MCQ
tan-1 $\left(\frac{x}{y}\right)$ - tan-1 $\frac{x-y}{x+y}$ का मान है:
  • A
    $\frac \pi 2$
  • B
    $\frac \pi 3$
  • C
    $-\frac{3 \pi}{4}$
  • D
    $\frac \pi 4$

Answer

$\tan ^{-1}\left(\frac{x}{y}\right)$$-\tan ^{-1}\left(\frac{x-y}{x+y}\right)$= $\tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1}$ $\left(\frac{\frac{x}{y}-1}{\frac{x}{y}+1}\right)$
= $\tan ^{-1}\left(\frac{x}{y}\right)$ - $\left(\tan ^{-1} \frac{x}{y}-\tan ^{-1} 1\right)$ [$\because $ tan-1 x - tan-1 y = tan-1 $(\frac{x-y}{1+x y})$]
= tan-1 1 = $\frac{\pi}{4}$

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