Tan -1x-y-tan -1x-y-x-y is equal toa -2b -3c -4d -3 -4

Tan^ 1 ( x/y ) tan^ 1 ( x y/x+y )=tan^ 1 ( x/y ) tan^ 1 ( x/y 1/x/y+1 ) =tan^ 1 ( x/y ) { tan^ 1x/y tan^ 11 } [ ∵ tan^ 1x tan^ 1y=tan^ 1 ( x y/1+xy ) ] =t ...

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Byju's Answer

Standard XII

Mathematics

Composition of Trigonometric Functions and Inverse Trigonometric Functions

tan -1x/y-tan...

Question

$t a n^{- 1} (\frac{x}{y}) - t a n^{- 1} (\frac{x - y}{x + y})$ is equal to
a) $\frac{π}{2}$
b) $\frac{π}{3}$
c) $\frac{π}{4}$
d) $\frac{- 3 π}{4}$

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Solution

$t a n^{- 1} (\frac{x}{y}) - t a n^{- 1} (\frac{x - y}{x + y}) = t a n^{- 1} (\frac{x}{y}) - t a n^{- 1} (\frac{\frac{x}{y} - 1}{\frac{x}{y} + 1}) = t a n^{- 1} (\frac{x}{y}) - {t a n^{- 1} \frac{x}{y} - t a n^{- 1} 1} [∵ t a n^{- 1} x - t a n^{- 1} y = t a n^{- 1} (\frac{x - y}{1 + x y})]$
$= t a n^{- 1} 1 = \frac{π}{4}$ , Hence, the correct option is (c).

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tan−1(xy)−tan−1(x−yx+y) is equal to a) π2 b) π3 c) π4 d) −3π4

tan−1(xy)−tan−1(x−yx+y)=tan−1(xy)−tan−1(xy−1xy+1)=tan−1(xy)−{tan−1xy−tan−11} [∵ tan−1x−tan−1y=tan−1(x−y1+xy)] =tan−11=π4, Hence, the correct option is (c).

$t a n^{- 1} (\frac{x}{y}) - t a n^{- 1} (\frac{x - y}{x + y})$ is equal to
a) $\frac{π}{2}$
b) $\frac{π}{3}$
c) $\frac{π}{4}$
d) $\frac{- 3 π}{4}$