Question

If $\sec \theta +\tan \theta =p$, show that $\sec \theta -\tan \theta =\frac{1}{p}$. Hence, find the values of $\cos \theta$ and $\sin \theta$.

Answer 1

Given, $\sec \theta +\tan \theta =p$
$\Rightarrow \frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }=p$
$\Rightarrow \frac{1+\sin \theta }{\cos \theta }=p$ ....(1)
Also given $\sec \theta -\tan \theta =\frac{1}{p}$
$\Rightarrow \frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta }=\frac{1}{p}$
$\Rightarrow \frac{1-\sin \theta }{\cos \theta }=\frac{1}{p}$ .....(2)
Divide equation (2) by equation (1), we get
$\frac{1+\sin \theta }{1-\sin \theta }=\frac{1}{p^2}$
Apply componendo dividendo, we get
$\frac{1-\sin \theta +1+\sin \theta }{1-\sin \theta -1-\sin \theta }=\frac{1+p^2}{1-p^2}$
$\Rightarrow \frac{2}{-2\sin \theta }=\frac{1+p^2}{1-p^2}$
$\Rightarrow -\sin \theta =\frac{1-p^2}{1-{+p}^2}$
$\Rightarrow \sin \theta =\frac{p^2-1}{p^2+1}$
$\Rightarrow {\cos }^2\theta =1-{\sin }^2\theta =1-{\left[\frac{p^2-1}{p^2+1}\right]}^2=\frac{{[p^2+1]}^2-{[p^2-1]}^2}{{[p^2+1]}^2}$
$\Rightarrow {\cos }^2\theta =\frac{p^4+1+2p^2-p^4-1+2p^2}{{(p^2+1)}^2}=\frac{4p^2}{{(p^2+1)}^2}$
$\Rightarrow \cos \theta =\frac{2p}{p^2+1}$