If 00<θ<1800 then 2+2+2+.....2(1+cosθ) there being 'n' number 2's equal to

2+2+2+.....2(1+cosθ) 2(1+cosθ)=2.2cos2 θ2 =2cos θ2 2+2+2+......+2(1+cosθ) =cosθ2n

If 00<θ<1800 then 2+2+2+.....2(1+cosθ) there being 'n' number 2's equal to

2+2+2+.....2(1+cosθ) 2(1+cosθ)=2.2cos2 θ2 =2cos θ2 2+2+2+......+2(1+cosθ) =cosθ2n

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$I f 0^{0} < θ < 180^{0} t h e n$ $\sqrt{2 + \sqrt{2 + \sqrt{2 + . . . . . \sqrt{2 (1 + \cos θ)}}}}$ there being 'n' number 2's equal to

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$2 \cos (\frac{θ}{2^{n - 1}})$

$2 \sin (\frac{θ}{2^{n}})$

$2 \cos (\frac{θ}{2^{n + 1}})$

$2 \cos (\frac{θ}{2^{n}})$

answer is A.

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Detailed Solution

$\begin{array}{rcl} \sqrt{2 + \sqrt{2 + \sqrt{2 + . . . . . \sqrt{2 (1 + cosθ)}}}} \\ \begin{array}{rcl} \sqrt{2 (1 + cosθ)} & = & \sqrt{2.2 \cos^{2} \frac{θ}{2}} \end{array} \\ \begin{array}{cl} = & 2 \cos \frac{θ}{2} \end{array} \\ \sqrt{2 + \sqrt{2 + \sqrt{2 + . . . . . . + \sqrt{2 (1 + cosθ)}}}} \\ \begin{array}{cl} = & \cos (\frac{θ}{2^{n}}) \end{array} \end{array}$