#2015_21 KVPY 2015 preparations!

From the given condition, we can say $2 sin \theta cos \theta = \frac14$ .

Simplifying the given equation, $64(sin \theta + cos \theta) - 8 \frac{(sin \theta + cos \theta )}{\frac{1}{8}}+ \frac{sin^2 \theta + cos^2 \theta}{\frac18}$ $64(sin \theta + cos \theta) - 64(sin \theta + cos \theta) +8$

From what is given:

$\begin{aligned} \theta & = \frac 12 \sin^{-1} \frac 14 \\ 2 \theta & = \sin^{-1} \frac 14 \\ \sin 2 \theta & = \frac 14 \\ 2\sin \theta \cos \theta & = \frac 14 \\ \sin \theta \cos \theta & = \frac 18 \end{aligned}$

Then we have:

$\begin{aligned} x & = 64\sin \theta + 64 \cos \theta - 8 \sec \theta - 8 \csc \theta + \tan \theta + \cot \theta \\ & = 64\sin \theta + 64 \cos \theta - \frac 8{\cos \theta} -\frac 8{\sin \theta} + \frac {\sin \theta}{\cos \theta} + \frac {\cos \theta}{\sin \theta} \\ & = \frac {64\sin^2 \theta \cos \theta + 64 \sin \theta \cos^2 \theta - 8 \sin \theta - 8 \cos \theta + \sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} & \small \color{#3D99F6} \text{Note that }\sin \theta \cos \theta = \frac 18 \\ & = \frac {8\sin \theta + 8 \cos \theta - 8 \sin \theta - 8 \cos \theta + 1}{\frac 18} \\ & = \boxed{8} \end{aligned}$