Solve tan -1x-y-tan -1x-y-x-y is equal toA -2 B- -3 C -4 D -3 -4

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Properties of Inequalities

Solve tan -1x...

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Solveis equal to

(A) (B). (C) (D)

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

$t a n^{- 1} \frac{x}{y} - t a n^{- 1} \frac{x - y}{x + y}$ is equal to (where x>y>0)

{tan}^{- 1} \frac{x}{y} - {tan}^{- 1} \frac{x - y}{x + y}

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

{s i n}^{- 1} (3 x / 5) + {s i n}^{- 1} (4 x / 5) = {s i n}^{- 1} x

{c o s}^{- 1} x + {s i n}^{- 1} (\frac{1}{2} x) = \frac{π}{6}

a \leq {t a n}^{- 1} (\frac{1 - x}{1 + x}) \leq b

0 \leq x \leq 1

(a, b) =

(0, π)

(0, π / 4)

(- π / 4, π / 4)

(π / 4, π / 2)

a \leq {({s i n}^{- 1} x)}^{3} + {({c o s}^{- 1} x)}^{3} \leq b

(\frac{π^{3}}{32}, \frac{7 π^{3}}{8})

y = s i n^{- 1} x + t a n^{- 1} x + s e c^{- 1} x

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