ExamBro
ExamBro
KCET · Maths · Inverse Trigonometric Functions

The simplified form of \( \tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1}\left(\frac{x-y}{x+y}\right) \) is equal to

  1. A \( 00 \)
  2. B \( \frac{\pi}{4} \)
  3. C \( \frac{\pi}{2} \)
  4. D II
Verified Solution

Answer & Solution

Correct Answer

(B) \( \frac{\pi}{4} \)

Step-by-step Solution

Detailed explanation

\( \tan^{-1}\left(\frac{x}{y}\right)-\tan ^{-1}\left(\frac{x-y}{x+y}\right) = \tan ^{-1}\left(\frac{\frac{x}{y}-\frac{x-y}{x+y}}{1+\frac{x}{y} \cdot \frac{x-y}{x+y}}\right) \) \( = \tan ^{-1}\left(\frac{x(x+y)-y(x-y)}{y(x+y)+x(x-y)}\right) \)