^{-1} \frac{(x+y)}{1-x y} .
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