Question

If $f(\theta )=\frac{1-\sin 2\theta +\cos 2\theta }{2\cos 2\theta }$ then value of $f\left(11^{\circ }\right)\cdot f\left(34^{\circ }\right)$ is_______.

Answer 1

Answer: 0.5

$f(\theta )=\frac{1-\sin 2\theta +\cos 2\theta }{2\cos 2\theta }$
$=\frac{(\cos \theta -\sin \theta {)}^2+\left({\cos }^2\theta -{\sin }^2\theta \right)}{2(\cos \theta -\sin \theta )(\cos \theta +\sin \theta )}$
$=\frac{\cos \theta }{\cos \theta +\sin \theta }=\frac{1}{1+\tan \theta }$
$f\left(11^{\circ }\right)\cdot f\left(34^{\circ }\right)=\frac{1}{\left(1+\tan 11^{\circ }\right)}\cdot \frac{1}{\left(1+\tan 34^{\circ }\right)}$
$=\frac{1}{\left(1+\tan 11^{\circ }\right)}\cdot \frac{1}{\left(1+\tan \left(45^{\circ }-11^{\circ }\right)\right)}$
$=\frac{1}{\left(1+\tan 11^{\circ }\right)}\cdot \frac{1}{1+\frac{1-\tan 11^{\circ }}{1+\tan 11^{\circ }}}$
$=\frac{1}{\left(1+\tan 11^{\circ }\right)}\cdot \frac{1+\tan 11^{\circ }}{2}=\frac{1}{2}$