- 1 ( y ) - - 1 ( x + y ) is equal to (A) 2 (B) 3 (C) 4 (D) 4

Option C is correct - 1 ( y ) - - 1 ( x + y ) = - 1 ( y - ( x + y ) 1 + ( y ) ( x + y ) ) = - 1 = - 1 ( + y 2 x 2 + y 2 ) = - 1 1 = 4 .

Inverse Trigonometric Functions — Class 12 Maths Solution

ncert misc SA NCERT Misc. , Q.17 , Page 52

Question

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

(A) $\frac{\pi }{2}$

(B) $\frac{\pi }{3}$

(D) $\frac{{ - 3\pi }}{4}$

Step-by-step Solution

Option C is correct

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

$= {\tan ^{ - 1}}\left[ {\frac{{x(x + Y) - y(x - y)}}{{y(x + y) + x(x - y)}}} \right] = {\tan ^{ - 1}}\left( {\frac{{{x^2} + {y^2}}}{{{x^2} + {y^2}}}} \right) = {\tan ^{ - 1}}1 = \frac{\pi }{4}.$

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