Question

If $\cos \theta +\sin \theta =p$ and $\sec \theta +cosec\theta =q$

Prove that $q(p^2-1)=2p$

Answer 1

$LHS=q(p^2-1)=(\sec \theta +cosec\theta )({(\cos \theta +\sin \theta )}^2-1)=(\frac{1}{\cos \theta }+\frac{1}{\sin \theta })({\cos }^2\theta +{\sin }^2\theta +2\sin \theta \cos \theta -1)=\frac{\sin \theta +\cos \theta }{\sin \theta \cos \theta }(1+2\sin \theta \cos \theta -1)(\because {\cos }^2\theta +{\sin }^2\theta =1)=2(\sin \theta +\cos \theta )=2p=RHS$