If sec θ + tan θ = p, prove that sin θ =

Solution:

Given that:

$\sec \theta+\tan \theta=p$, then we have to prove that $\sin \theta=\frac{p^{2}-1}{p^{2}+1}$

We can rewrite the given data as

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

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

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

Now we take the right hand side

$R H S=\frac{p^{2}-1}{p^{2}+1}$

Now we are putting the value of p in the above expression, we get

$R H S=\frac{\left(\frac{1+\sin \theta}{\cos \theta}\right)^{2}-1}{\left(\frac{1+\sin \theta}{\cos \theta}\right)^{2}+1}$

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

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

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

$=\frac{2 \sin \theta(1+\sin \theta)}{2(1+\sin \theta)}$

$=\sin \theta$

$=L H S$

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