Question

If $\frac{\cos \alpha }{\cos \beta }=m,\frac{\cos \alpha }{\sin \beta }=n,$ then show that $(m^2+n^2){\cos }^2\beta =m^2{n}^2$

Answer 1

$\frac{\cos \alpha }{\cos \beta }=m$ and $\frac{\cos \alpha }{\cos \beta }=n$

$\frac{\cos \alpha }{m}=\cos \beta .....\left(1\right)$

$\frac{\cos \alpha }{n}=\sin \beta ....\left(2\right)$

Squaring and adding $\left(1\right)$ & $\left(2\right)$

${\cos }^2\beta =\frac{{\cos }^2\alpha }{m^2}$

${\sin }^2\beta =\frac{{\cos }^2\alpha }{n^2}$

Adding the $2$ equations

${\cos }^2\beta +{\sin }^2\beta ={\cos }^2\alpha (\frac{1}{m^2}+\frac{1}{n^2})$

$1=\frac{{\cos }^2\alpha (m^2+n^2)}{m^2{n}^2}$

$\Rightarrow m^2{n}^2={\cos }^2\alpha (m^2+n^2)$

Hence proved