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
- A \( 00 \)
- B \( \frac{\pi}{4} \)
- C \( \frac{\pi}{2} \)
- D II
Answer & Solution
Correct Answer
(B) \( \frac{\pi}{4} \)
Step-by-step Solution
Detailed explanation
\( \tan^{-1}\left(\frac{x}{y}\right)-\tan ^{-1}\left(\frac{x-y}{x+y}\right) = \tan ^{-1}\left(\frac{\frac{x}{y}-\frac{x-y}{x+y}}{1+\frac{x}{y} \cdot \frac{x-y}{x+y}}\right) \) \( = \tan ^{-1}\left(\frac{x(x+y)-y(x-y)}{y(x+y)+x(x-y)}\right) \)
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