If esin2⁡x+sin4⁡x+sin6⁡x+∞loge 2=8 and 0<x<π2 then cos⁡xcos⁡x+sin⁡x=

sin2x+sin4x+...∞=sin2x1−sin2x=tan2x⇒tan2x=3⇒tanx=3G.E. cosxcosx+sinx=11+tanx=11+3×3−13−1=3−12

If esin2⁡x+sin4⁡x+sin6⁡x+∞loge 2=8 and 0<x<π2 then cos⁡xcos⁡x+sin⁡x=

sin2x+sin4x+...∞=sin2x1−sin2x=tan2x⇒tan2x=3⇒tanx=3G.E. cosxcosx+sinx=11+tanx=11+3×3−13−1=3−12

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$If e^{(\sin^{2} x + \sin^{4} x + \sin^{6} x + \infty) \log_{e} 2} = 8 and 0 < x < \frac{π}{2} then \frac{\cos x}{\cos x + \sin x} =$

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$\frac{\sqrt{3} + 1}{2}$

$\frac{\sqrt{3} - 1}{2}$

$\frac{2}{\sqrt{3} - 1}$

$\frac{2}{\sqrt{3} + 1}$

answer is B.

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$\sin^{2} x + \sin^{4} x + ... \infty = \frac{\sin^{2} x}{1 - \sin^{2} x} = \tan^{2} x$

$\Rightarrow \tan^{2} x = 3$

$\Rightarrow \tan x = \sqrt{3}$

G.E. $\frac{\cos x}{\cos x + \sin x} = \frac{1}{1 + \tan x} = \frac{1}{1 + \sqrt{3}} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{\sqrt{3} - 1}{2}$

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If esin2⁡x+sin4⁡x+sin6⁡x+∞loge 2=8 and 0<x<π2 then cos⁡xcos⁡x+sin⁡x=

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$If e^{(\sin^{2} x + \sin^{4} x + \sin^{6} x + \infty) \log_{e} 2} = 8 and 0 < x < \frac{π}{2} then \frac{\cos x}{\cos x + \sin x} =$