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COMEDK · Maths · 23. Inverse Trigonometric Functions

If \(a>b>0\), then the value of \(\tan ^{-1}\left(\frac{a}{b}\right)+\tan ^{-1}\left(\frac{a+b}{a-b}\right)\) depends on

  1. A neither \(a\) nor \(b\)
  2. B \(a\) and not \(b\)
  3. C \(b\) and not \(a\)
  4. D both \(a\) and \(b\)
Verified Solution

Answer & Solution

Correct Answer

(A) neither \(a\) nor \(b\)

Step-by-step Solution

Detailed explanation

We have, \(\tan ^{-1}\left(\frac{a}{b}\right)+\tan ^{-1}\left(\frac{a+b}{a-b}\right)\) \(=\tan ^{-1}\left[\frac{\frac{a}{b}+\frac{a+b}{a-b}}{1-\left(\frac{a}{b}\right)\left(\frac{a+b}{a-b}\right)}\right]\)…