Question

If $\sec \theta +\tan \theta =p$, show that $\frac{p^2-1}{p^2+1}=\sin \theta$

Answer 1

If $sec\theta +tan\theta =P$

Consider $\frac{P^2-1}{P+1}=\frac{{(sec\theta +tan\theta )}^2-1}{sec\theta +tan\theta +1}$
$=\frac{se{c}^2\theta +ta{n}^2\theta +2sec\theta tan\theta -1}{se{c}^2\theta +ta{n}^2\theta +2sec\theta tan\theta +1}$ [Since, $(a+b{)}^2=a^2+b^2+2ab$]

$=\frac{2ta{n}^2\theta +2sec\theta tan\theta }{2se{c}^2\theta +2sec\theta tan\theta }$ [Since, $ta{n}^2\theta =se{c}^2\theta -1$]

$=\frac{2tan\theta (tan\theta +sec\theta )}{2sec\theta (sec\theta +tan\theta )}$
$=\frac{sin\theta }{cos\theta }\times \frac{1}{\frac{1}{\cos \theta }}$
$=\frac{sin\theta }{cos\theta }\times \frac{cos\theta }{1}=sin\theta$

Hence, proved.