Question

please solve it

NO SPAMMED
$\to \sf \cot ^{ - 1} \left \{ \dfrac{ \sqrt{1 + \sin \: x } \: \: + \sqrt{1 - \sin \: x } }{ \sqrt{1 + \sin \: x} \: \: - \sqrt{1 - \sin \: x } } \right \} \: = \dfrac{x}{2} ; \: x \in \: \bigg(0, \dfrac{ \pi}{4} \bigg)$
​

Answer 1

TO PROVE:

$\sf cot ^{ - 1} \left(\dfrac{ \sqrt{1 + \sin \: x } + \sqrt{1 - \sin \: x } }{ \sqrt{1 + \sin \: x} - \sqrt{1 - \sin \: x } } \right )= \dfrac{x}{2}$

PROOF:

$\leadsto\sf Take \ \sqrt{1 + sin\ x}$

$\boxed{ \large{ \bold{ \gray{sin \ 2x = 2 \ sin \ x \ cos \ x}}}}$

$\leadsto \sf sin \ x = 2 \ sin \ \left(\dfrac{x}{2}\right) \ cos \ \left( \dfrac{x}{2} \right)$

Add one on both sides.

$\leadsto\sf 1 + sin \ x = 1 + 2 \ sin \ \left(\dfrac{x}{2}\right) \ cos \ \left( \dfrac{x}{2} \right)$

$\boxed{ \large{ \bold{ \gray{sin ^{2} \ \frac{x}{2} + {cos}^{2} \ \frac{x}{2} =1}}}}$

1 + 2 sin(x/2) cos (x/2) = sin² (x/2) + cos² (x/2) + 2 sin(x/2) cos (x/2)

$\boxed{ \large{ \bold{ \gray{ A^2+ 2AB + B^2 = (A+B)^2}}}}$

1 - sin x = [sin (x/2) - cos (x/2)]²

$\leadsto\sf 1 + sin \ x = \left( sin \ \left(\dfrac{x}{2}\right) + \ cos \ \left( \dfrac{x}{2} \right) \right) ^{2}$

$\leadsto\sf \sqrt{1 + sin \ x} = sin \ \left(\dfrac{x}{2}\right) + \ cos \ \left( \dfrac{x}{2} \right)$

$\sf Take \ \sqrt{1 - sin\ x}$

$\boxed{ \large{ \bold{ \gray{sin \ 2x = 2 \ sin \ x \ cos \ x}}}}$

$\leadsto \sf - sin \ x = - 2 \ sin \ \left(\dfrac{x}{2}\right) \ cos \ \left( \dfrac{x}{2} \right)$

Add one on both sides

$\leadsto\sf 1 - sin \ x = 1 - 2 \ sin \ \left(\dfrac{x}{2}\right) \ cos \ \left( \dfrac{x}{2} \right)$

$\boxed{ \large{ \bold{ \gray{sin ^{2} \ \frac{x}{2} + {cos}^{2} \ \frac{x}{2} =1}}}}$

1 - 2 sin(x/2) cos (x/2) = sin² (x/2) + cos² (x/2) - 2 sin(x/2) cos (x/2)

$\boxed{ \large{ \bold{ \gray{ A^2 - 2AB + B^2 = (A - B)^2}}}}$

1 - sin x = [sin (x/2) - cos (x/2)]²

This can also be written as:

1 - sin x = [cos (x/2) - sin (x/2)]²

$\leadsto\sf \sqrt{1 - sin \ x }= \ cos \ \left( \dfrac{x}{2} \right) - sin \ \left(\dfrac{x}{2}\right)$

$\leadsto \sf Take \ \sqrt{1 + sin\ x} + \sqrt{1 - sin\ x}$

sin (x/2) + cos (x/2) + cos (x/2) - sin (x/2)

$\leadsto\sf \sqrt{1 + sin\ x} + \sqrt{1 - sin\ x} = 2 cos \ \left(\dfrac{x}{2} \right)$

$\sf Take \ \sqrt{1 + sin\ x} - \sqrt{1 - sin\ x}$

sin (x/2) + cos (x/2) - cos (x/2) + sin (x/2)

$\leadsto\sf \sqrt{1 + sin\ x} - \sqrt{1 - sin\ x} = 2 sin \ \left(\dfrac{x}{2} \right)$

Substitute the respective values.

$\leadsto\sf \cot ^{ - 1} \left( \dfrac{2cos \ \left(\dfrac{x}{2} \right )}{ 2sin \ \left(\dfrac{x}{2}\right ) }\right )$

$\leadsto\sf \cancel{cot}^{ - 1} \left( \cancel{cot} \ \left( \dfrac{x}{2} \right)\right)$

$\leadsto\sf cot ^{ - 1} \left(\dfrac{ \sqrt{1 + \sin \: x } + \sqrt{1 - \sin \: x } }{ \sqrt{1 + \sin \: x} - \sqrt{1 - \sin \: x } } \right )= \dfrac{x}{2}$

HENCE PROVED.

Answer 2

$\huge\bold{\green{\underbrace{\orange{SOLUTION:-}}}}$

PROOF :-

✍️ L.H.S ;

$\to \sf \cot ^{ - 1} \left \{ \dfrac{ \sqrt{1 + \sin \: x } \: \: + \sqrt{1 - \sin \: x } }{ \sqrt{1 + \sin \: x} \: \: - \sqrt{1 - \sin \: x } } \right \}$

✍️ First of all, we simplify

$\bold{\dfrac{\sqrt{1\:+\:\sin{x}}\:+\:\sqrt{1\:-\:\sin{x}}}{\sqrt{1\:+\:\sin{x}}\:-\:\sqrt{1\:-\:\sin{x}}}\:}$ .

$\rm{=\:\dfrac{\sqrt{1\:+\:\sin{x}}\:+\:\sqrt{1\:-\:\sin{x}}}{\sqrt{1\:+\:\sin{x}}\:-\:\sqrt{1\:-\:\sin{x}}}\:\times\:{\dfrac{\sqrt{1\:+\:\sin{x}}\:+\:\sqrt{1\:-\:\sin{x}}}{\sqrt{1\:+\:\sin{x}}\:+\:\sqrt{1\:-\:\sin{x}}}}\:}$

$\rm{=\:\dfrac{(\sqrt{1\:+\:\sin{x}}\:+\:\sqrt{1\:-\:\sin{x}})^2}{1\:+\:\sin{x}\:1\:-\:\sin{x}}\:}$

$\rm{=\:\dfrac{1\:+\:\sin{x}\:+\:1\:-\:\sin{x}\:+\:2\:\sqrt{(1\:+\:\sin{x})\:(1\:-\:\sin{x})}\:}{2\:\sin{x}}\:}$

$\rm{=\:\dfrac{2\:+\:2\:\sqrt{1\:-\:\sin^2{x}}}{2\:\sin{x}}\:}$

$\rm{=\:\dfrac{1\:+\:\cos{x}}{\sin{x}}\:}$

$\rm{=\:\dfrac{1\:+\:2\:\cos^2x/2\:-\:1}{2\:\sin{x/2}\:.\:\cos{x/2}\:}\:}$

$\rm{\purple{=\:\cot{x/2}\:}}$

✍️ So, we also write

$\rm{\cot^{-1}\:({\dfrac{\sqrt{1\:+\:\sin{x}}\:+\:\sqrt{1\:-\:\sin{x}}}{\sqrt{1\:+\:\sin{x}}\:-\:\sqrt{1\:-\:\sin{x}}}})\:}$

$\rm{=\:\cot^{-1}\:(\cot{x/2})\:}$

$\rm{=\:\dfrac{x}{2}\:}$ = R.H.S .

✍️ ‘x/2’ also present in the range of

(0 , π/2) .

✍️ So, L.H.S = R.H.S .

✍️ Hence, Proved .

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Formula which are applied in the above is;

$\checkmark\:\large\mathcal{\orange{\boxed{(a\:+\:b)\:(a\:-\:b)\:=\:a^2\:-\:b^2\:}}}$

$\checkmark\:\large\mathcal{\green{\boxed{(a\:+\:b)^2\:=\:a^2\:+\:b^2\:+\:2\:a\:b\:}}}$

$\checkmark\:\large\mathcal{\boxed{\sin^2{x}\:+\:\cos^2{x}\:=\:1\:\:\implies\:1\:-\:\sin^2{x}\:=\:\cos^2{x}\:}}$

$\checkmark\:\large\mathcal{\red{\boxed{\sin{2x}\:=\:2\:\sin{x}\:.\:\cos{x}\:}}}$

$\checkmark\:\large\mathcal{\purple{\boxed{\dfrac{\cos{x}}{\sin{x}}\:=\:\cot{x}\:}}}$

$\checkmark\:\large\mathcal{\blue{\boxed{\cot^{-1}(\cot\:{x})\:=\:x\:}}}$

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