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$\begin{gathered} \boxed{ \begin{gathered}\sf \bold{solve : } \\\sf \bold{ \: \to \sf \cot ^{ - 1} \left \{ \dfrac{ \sqrt{1 + \sin \: x } \: \: + \sqrt{1 - \sin \: x } }{ \sqrt{1 + \sin \: x} \: : - \sqrt{1 - \sin \: x } } \right \} \: = \dfrac{x}{2} ; \: x \in \: \bigg(0, \dfrac{ \pi}{4} \bigg)}\end{gathered}}\end{gathered}$

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Answer 1

Answer:

Solution:-

Given that,

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

On rationalizing, we get

$\rm = \cot ^{ - 1} \left \{ \dfrac{ \sqrt{1 + sinx } + \sqrt{1 - sinx } }{ \sqrt{1 + sinx} - \sqrt{1 - sinx }} \times\dfrac{ \sqrt{1 + sinx} + \sqrt{1 - sinx} }{ \sqrt{1 + sinx} + \sqrt{1 - sinx} } \right \}$

We know,

$\green{ \boxed{ \sf{ \:(x + y)(x - y) = {x}^{2} - {y}^{2}}}}$

So, using this identity, we get

$\rm = \cot ^{ - 1} \left \{ \dfrac{\bigg( \sqrt{1 + sinx } + \sqrt{1 - sinx } \bigg)^{2}}{ (\sqrt{1 + sinx})^{2} - (\sqrt{1 - sinx })^{2}} \right \}$

We know,

$\green{ \boxed{ \bf{ \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}}$

So, using this, we get

$\rm = \cot ^{ - 1} \left \{ \dfrac{1 + sinx + 1 - sinx + 2 \sqrt{(1 + sinx)(1 - sin)} }{(1 + sinx) - (1 - sinx) }\right \}$

$\rm = \cot ^{ - 1} \left \{ \dfrac{2 + 2 \sqrt{1 - {sin}^{2} x} }{1 + sinx - 1 + sinx}\right \}$

We know,

$\green{ \boxed{ \sf{ \: {sin}^{2}x + {cos}^{2}x = 1}}}$

So, using this, we get

$\rm = \cot ^{ - 1} \left \{ \dfrac{2 + 2 \sqrt{{cos}^{2} x} }{2sinx}\right \}$

$\rm = \cot ^{ - 1} \left \{ \dfrac{2 + 2 |cosx| }{2sinx}\right \}$

As it is given that,

$\rm :\longmapsto\:x \: \in \: \bigg(0, \: \dfrac{\pi}{4} \bigg)\bf\implies \: |cosx| = cos$

So, using this, we get

$\rm = \cot ^{ - 1} \left \{ \dfrac{2 + 2 cosx}{2sinx}\right \}$

$\rm = \cot ^{ - 1} \left \{ \dfrac{2(1 + cosx)}{2sinx}\right \}$

$\rm = \cot ^{ - 1} \left \{ \dfrac{1 + cosx}{sinx}\right \}$

$\rm = \cot ^{ - 1} \left \{ \dfrac{2 {cos}^{2} \bigg[\dfrac{x}{2} \bigg]}{2sin\bigg[\dfrac{x}{2} \bigg]cos\bigg[\dfrac{x}{2} \bigg]}\right \}$

$\rm = \cot ^{ - 1} \left \{ \dfrac{cos\bigg[\dfrac{x}{2} \bigg]}{sin\bigg[\dfrac{x}{2} \bigg]}\right \}$

$\rm = \cot ^{ - 1} \left \{ cot\bigg[\dfrac{x}{2} \bigg]\right \}$

Hence,

$\rm :\longmapsto\: \red{ \boxed{ \sf{ \:\cot ^{ - 1} \left \{ \dfrac{ \sqrt{1 + \sin \: x } \: + \sqrt{1 - \sin \: x } }{ \sqrt{1 + \sin \: x} - \sqrt{1 - \sin \: x } } \right \} \: = \frac{x}{2}}}}$

Hence Verified