If tan -1 x -tan -1 y -4- xy -1- then write the value of x - y - xy

We have tan^ 1x+tan^ 1y=π/4, ⇒ tan^ 1 ( x+y/1 xy )=π/4 ⇒x+y/1 xy=tanπ/4 ⇒x+y/1 xy=1 ∴ x+y+xy=1. ...

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Summation by Sigma Method

If tan -1 x +...

Question

If $t a n^{- 1} x + t a n^{- 1} y = \frac{π}{4}, x y < 1$ . then write the value of $x + y + x y$ .

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If $t a n^{- 1} x + t a n^{- 1} y = \frac{π}{4}$ where $x y$ < 1, find the value of $x + y + x y$ .

{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

{t a n}^{- 1} (\frac{1}{2} t a n 2 A) + {t a n}^{- 1} (c o t A) + {t a n}^{- 1} ({c o t}^{3} A)

0

π / 4 < A < π / 2

= π

0 < A < π / 4

{sin}^{- 1} x, {cos}^{- 1} x, {tan}^{- 1} x

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

x y < 1

π + {tan}^{- 1} \frac{x + y}{1 - x y}

x y > 1

{tan}^{- 1} \frac{1 - x}{1 + x} {tan}^{- 1} \frac{1 - y}{1 + y} = {sin}^{- 1} \frac{y - x}{\sqrt{1 + x^{2}} \sqrt{1 + y^{2}}}

{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

{t a n}^{- 1} \frac{1}{4} + 2 {t a n}^{- 1} \frac{1}{5} + {t a n}^{- 1} \frac{1}{6} + {t a n}^{- 1} \frac{1}{x} = \frac{π}{4}

{t a n}^{- 1} (x - 1) + {t a n}^{- 1} x + {t a n}^{- 1} (x + 1) = {t a n}^{- 1} 3 x

{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

{c o s}^{- 1} p + {c o s}^{- 1} q + {c o s}^{- 1} r = π

p^{2} + q^{2} + r^{2} + 2 p q r = 1

{s i n}^{- 1} x + {s i n}^{- 1} y + {s i n}^{- 1} z = π

x^{4} + y^{4} + z^{4} + 4 x^{2} y^{2} z^{2} = 2 (x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2})

{t a n}^{- 1} x + {t a n}^{- 1} y + {t a n}^{- 1} z = π

π / 2

x + y + z = x y z

x y + y z + z x = 1

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