Question

If $f\left(x\right)=\frac{1-\sin 2\theta +\cos 2\theta }{2\cos 2\theta }$ , find the value of $8f\left({11}^{\circ }\right)\times f\left({34}^{\circ }\right)$ is

Answer 1

Given $f\left(x\right)=\frac{1-\sin 2\theta +\cos 2\theta }{2\cos 2\theta }$
Consider, $8f\left({11}^{\circ }\right)\cdot f\left({34}^{\circ }\right)$
$=8\left(\frac{1-\sin 22+\cos 22}{2\cos 22}\right)\left(\frac{1-\sin 68+\cos 68}{2\cos 68}\right)$
$=8\left(\frac{1-\sin 22+\cos 22}{2\cos 22}\right)\left(\frac{1-\sin (90-22)+\cos (90-22)}{2\cos (90-22)}\right)$
$=8\left(\frac{1-\sin 22+\cos 22}{2\cos 22}\right)\left(\frac{1-\cos 22+\sin 22}{2\sin 22}\right)$
$=8\left(\frac{1+(\cos 22-\sin 22)}{2\cos 22}\right)\left(\frac{1-(\cos 22-\sin 22)}{2\sin 22}\right)$
$=8\left(\frac{1-(\cos 22-\sin 22{)}^2}{2\sin 44}\right)$
$=4$