If tan -1 x -tan -1 y -4 where xy -1- find the value of x - y - xy-

Tan^ 1 ( x+y/1 xy ) = π/4 ⇒x+y/1 xy=1 ⇒ x + y + xy = 1 ...

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Trigonometric Ratios of Compound Angles

If tan -1 x +...

Question

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

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

{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

{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} (t a n \frac{3 π}{4})

{t a n}^{- 1} (t a n \frac{2 π}{3})

c o s [{c o s}^{- 1} (\frac{- \sqrt{3}}{2}) = \frac{π}{6}]

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

0 \leq x \leq 1

{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} 2 x + {t a n}^{- 1} 3 x = n π + (π / 4)

{t a n}^{- 1} x + {c o s}^{- 1} (\frac{y}{\sqrt{1 + y^{2}}}) = {s i n}^{- 1} (\frac{3}{\sqrt{10}})

{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}}}

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