Sin -1 x -sin -1 1 x -cos -1 x -cos -1 1 x -

The correct option is A π We know, sin ^ 1 θ #160; +cos ^ 1 θ #160; =π #160;/ 2 #160;∴sin ^ 1 x +sin ^ 1 1 x #160; +cos ^ 1 x ...

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Continuity of a Function

sin -1 x +sin...

Question

${sin}^{- 1} x + {sin}^{- 1} \frac{1}{x} + {cos}^{- 1} x + {cos}^{- 1} \frac{1}{x} =$

π

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\frac{π}{2}

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\frac{3 π}{2}

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None of these

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Solution

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Similar questions

c o s^{- 1} x = 2 s i n^{- 1} \sqrt{\frac{1 - x}{2}} = 2 c o s^{- 1} \sqrt{\frac{1 + x}{2}}

s i n^{- 1} (- x) = - s i n^{- 1} x, c o s^{- 1} (- x) = π - c o s^{- 1} x (- 1 \leq x \leq 1)

0 \leq {cos}^{- 1} x \leq π

- \frac{π}{2} \leq {sin}^{- 1} x \leq \frac{π}{2}

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

x = \frac{1}{5}

{cos}^{- 1} x + {cos}^{- 1} y = \frac{π}{2}

{cos}^{- 1} x = {sin}^{- 1} y

A = \frac{1}{π} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} {sin}^{- 1} (π x) & {tan}^{- 1} (\frac{π}{x}) {sin}^{- 1} (\frac{π}{x}) & {tan}^{- 1} (π x) \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦, B = \frac{1}{π} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} - {cos}^{- 1} (π x) & {tan}^{- 1} (\frac{π}{x}) {sin}^{- 1} (\frac{π}{x}) & - {tan}^{- 1} (π x) \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

A - B

{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} x + {c o s}^{- 1} [\frac{x}{2} + \frac{\sqrt{(3 - 3 x^{2})}}{2}]

(\frac{1}{2} \leq x \leq 1)

c o s (2 {c o s}^{- 1} x + {s i n}^{- 1} x)

x = 1 / 5

0 \leq {c o s}^{- 1} x \leq π

- π / 2 \leq {s i n}^{- 1} x \leq π / 2

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