Question

If $\frac{\cos \theta }{a}=\frac{\sin \theta }{b}then\frac{a}{\sec 2\theta }+\frac{b}{\text{cosec}2\theta }=$

Answer 1

$\frac{\cos \theta }{a}=\frac{\sin \theta }{b}$ [ Given ]

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

$\therefore$ $\tan \theta =\frac{b}{a}$ ---- ( 1 )

$\frac{a}{\sec 2\theta }+\frac{b}{cosec2\theta }$

$\Rightarrow$ $\frac{a}{\frac{1}{\cos 2\theta }}+\frac{b}{\frac{1}{\sin 2\theta }}$

$\Rightarrow$ $a\cos 2\theta +b\sin 2\theta$

$\Rightarrow$ $a\times \frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }+b\times \frac{2\tan \theta }{1+{\tan }^2\theta }$

$\Rightarrow$ $a\times \frac{1-\frac{b^2}{a^2}}{1+\frac{b^2}{a^2}}+b\times \frac{2\times \frac{b}{a}}{1+\frac{b^2}{a^2}}$

$\Rightarrow$ $a\times \frac{a^2-b^2}{a^2+b^2}+b\times \frac{2ab}{a^2+b^2}$

$\Rightarrow$ $\frac{a^3-a{b}^2}{a^2+b^2}+\frac{2a{b}^2}{a^2+b^2}$

$\Rightarrow$ $\frac{a^3-a{b}^2+2a{b}^2}{a^2+b^2}$

$\Rightarrow$ $\frac{a^3+a{b}^2}{a^2+b^2}$

$\Rightarrow$ $\frac{a(a^2+b^2)}{a^2+b^2}$

$\Rightarrow$ $a$