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

esin2⁡x+sin4⁡x+sin6⁡x+….∞log⁡2=8 ⇒sin2⁡x+sin4⁡x+sin6⁡x+….∞log⁡2=3 log 2sin2⁡x+sin4⁡x+…∞=sin2⁡x1−sin2⁡x=tan2⁡x=3∴G.E=11+tan⁡x=3−12

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

esin2⁡x+sin4⁡x+sin6⁡x+….∞log⁡2=8 ⇒sin2⁡x+sin4⁡x+sin6⁡x+….∞log⁡2=3 log 2sin2⁡x+sin4⁡x+…∞=sin2⁡x1−sin2⁡x=tan2⁡x=3∴G.E=11+tan⁡x=3−12

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If $e^{(\sin^{2} x + \sin^{4} x + \sin^{6} x + \dots . \infty) \log 2} = 8 and 0 < x < \frac{π}{2}$ 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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Detailed Solution

$e^{(\sin^{2} x + \sin^{4} x + \sin^{6} x + \dots . \infty) \log 2} = 8 \Rightarrow (\sin^{2} x + \sin^{4} x + \sin^{6} x + \dots . \infty) \log 2 = 3 \log 2$

$\begin{array}{l} \sin^{2} x + \sin^{4} x + \dots \infty = \frac{\sin^{2} x}{1 - \sin^{2} x} = \tan^{2} x = 3 \\ ∴ G . E = \frac{1}{1 + \tan x} = \frac{\sqrt{3} - 1}{2} \end{array}$