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KCET · Maths · Inverse Trigonometric Functions

The value of \(\tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1}\left(\frac{x-y}{x+y}\right)\) (where, \(x, y>0\) ) is

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

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

\(\tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1}\left(\frac{x-y}{x+y}\right), \quad(\because x, y>0)\)
\(\quad=\tan ^{-1}\left\{\frac{\frac{x}{y}-\left(\frac{x-y}{x+y}\right)}{1+\frac{x(x-y)}{y(x+y)}}\right\}\)
\(\begin{aligned} & {\left[\because \tan ^{-1} A-\tan ^{-1} B=\tan ^{-1}\left\{\frac{A-B}{1+A B}\right\}\right] } \\=& \tan ^{-1}\left(\frac{x^{2}+x y-x y+y^{2}}{x y+y^{2}+x^{2}-x y}\right) \\=& \tan ^{-1}\left(\frac{x^{2}+y^{2}}{x^{2}+y^{2}}\right) \\=& \tan ^{-1}(1)=\tan ^{-1}\left(\tan \frac{\pi}{4}\right)=\frac{\pi}{4} \end{aligned}\)