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Sign of Trigonometric Ratios in Different Quadrants

Solve: sins...

Question

Solve: $sin {{sin}^{- 1} (\frac{1 - \sqrt{3}}{2 \sqrt{2}}) + {cos}^{- 1} (\frac{\sqrt{3} + 1}{2 \sqrt{2}})}$

A

\frac{1}{2}

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B

\frac{\sqrt{3}}{2}

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C

0

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D

\frac{1}{\sqrt{2}}

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Solution

The correct option is C $0$
$sin {{sin}^{- 1} (\frac{1 - \sqrt{3}}{2 \sqrt{2}}) + {cos}^{- 1} (\frac{\sqrt{3} + 1}{2 \sqrt{2}})}$
$= sin {- {sin}^{- 1} (\frac{\sqrt{3} - 1}{2 \sqrt{2}}) + {cos}^{- 1} (\frac{\sqrt{3} + 1}{2 \sqrt{2}})}$ [Since ${sin}^{- 1} (- x) = - {sin}^{- 1} x$ ]
$= sin {- {sin}^{- 1} (\frac{\sqrt{3} - 1}{2 \sqrt{2}}) + \frac{π}{2} - {sin}^{- 1} (\frac{\sqrt{3} + 1}{2 \sqrt{2}})}$
$= sin {\frac{π}{2} - {sin}^{- 1} (\frac{\sqrt{3} - 1}{2 \sqrt{2}}) - {sin}^{- 1} (\frac{\sqrt{3} + 1}{2 \sqrt{2}})}$
$= sin {\frac{π}{2} - \frac{π}{2}} = 0$

$∵ [{sin}^{- 1} (\frac{\sqrt{3} - 1}{2 \sqrt{2}}) + {sin}^{- 1} (\frac{\sqrt{3} + 1}{2 \sqrt{2}}) = {sin}^{- 1} (\frac{\sqrt{3} - 1}{2 \sqrt{2}} . \frac{\sqrt{3} - 1}{2 \sqrt{2}} + \frac{\sqrt{3} + 1}{2 \sqrt{2}} . \frac{\sqrt{3} + 1}{2 \sqrt{2}}) = {sin}^{- 1}]$

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