Question

If $\sec \theta =\frac{\sqrt{p^2+q^2}}{q}$, then the value of $\frac{p\sin \theta -q\cos \theta }{p\sin \theta +q\cos \theta }$

Answer 1

Given $\text{sec}\theta =\frac{\sqrt{p^2+q^2}}{q}$.

We know that,

${\text{sec}}^2\theta ={\tan }^2\theta +1$

${\tan }^2\theta +1=\frac{p^2+q^2}{q^2}$

${\tan }^2\theta =\frac{p^2}{q^2}$

$\tan \theta =\frac{-p}{q},\frac{p}{q}$

$\frac{p\sin \theta -q\cos \theta }{p\sin \theta +q\cos \theta }=\frac{p\tan \theta -q}{p\tan \theta +q}=\frac{\frac{p^2}{q}-q}{\frac{p^2}{q}+q}or\frac{\frac{-p^2}{q}-q}{\frac{-p^2}{q}+q}$

$=\frac{p^2-q^2}{p^2+q^2}or\frac{p^2+q^2}{p^2-q^2}$