Cos- tan -1 sin -1 x - - x 2 -1 - x 2 -2

cos[ tan^ 1{sin( ^ 1x)}] #160; Let ^ 1x = y ⇒ y = x sin y = 1√(1 + x^2) ⇒ y = sin ^ 1( 1√(1 + x^2)) ...

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Trigonometric Identities

cos[ tan -1 s...

Question

$cos [{tan}^{- 1} {sin ({cot}^{- 1} x)}] = \sqrt{\frac{x^{2} + 1}{x^{2} + 2}}$

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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

3 {s i n}^{- 1} \frac{2 x}{1 + x^{2}} - 4 {c o s}^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 {t a n}^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}

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

{t a n}^{- 1} \sqrt{x^{2} + x} + {s i n}^{- 1} \sqrt{x^{2} + x + 1} = \frac{π}{2}

sin [{t a n}^{- 1} \frac{1 - x^{2}}{2 x} + {cos}^{- 1} \frac{1 - x^{2}}{1 + x^{2}}] = 1

{sin}^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - . . . . .) + {cos}^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - . . . . . . .) = \frac{π}{2}

0 < | x | < \sqrt{2}

(1) {sin}^{- 1} (\frac{2 x}{1 + x^{2}}) = 2 {tan}^{- 1} x, | x | \leq 1

(2) {cos}^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) = 2 {tan}^{- 1} x, x \geq 0

(3) {tan}^{- 1} (\frac{2 x}{1 - x^{2}}) = 2 {tan}^{- 1} x, - 1 < x < 1

{cos}^{- 1} (x^{2} + \frac{1}{x^{2}} - 1) + {sin}^{- 1} (x^{2} - \frac{1}{x^{2}}) + {tan}^{- 1} (x^{2})

(where x \in R - {0})

s i n^{- 1} [x - \frac{x^{2}}{2} + \frac{x^{3}}{4} . . . .] = π / 2 - c o s^{- 1} [x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} . . . .]

0 < | x | < \sqrt{2}

t a n^{- 1} \sqrt{x (x + 1)} + s i n^{- 1} \sqrt{x^{2} + x + 1} = π / 2

- \sqrt{2} < x < 0

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