Question

Prove that,

{c o t}^{- 1} (\frac{\sqrt{1 + s i n x} + \sqrt{1 - s i n x}}{\sqrt{1 + s i n x} - \sqrt{1 - s i n x}}) = \frac{x}{2}, x \in (0, \frac{π}{4})

Accepted Answer

LHS $= \cot^{- 1} (\frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}})$
$= \cot^{- 1} (\frac{\sqrt{1 + \cos (\frac{π}{2} - x)} + \sqrt{1 - \cos (\frac{π}{2} - x)}}{\sqrt{1 + \cos (\frac{π}{2} - x)} - \sqrt{1 - \cos (\frac{π}{2} - x)}})$ $(∵ \cos (\frac{π}{2} - x) = \sin x)$
Let, $\frac{π}{2} - x = y$
$= \cot^{- 1} (\frac{\sqrt{1 + \cos y} + \sqrt{1 - \cos y}}{\sqrt{1 + \cos y} - \sqrt{1 - \cos y}})$
$= \cot^{- 1} (\frac{\sqrt{2 \cos^{2} \frac{y}{2}} + \sqrt{2 \sin^{2} \frac{y}{2}}}{\sqrt{2 \cos^{2} \frac{y}{2}} - \sqrt{2 \sin^{2} \frac{y}{2}}})$
$= \cot^{- 1} (\frac{\sqrt{2} \cos \frac{y}{2} + \sqrt{2} \sin \frac{y}{2}}{\sqrt{2} \cos \frac{y}{2} - \sqrt{2} \sin \frac{y}{2}})$
Divide both numerator and denominator by $\sqrt{2} \cos \frac{y}{2}$ , we get

$= \cot^{- 1} (\frac{1 + \tan \frac{y}{2}}{1 - \tan \frac{y}{2}})$ $= \cot^{- 1} (\frac{\tan \frac{π}{4} + \tan \frac{y}{2}}{1 - \tan \frac{π}{4} \tan \frac{y}{2}})$
$= \cot^{- 1} [\tan (\frac{π}{4} + \frac{y}{2})]$
$= \cot^{- 1} [\cot \{\frac{π}{2} - (\frac{π}{4} + \frac{y}{2})\}]$
$= \frac{π}{2} - (\frac{π}{4} + \frac{y}{2}) = \frac{π}{4} - \frac{y}{2}$
$= \frac{π}{4} - \frac{1}{2} (\frac{π}{2} - x)$
$= \frac{x}{2} =$ RHS
Therefore,
LHS = RHS
Hence Proved.

Karnataka Board Solutions for Chapter: Inverse Trigonometric Functions, Exercise 3: Miscellaneous Exercise on Chapter 2

Karnataka Board Mathematics Solutions for Exercise - Karnataka Board Solutions for Chapter: Inverse Trigonometric Functions, Exercise 3: Miscellaneous Exercise on Chapter 2

Questions from Karnataka Board Solutions for Chapter: Inverse Trigonometric Functions, Exercise 3: Miscellaneous Exercise on Chapter 2 with Hints & Solutions

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