Question

If $\sec \theta +\tan \theta =p$ then prove that $\frac{p^2-1}{p^2+1}=\sin \theta$.

• A. True
• B. False

Answer 1

The correct option is A True

Given :

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

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

LHS :-

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

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

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

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

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