Question

$sinx + cosx \div \sqrt{1 + sin2x}$
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Answer 1

(1) Question :-

if your question is -

Differentiate it :

$\leadsto \sf \frac{ \sin x + \cos x}{ \sqrt{1 + \sin2x} }$

Answer :-

$\leadsto \sf \frac{ \cos x - \sin x}{ \sin x + \cos x} + \frac{ \cos 2x}{1 + \sin2x} \:$

Explanation :-

Here we used quotient rule of differentiation :

$\leadsto \sf \frac{ \sin x + \cos x}{ \sqrt{1 + \sin2x} } \\ \\ \sf differentiate \: with \: respect \: to \: x \\ \\ \leadsto \sf \frac{ \sqrt{1 + \sin2x}( \cos x - \sin x) - \frac{ \sin x + \cos x}{2 \sqrt{1 + \sin2x} }( 2 \cos2x) }{ {( \sqrt{1 + \sin2x} )}^{2} } \\ \\ \leadsto \: \sf \frac{ \sqrt{1 + \sin2x}( \cos x - \sin x) - \frac{ \sin x + \cos x}{2 \sqrt{ {( \sin x + \cos x)}^{2} } }( 2 \cos2x) }{ {( \sqrt{1 + \sin2x} )}^{2} } \\ \\ \leadsto \: \: \sf \frac{ \sqrt{1 + \sin2x}( \cos x - \sin x) - ( \cos2x) }{ {( \sqrt{1 + \sin2x} )}^{2} } \\ \\ or \\ \\ \leadsto \sf \frac{ \cos x - \sin x}{ \sqrt{1 + \sin2x} } - \frac{ \cos2x}{1 + \sin2x} \\ \\ \leadsto \sf \frac{ \cos x - \sin x}{ \sqrt{ {( \sin x + \cos x)}^{2} } } + \frac{ \cos2x}{1 + \sin2x} \\ \\ \leadsto \sf \frac{ \cos x - \sin x}{ \sin x + \cos x} + \frac{ \cos 2x}{1 + \sin2x}$

hope this will help you .

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If your question is -

(2) Question :-

simplify it

$\leadsto \sf \: \frac{ \sin x + \cos x}{ \sqrt{1 + \sin2x} }$

Answer :-

$\leadsto \sf \: \frac{ \sin x + \cos x}{ \sqrt{1 + \sin2x} } = 1$

Explanation :-

We have

$\leadsto \sf \: \frac{ \sin x + \cos x}{ \sqrt{1 + \sin2x} } \\ \: \: \\ \because \sf \sin2x = 2 \sin x \cos x \\ \\ \leadsto \sf \frac{ \sin x + \cos x}{ \sqrt{1 + 2 \sin x \cos x} } \\ \\ \because \: { \sin}^{2} \theta + { \cos}^{2} \theta = 1 \\ \\ \leadsto \sf \frac{ \sin x + \cos x}{ \sqrt{ { \sin}^{2}x + { \cos}^{2}x + 2 \sin x \cos x } } \\ \\ \leadsto \sf \frac{ \sin x + \cos x}{ \sqrt{ {( \sin x + \cos x) }^{2} } } \\ \\ \leadsto \: \frac{ \sin x + \cos x}{ \sin x + \cos x} = 1$