If sin theta + cos theta = p and tan theta + | vedclass

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(B) Given:$\sin 	heta + \cos 	heta = p$  --- $(i)$$	an 	heta + \cot 	heta = q$  --- $(ii)$Squaring equation $(i)$:$(\sin 	heta + \cos 	heta)^2

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$2$

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(B) Given:$\sin 	heta + \cos 	heta = p$  --- $(i)$$	an 	heta + \cot 	heta = q$  --- $(ii)$Squaring equation $(i)$:$(\sin 	heta + \cos 	heta)^2 = p^2$$\sin^2 	heta + \cos^2 	heta + 2 \sin 	heta \cos 	heta = p^2$Since $\sin^2 	heta + \cos^2 	heta = 1$ and $2 \sin 	heta \cos 	heta = \sin 2	heta$,we have:$1 + \sin 2	heta = p^2$$\sin 2	heta = p^2 - 1$  --- $(iii)$Now,simplifying equation $(ii)$:$	an 	heta + \cot 	heta = q$$\frac{\sin 	heta}{\cos 	heta} + \frac{\cos 	heta}{\sin 	heta} = q$$\frac{\sin^2 	heta + \cos^2 	heta}{\sin 	heta \cos 	heta} = q$$\frac{1}{\sin 	heta \cos 	heta} = q$Multiply numerator and denominator by $2$:$\frac{2}{2 \sin 	heta \cos 	heta} = q$$\frac{2}{\sin 2	heta} = q$$\sin 2	heta = \frac{2}{q}$  --- $(iv)$Equating $(iii)$ and $(iv)$:$p^2 - 1 = \frac{2}{q}$$q(p^2 - 1) = 2$

If $\sin \theta + \cos \theta = p$ and $\tan \theta + \cot \theta = q$,then $q(p^2 - 1)$ is equal to

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