Question

If $\sec \theta +\tan \theta =p$. Find the value of $\csc \theta$ and $\sin \theta$.

Answer 1

We have ${\sec }^2\theta -{\tan }^2\theta =1$

$(\sec \theta -\tan \theta )(\sec \theta +\tan \theta )=1$

$(\sec \theta -\tan \theta )p=1$ $(\because \sec \theta +\tan \theta =p)$

$(\sec \theta -\tan \theta )=\frac{1}{p}$

$\sec \theta +\tan \theta =p$

$\sec \theta -\tan \theta =\frac{1}{p}$

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

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

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

$\sin \theta =\sqrt{1-{\cos }^2\theta }$

${\cos }^2\theta ={\left(\frac{2p}{p^2+1}\right)}^2$

${\cos }^2\theta =\frac{4p^2}{p^4+1+2p^2}$

$1-{\cos }^2\theta =1-\frac{4p^2}{p^4+1+2p^2}$

$=\frac{p^4+1+2p^2-4p^2}{{(p^2+1)}^2}$

$=\frac{p^4+1-2p^2}{{(p^2+1)}^2}$

$=\frac{{(p^2-1)}^2}{{(p^2+1)}^2}$

$\sqrt{1-{\cos }^2\theta }=\frac{p^2-1}{p^2+1}$

$\therefore \sin \theta =\frac{p^2-1}{p^2+1}$

$\therefore \text{cosec}\theta =\frac{1}{\sin \theta }=\frac{p^2+1}{p^2-1}$