(b) sin $\frac{\pi }{2^n}$ + cos $\frac{\pi }{2^n}$ = $\frac{\sqrt{n}}{2}$
      $\Rightarrow$ $\sqrt{2}$ $(\sin \frac{\pi }{2^n}.\cos \frac{\pi }{4}+\cos \frac{\pi }{2^n}.\sin \frac{\pi }{4})$ = $\frac{\sqrt{n}}{2}$
      $\Rightarrow$ $\sqrt{2}$ sin$(\frac{\pi }{4}+\frac{\pi }{2^n})$ = $\frac{\sqrt{n}}{2}$
   Since sin$(\frac{\pi }{4}+\frac{\pi }{2^n})\le 1$
       $\therefore$ $\frac{\sqrt{n}}{2}$ $\le$ $\sqrt{2}$ $\Rightarrow$ $\sqrt{2}$ $\le$ 2$\sqrt{2}$ $\Rightarrow$ n $\le$ 8 .
  Again  $\frac{\sqrt{n}}{2}$ = $\sqrt{2}$ sin$(\frac{\pi }{4}+\frac{\pi }{2^n})$ > $\sqrt{2}$.$\frac{1}{\sqrt{2}}$ = 1
$(\because \sin (\frac{\pi }{4}+\frac{\pi }{2^n})\sin \frac{\pi }{4})$
       $\therefore$ n > 4, Hence, 4 < n $\le$ 8.

(b) We have k = sin$\frac{\pi }{18}$.sin$\frac{5\pi }{18}$.sin$\frac{7\pi }{18}$
                       = cos$(\frac{\pi }{2}-\frac{\pi }{18})$ cos$(\frac{\pi }{2}-\frac{5\pi }{18})$ cos$(\frac{\pi }{2}-\frac{7\pi }{18})$
                       = cos$\frac{\pi }{9}$ cos$\frac{2\pi }{9}$ cos$\frac{4\pi }{9}$ = $\frac{sin{2}^3\frac{\pi }{9}}{2^{3}sin\frac{\pi }{9}}$ = $\frac{sin\frac{8\pi }{9}}{8sin\frac{\pi }{9}}$
                       = $\frac{sin(\pi -\frac{\pi }{9})}{8sin\frac{\pi }{9}}$ = $\frac{1}{8}$

(c) Given $\alpha$ < $\beta$ < $\gamma$ < $\delta$ and sin $\alpha$ = sin $\beta$ = sin $\gamma$ = sin $\delta$ = k. Also $\alpha$, $\beta$, $\gamma$, $\delta$ are smallest positive angles satisfying above two conditions.
$\therefore$ We can take $\beta$ = $\pi$ - $\alpha$, $\gamma$ = 2$\pi$ + $\alpha$, $\delta$ = 3$\pi$ - $\alpha$
 Given expression
 = 4 sin $\frac{\alpha }{2}$ + 3 sin$(\frac{\pi }{2}-\frac{\alpha }{2})$ + 2 sin$(\pi +\frac{\alpha }{2})$ + 3 sin$(\frac{3\pi }{2}-\frac{\alpha }{2})$
 = 4 sin $\frac{\alpha }{2}$ + 3 cos $\frac{\alpha }{2}$ - 2 sin $\frac{\alpha }{2}$ - cos $\frac{\alpha }{2}$ = 2 $(\sin \frac{\alpha }{2}+\cos \frac{\alpha }{2})$
 = 2 $\sqrt{{(\sin \frac{1}{2}\alpha +\cos \frac{1}{2}\alpha )}^2}$ = 2 $\sqrt{1+\sin \alpha }$ = 2$\sqrt{1+k}$.

(a,b) If L=M, then L${}^2$ = LM or ML = M${}^2$
        Both LM = ML = 1 as sec${}^2$A - tan${}^2$A = 1
        $\therefore$L${}^2$ = M${}^2$ = 1.

(d)  1 + 2(1 - cos$\theta$) + 3(1 - cos$\theta$)${}^2$ + 4(1 - cos$\theta$)${}^3$+........ to $\infty$
      = (1 - x)${}^{-1}$, [Let (1 - cos$\theta$) = x]
      $\therefore$ Series  = (1 - 1 + cos$\theta$)${}^{-2}$ = sec${}^2$$\theta$
     = (1 + tan${}^2$$\theta$) = 1 + $\frac{3}{2}$ = $\frac{5}{2}$

(a) We have sin $\alpha$ sec $x_1$ sec $x_2$ + sin $\alpha$ sec $x_2$ sec $x_3$ +....... + sin $\alpha$ sec $x_{n-1}$, sec $x_n$)
                    =$\frac{\sin (x_2-x_1)}{\cos x_1\cos x_2}$ + $\frac{\sin (x_3-x_2)}{\cos x_2\cos x_3}$ + ........ + $\frac{\sin (x_n-x_{n-1})}{\cos x_{n-1}\cos x_n}$
                    = tan x${}_2$ - tan x${}_1$ + tan x${}_3$ - tan x${}_2$ + .......... + tan x${}_n$ - tan x${}_{n-}$
                    = tan x${}_n$ - tan x${}_1$ = $\frac{\sin (x_n-x_{n-1})}{\cos x_{n-1}\cos x_1}$ = $\frac{\sin (n-1)\alpha }{\cos x_n\cos x_1}$  {($\because$ x${}_n$ = x${}_1$ + (n - 1)$\alpha$)}
                   

Option C is the correct answer.