If tan-1 x - tan-1 y - 4 -5- then cot-1x - cot-1 y is equal to

The correct option is A π5 tan ^ 1 x +tan ^ 1 y = 4π #160; 5 Now, tan ^ 1 x + ^ 1 x =π #160; 2 tan ^ 1 x =π #160; 2 ...

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Logarithmic Differentiation

If tan-1 x ...

Question

If $t a n^{- 1} x + t a n^{- 1} y = \frac{4 π}{5}$ , then $c o t^{- 1} x + c o t^{- 1} y$ is equal to

\frac{π}{5}

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\frac{3 π}{5}

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π

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\frac{2 π}{5}

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{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} \frac{12}{13} + {c o s}^{- 1} \frac{4}{5} + {t a n}^{- 1} \frac{63}{16} = π

2 {c o s}^{- 1} \frac{3}{\sqrt{13}} + {c o t}^{- 1} \frac{16}{63} + \frac{1}{2} {c o s}^{- 1} \frac{7}{25} = π

{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}

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c o s (2 {s i n}^{- 1} x) = 1 / 9

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

s i n ({s i n}^{- 1} \frac{1}{5} + {c o s}^{- 1} x) = 1

1 / 5

If $t a n^{- 1} x + t a n^{- 1} y = \frac{4 π}{5}, t h e n c o t^{- 1} x + c o t^{- 1} y$ equals to

(a) $\frac{π}{5}$ (b) $\frac{2 π}{5}$ (c) $\frac{3 π}{5}$ (d) $π$

{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 t}^{- 1} x + {s i n}^{- 1} \frac{1}{\sqrt (5)} = \frac{π}{4}

2 {t a n}^{- 1} (c o s x) = t a n {t a n}^{- 1} (2 c o s e c x)

t a n ({c o s}^{- 1} x) = s i n ({c o t}^{- 1} \frac{1}{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

{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})

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