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

Question

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

Accepted Answer

Step 1: LHS $= q(p^2 - 1) = (\sec 	heta + \csc 	heta)((\sin 	heta + \cos 	heta)^2 - 1)$
Step 2: $= \frac{\sin 	heta + \cos 	heta}{\sin 	heta \cos 	heta} 	imes 2 \sin 	heta \cos 	heta$
Step 3: $= 2(\sin 	heta + \cos 	heta) = 2p = $ RHS

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