17.\tan^{-1}\left (\frac{x}{y} \right )-\tan^{-1}\frac{x-y}{x+y} is equal to 

        (A)    \frac{\pi}{2}

        (B)    \frac{\pi}{3}

        (C)    \frac{\pi}{4}

        (D)    \frac{3\pi}{4}

 

 

Answers (1)

Applying formula: \left [ \tan^{-1} x - \tan^{-1}y = \tan^{-1} \left ( \frac{x-y}{1+xy} \right ) \right ].

We get, 

\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+\left ( \frac{x}{y} \right ) \left ( \frac{x-y}{x+y} \right ) } \right ]

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

= \tan^{-1}\left ( \frac{x^2+xy - xy + y^2}{xy + y^2 + x^2 - xy} \right )

= \tan^{-1}\left ( \frac{x^2 + y^2}{ y^2 + x^2 } \right ) = \tan^{-1} 1 = \frac{\pi}{4}

Hence, the correct answer is C.

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