If e ^{left(cos ^{2} x+cos ^{4} x+cos ^{6} x+ | vedclass

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Question

$e ^{\left(\cos ^{2} 	heta+\cos ^{4} 	heta+\ldots . . \inftyight) \ell n ^{2}}=2^{\cos ^{2} 	heta+\cos ^{4} 	heta+\ldots \infty}$

$=2^{\cot ^

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

$e ^{\left(\cos ^{2} 	heta+\cos ^{4} 	heta+\ldots . . \inftyight) \ell n ^{2}}=2^{\cos ^{2} 	heta+\cos ^{4} 	heta+\ldots \infty}$

$=2^{\cot ^{2} 	heta}$

Now $t^{2}-9 t+9=0 \Rightarrow t=1,8$

$\Rightarrow \quad 2^{\cot ^{2} 	heta}=1,8 \Rightarrow \cot ^{2} 	heta=0,3$

$0<	heta<\frac{\pi}{2} \Rightarrow \cot 	heta=\sqrt{3}$

$\Rightarrow \quad \frac{2 \sin 	heta}{\sin 	heta+\sqrt{3} \sin 	heta}=\frac{2}{1+\sqrt{3} \cot 	heta}=\frac{2}{4}=\frac{1}{2}$

If $e ^{\left(\cos ^{2} x+\cos ^{4} x+\cos ^{6} x+\ldots \ldots \infty\right) \log _{e} 2}$ satisfies the equation $t ^{2}-9 t +8=0,$ then the value of $\frac{2 \sin x}{\sin x+\sqrt{3} \cos x}\left(0 < x < \frac{\pi}{2}\right)$ is

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