Tan -1x-y-tan -1x-y-x-y is equal to where -x-y-0A- - - 4B- - - 4 C- 3 - - 4D- 0

The correct option is B π/4tan^ 1x/y tan^ 1x y/x+y =tan^ 1x/y tan^ 1[1 y/x/1+y/x ] =tan^ 1x/y tan^ 1 tan^ 1(1)+tan^ 1y/x =tan^ 1x/y+cot^ 1x/y tan^ 1(tanπ/4 ) = ...

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Byju's Answer

Standard XII

Mathematics

Standard Formulae - 3

tan -1x/y-tan...

Question

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

-π/4

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π/4

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3π/4

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Solution

The correct option is B π/4
$t a n^{- 1} \frac{x}{y} - t a n^{- 1} \frac{x - y}{x + y}$
$= t a n^{- 1} \frac{x}{y} - t a n^{- 1} [\frac{1 - y / x}{1 + y / x}]$
$= t a n^{- 1} \frac{x}{y} - t a n^{- 1} - t a n^{- 1} (1) + t a n^{- 1} \frac{y}{x}$
$= t a n^{- 1} \frac{x}{y} + c o t^{- 1} \frac{x}{y} - t a n^{- 1} (t a n \frac{π}{4})$
$= \frac{π}{2} - \frac{π}{4}$
$= \frac{π}{4}$

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Similar questions

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

Solveis equal to

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

Q. If

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

, then set of values of y is

Q. The simplified form of

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

is equal to

Q. Inverse circular functions,Principal values of

{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

.
(a)

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

(b)

{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

where

0 \leq x \leq 1

then

(a, b) =

(a)

(0, π)

(b)

(0, π / 4)

(c)

(- π / 4, π / 4)

(d)

(π / 4, π / 2)

(d) If

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

then (a,b) is equal to

(\frac{π^{3}}{32}, \frac{7 π^{3}}{8})

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tan−1xy−tan−1x−yx+y is equal to (where x>y>0)

The correct option is B π/4tan−1xy−tan−1x−yx+y =tan−1xy−tan−1[1−y/x1+y/x] =tan−1xy−tan−1−tan−1(1)+tan−1yx =tan−1xy+cot−1xy−tan−1(tanπ4) =π2−π4 =π4

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