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Question

If $$\displaystyle e^{\displaystyle (\sin^2 x+\sin^4 x+\sin^6 x + ..... \infty) \ln 3}$$, $$\displaystyle x \epsilon (0, \frac {\pi}{2})$$ satisfies the equation $$\displaystyle t^2 - 28t + 27 = 0$$ then value of $$\displaystyle (\cos \: x + \sin \: x)^{-1}$$ equals


A
31
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B
2(31)
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C
3+1
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D
1
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Solution

The correct option is A $$\displaystyle \sqrt 3 - 1$$

\sin ce $$\displaystyle  \sin ^2x < 1$$ for $$x\epsilon (0, \pi/2)$$
So, $$ \displaystyle e^{ (\sin ^2x + \sin ^4x + \sin ^6x + .......\infty)\ln3} $$ $$= e^{ (\tfrac{\sin ^2x}{1 -  \sin ^2x})\ln3} $$
$$\displaystyle = e^{(\tan ^2x)\ln3} $$ $$= e^{\ln3^{\tan ^2x}} $$ $$= 3^{\tan ^2x} $$ 
Now roots of given quadratic equation is 1 and 27 and  also $$x\epsilon (0, \pi/2)$$, So
$$ \displaystyle 3^{\tan ^2x} = 27$$ $$\Rightarrow \tan x = \sqrt{3}$$
$$\Rightarrow \sin x =\dfrac{\sqrt{3}}{2}$$ and $$ \cos x =\dfrac{1}{2}$$
So, $$\displaystyle ( \sin x + \cos x )^{-1} = \dfrac{2}{\sqrt{3} + 1} $$ $$= \dfrac{2}{\sqrt{3} + 1} .\dfrac{\sqrt{3} - 1}{\sqrt{3} - 1} $$ $$= \sqrt{3} -1 $$
Hence, option 'A' is correct.


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