If $\cot heta + \cos heta = p$ and $\cot heta - \cos heta = q$, prove that $p^2 - q^2 = 4\sqrt{pq}$.

Question

If $\cot	heta + \cos	heta = p$ and $\cot	heta - \cos	heta = q$, prove that $p^2 - q^2 = 4\sqrt{pq}$.

Accepted Answer

Step 1: LHS $= p^2 - q^2 = (\cot	heta + \cos	heta)^2 - (\cot	heta - \cos	heta)^2 = 4\cot	heta\cos	heta$
Step 2: RHS $= 4\sqrt{pq} = 4\sqrt{(\cot	heta + \cos	heta)(\cot	heta - \cos	heta)} = 4\sqrt{\cot^2	heta - \cos^2	heta}$
Step 3: $= 4\sqrt{\cos^2	heta(\cosec^2	heta - 1)} = 4\sqrt{\cos^2	heta \cdot \cot^2	heta} = 4\cot	heta\cos	heta$
$	herefore$ LHS $=$ RHS.

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