Question

$\huge\fbox\colorbox{beige}{❥Question♬}$
Prove that
$ta {n}^{ - 1}( \frac{ \sqrt{1 + x} - \sqrt{1 - x} }{ \sqrt{1 + x} + \sqrt{1 - x} } ) = \frac{\pi}{4} \: \\ - \frac { 1}{2} \: \: co {s}^{ - 1} x \: - \frac{1}{ \sqrt{2} } \leqslant x \leqslant 1$
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Answer 1

Answer:-

$\tan^{ - 1} ( \frac{ \sqrt{(1 + x - } \sqrt{1 - x) \times 2} }{2} ) \\ = \frac{ - \cos ^{ - 1}(x) }{2} \\ = \frac{ \sqrt{2} }{2} \leqslant x \leqslant 1$

Answer 2

$\large\underline{\sf{Solution-}}$

Consider, LHS

$\rm :\longmapsto\: {tan}^{ - 1}\bigg( \dfrac{ \sqrt{1 + x} - \sqrt{1 - x} }{ \sqrt{1 + x} + \sqrt{1 - x} } \bigg)$

To evaluate, this we use method of Substitution

So, Substitute

$\red{\rm :\longmapsto\:x = cos2y}$

$\rm\implies \:\boxed{\tt{ y = \frac{1}{2} {cos}^{ - 1}x}}$

So, above expression can be rewritten as

$\rm \: = \: {tan}^{ - 1}\bigg( \dfrac{ \sqrt{1 + cos2y} - \sqrt{1 - cos2y} }{ \sqrt{1 + cos2y} + \sqrt{1 - cos2y} } \bigg)$

We know,

$\boxed{\tt{ 1 + cos2x = {2cos}^{2}x}} \: \: and \: \: \boxed{\tt{ 1 - cos2x = {2sin}^{2}x}}$

So, using these, we get

$\rm \: = \: {tan}^{ - 1}\bigg(\dfrac{ \sqrt{ {2cos}^{2} y} - \sqrt{ {2sin}^{2} y} }{ \sqrt{ {2cos}^{2}y } + \sqrt{ {2sin}^{2} y} }\bigg)$

$\rm \: = \: {tan}^{ - 1}\bigg(\dfrac{ \sqrt{2}cosy - \sqrt{2} siny }{ \sqrt{2} cosy + \sqrt{2}siny } \bigg)$

$\rm \: = \: {tan}^{ - 1}\bigg(\dfrac{ cosy - siny }{ cosy + siny } \bigg)$

$\rm \: = \: {tan}^{ - 1}\bigg(\dfrac{ cosy\bigg[1 - \dfrac{siny}{cosy} \bigg]}{ cosy\bigg[1 + \dfrac{siny}{cosy} \bigg] } \bigg)$

$\rm \: = \: {tan}^{ - 1}\bigg[\dfrac{1 - tany}{1 + tany} \bigg]$

$\rm \: = \: {tan}^{ - 1}\bigg[tan\bigg(\dfrac{\pi}{4} - y\bigg) \bigg]$

$\rm \: = \: \dfrac{\pi}{4} - y$

$\rm \: = \: \dfrac{\pi}{4} - \dfrac{1}{2} {cos}^{ - 1}x$

Hence,

$\boxed{\tt{ {tan}^{ - 1}\bigg( \dfrac{ \sqrt{1 + x} - \sqrt{1 - x} }{ \sqrt{1 + x} + \sqrt{1 - x} } \bigg) = \frac{\pi}{4} - \frac{1}{2} {cos}^{ - 1}x}}$

Note :-

$\rm :\longmapsto\: - \dfrac{1}{ \sqrt{2} } \leqslant x \leqslant 1$

$\rm :\longmapsto\: - \dfrac{1}{ \sqrt{2} } \leqslant cos2y \leqslant 1$

$\rm :\longmapsto\:0 \leqslant 2y \leqslant \dfrac{3\pi}{4}$

$\rm :\longmapsto\:0 \leqslant y \leqslant \dfrac{3\pi}{8}$

$\rm :\longmapsto\:0 \geqslant - y \geqslant - \dfrac{3\pi}{8}$

$\rm :\longmapsto\:\dfrac{\pi}{4} \geqslant \dfrac{\pi}{4} - y \geqslant \dfrac{\pi}{4}- \dfrac{3\pi}{8}$

$\rm :\longmapsto\:\dfrac{\pi}{4} \geqslant \dfrac{\pi}{4} - y \geqslant - \dfrac{\pi}{4}$

$\rm :\longmapsto\: - \: \dfrac{\pi}{4} \: \leqslant \dfrac{\pi}{4} - y \: \leqslant \: \dfrac{\pi}{4}$

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Explore More

$\boxed{\tt{ {tan}^{ - 1}x + {tan}^{ - 1}y = {tan}^{ - 1}\bigg[\dfrac{x + y}{1 - xy} \bigg]}}$

$\boxed{\tt{ {tan}^{ - 1}x - {tan}^{ - 1}y = {tan}^{ - 1}\bigg[\dfrac{x - y}{1 + xy} \bigg]}}$

$\boxed{\tt{ {2tan}^{ - 1}x = {sin}^{ - 1}\bigg[\dfrac{2x}{1 + {x}^{2} } \bigg]}}$

$\boxed{\tt{ {2tan}^{ - 1}x = {tan}^{ - 1}\bigg[\dfrac{2x}{1 - {x}^{2} } \bigg]}}$

$\boxed{\tt{ {2tan}^{ - 1}x = {cos}^{ - 1}\bigg[\dfrac{1 - {x}^{2} }{1 + {x}^{2} } \bigg]}}$