If cot theta + cos theta = p and cot theta - | vedclass

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(A) Given:$p = \cot 	heta + \cos 	heta$$q = \cot 	heta - \cos 	heta$Step $1$: Calculate $p^2 - q^2$$p^2 - q^2 = (p + q)(p - q)$$p + q = (\cot 	he

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$16$

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(A) Given:$p = \cot 	heta + \cos 	heta$$q = \cot 	heta - \cos 	heta$Step $1$: Calculate $p^2 - q^2$$p^2 - q^2 = (p + q)(p - q)$$p + q = (\cot 	heta + \cos 	heta) + (\cot 	heta - \cos 	heta) = 2 \cot 	heta$$p - q = (\cot 	heta + \cos 	heta) - (\cot 	heta - \cos 	heta) = 2 \cos 	heta$$p^2 - q^2 = (2 \cot 	heta)(2 \cos 	heta) = 4 \cot 	heta \cos 	heta$Step $2$: Calculate $(p^2 - q^2)^2$$(p^2 - q^2)^2 = (4 \cot 	heta \cos 	heta)^2 = 16 \cot^2 	heta \cos^2 	heta$Step $3$: Express $pq$ in terms of $	heta$$pq = (\cot 	heta + \cos 	heta)(\cot 	heta - \cos 	heta) = \cot^2 	heta - \cos^2 	heta$$pq = \frac{\cos^2 	heta}{\sin^2 	heta} - \cos^2 	heta = \cos^2 	heta \left( \frac{1 - \sin^2 	heta}{\sin^2 	heta} ight) = \cos^2 	heta \left( \frac{\cos^2 	heta}{\sin^2 	heta} ight) = \cot^2 	heta \cos^2 	heta$Step $4$: Substitute $pq$ into the expression$(p^2 - q^2)^2 = 16 (\cot^2 	heta \cos^2 	heta) = 16pq$

If $\cot \theta + \cos \theta = p$ and $\cot \theta - \cos \theta = q$,then $(p^2 - q^2)^2$ in terms of $p$ and $q$ is: (in $pq$)

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