Question

If $\sin \theta +\cos \theta =p$ and $\sec \theta +\csc \theta =q$, show that $q(p^2-1)=2p$.a

Answer 1

$\sin \theta +\cos \theta =p$

$\sec \theta +\csc \theta =q$

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

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

$\therefore p^2-1=2\sin \theta \cos \theta$

$\therefore q(p^2-1)=\left(2\sin \theta \cos \theta \right)(\sec \theta +\csc \theta )$

$\therefore q(p^2-1)=2\sin \theta +2\cos \theta$

$\therefore q(p^2-1)=2(\sin \theta +\cos \theta )=LHS$

$\therefore 2p=RHS=2(\sin \theta +\cos \theta )$

$\therefore LHS=RHS$

Hence proved