If $\frac{\cos \theta }{a}=\frac{\sin \theta }{b},$ then $\frac{a}{\sec 2\theta }+\frac{b}{\mathrm{cosec}2\theta }$ is equal to

Solution:

We have,

    $\frac{\cos \theta }{a}=\frac{\sin \theta }{b}$

$\begin{array}{r}=\frac{\cos \theta }{a}=\frac{\sin \theta }{b}=\sqrt{\frac{{\cos }^2\theta +{\sin }^2\theta }{a^2+b^2}\frac{-}{b}}\\ =\cos \theta =\frac{a}{\sqrt{a^2+b^2}}\text{ and }\sin \theta =\frac{b}{\sqrt{a^2+b^2}}\end{array}$

     $\frac{a}{\sec 2\theta }+\frac{b}{\mathrm{cosec}2\theta }=a\cos 2\theta +b\sin 2\theta$

$\begin{array}{l}= \frac{a}{\sec 2\theta }+\frac{b}{\mathrm{cosec}2\theta }=a\left({\cos }^2\theta -{\sin }^2\theta \right)+2b\sin \theta \cos \theta \\ = \frac{a}{\sec 20}+\frac{b}{\mathrm{cosec}2\theta }\approx a\frac{\left(a^2-b^2\right)}{a^2+b^2}+\frac{2a{b}^2}{a^2+b^2}=a\end{array}$