First slide

Introduction to ITF

Question

$The value of \sin^{- 1} \{\cot (\sin^{- 1} \sqrt{\frac{2 - \sqrt{3}}{4}} + \cos^{- 1} \frac{\sqrt{12}}{4} + \sec^{- 1} \sqrt{2})\} is equal to$

Moderate

A

1
B

0
C

-1
D

2

Solution

$Given that \sin^{- 1} \{\cot (\sin^{- 1} \sqrt{\frac{2 - \sqrt{3}}{4}} + \cos^{- 1} \frac{\sqrt{12}}{4} + \sec^{- 1} \sqrt{2})\}$
$Consider, \sqrt{\frac{2 - \sqrt{3}}{4}} = \sqrt{\frac{4 - 2 \sqrt{3}}{8}} = \sqrt{\frac{(\sqrt{3} - 1)^{2}}{8}} = \frac{\sqrt{3} - 1}{2 \sqrt{2}}$
$Now, \sin^{- 1} \sqrt{\frac{2 - \sqrt{3}}{4}} = \sin^{- 1} \frac{\sqrt{3} - 1}{2 \sqrt{2}} = \frac{π}{12} [15 °]$
$Also \cos^{- 1} \frac{\sqrt{12}}{4} = \cos^{- 1} (\frac{\sqrt{3}}{2}) = \frac{π}{6}$
$\sec^{- 1} \sqrt{2} = \frac{π}{4}$
$∴$ $\sin^{- 1} \{\cot (\sin^{- 1} \sqrt{\frac{2 - \sqrt{3}}{4}} + \cos^{- 1} \frac{\sqrt{12}}{4} + \sec^{- 1} \sqrt{2})\}$
$\begin{array}{l} = \sin^{- 1} \cot (\frac{π}{12} + \frac{π}{6} + \frac{π}{4}) \\ = \sin^{- 1} \cot \frac{π}{2} \\ = \sin^{- 1} (0) \\ = 0 \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App