The principal value of sin -1 tan - 5- 4 is

The correct option is D π 2 Principal value of sin^ 1tan ( 5 π4) =sin^ 1( tan(π+π4)) =sin^ 1( 1) = π2 ...

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Modulus of a Complex Number

The principal...

Question

The principal value of ${sin}^{- 1} {tan (- \frac{5 π}{4})}$ is

\frac{π}{4}

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\frac{- π}{4}

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\frac{π}{2}

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- \frac{π}{2}

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{sin}^{- 1} \frac{1}{2} + {cos}^{- 1} \frac{1}{2} + {tan}^{- 1} \frac{1}{\sqrt{3}}

(a) \frac{2 π}{3} (b) \frac{π}{2}

(c) \frac{3 π}{4} (d) \frac{5 π}{6}

{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} \frac{1}{2} + {t a n}^{- 1} \frac{1}{3} = \frac{π}{4}

{t a n}^{- 1} \frac{1}{2} + {t a n}^{- 1} \frac{1}{5} + {t a n}^{- 1} \frac{1}{8} = \frac{π}{4}

{t a n}^{- 1} \frac{3}{4} + {t a n}^{- 1} \frac{3}{5} - {t a n}^{- 1} \frac{8}{19} = \frac{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

{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})

f (x) = {sin}^{4} x + {cos}^{4} x

f

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