Question

If $\frac{\cos \alpha }{\cos \theta }+\frac{\sin \alpha }{\sin \theta }=\frac{\cos \beta }{\cos \theta }+\frac{\sin \beta }{\sin \theta }=1$ where $\alpha$ and $\beta$ do not differ by an even multiple of $\pi$, then show $\frac{\cos \alpha \cos \beta }{{\cos }^2\theta }+\frac{\sin \alpha \sin \beta }{{\sin }^2\theta }+1=0$.

Answer 1

$\frac{\cos \alpha }{\cos \theta }+\frac{\sin \alpha }{\sin \theta }=\frac{\cos \beta }{\cos \theta }+\frac{\sin \beta }{\sin \theta }=1$

From $\frac{\cos \varphi }{\cos \theta }+\frac{\sin \varphi }{\sin \theta }=1$

We have $\frac{\cos \varphi }{\cos \theta }=1-\frac{\sin \varphi }{\sin \theta }$

Squaring both sides
$\frac{{\cos }^2\varphi }{{\cos }^2\theta }=1+\frac{{\sin }^2\varphi }{{\sin }^2\theta }-\frac{2\sin \varphi }{\sin \theta }$

$\Rightarrow \frac{{\sin }^2\varphi }{{\sin }^2\theta }+\frac{{\sin }^2\varphi }{{\cos }^2\theta }-\frac{2{\sin }^2\varphi }{\sin \theta }+1-\frac{1}{{\cos }^2\theta }=0$

$\Rightarrow (\frac{1}{{\sin }^2\theta }+\frac{1}{{\cos }^2\theta }){\sin }^2\varphi -\left(\frac{2}{{\sin }^2\theta }\right)\sin \varphi -{\tan }^2\theta =0$

This is quadratic in $\sin \varphi$ where $\sin \alpha ,\sin \beta$ as the root so

$\sin \alpha ,\sin \beta =\frac{-{\tan }^2\theta }{(\frac{1}{{\sin }^2\theta }+\frac{1}{{\cos }^2\theta })}=-{\sin }^4\theta$

Also $\frac{\sin \varphi }{\sin \theta }=1-\frac{\cos \varphi }{\cos \theta }$

Squaring both sides
$\frac{{\sin }^2\varphi }{{\sin }^2\theta }=1+\frac{{\cos }^2\varphi }{{\cos }^2\theta }-\frac{2\cos \varphi }{\cos \theta }$

$(\frac{1}{{\sin }^2\theta }+\frac{1}{{\cos }^2\theta }){\cos }^2\varphi -\left(\frac{2}{\cos \theta }\right)\cos \varphi -{\cot }^2\theta =0$

This is quadratic in $\cos \varphi$ with $\cos \alpha ,\cos \beta$ as the roots so

$\cos \alpha \cos \beta =\frac{-{\cot }^2\theta }{(\frac{1}{{\cos }^2\theta }+\frac{1}{{\sin }^2\theta })}={\cos }^4\theta$

$\therefore \frac{\cos \alpha \cos \beta }{{\cos }^2\theta }+\frac{\sin \alpha \sin \beta }{{\sin }^2\theta }=1$