If sec theta + tan theta = p, then tan theta | vedclass

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(B) Given that $\sec 	heta + 	an 	heta = p$  $(i)$We know that $\sec^2 	heta - 	an^2 	heta = 1$,which can be written as $(\sec 	heta - 	an 	h

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$\frac{p^2 - 1}{2p}$

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(B) Given that $\sec 	heta + 	an 	heta = p$  $(i)$We know that $\sec^2 	heta - 	an^2 	heta = 1$,which can be written as $(\sec 	heta - 	an 	heta)(\sec 	heta + 	an 	heta) = 1$.Substituting $(i)$ into this identity,we get $\sec 	heta - 	an 	heta = \frac{1}{p}$  $(ii)$Subtracting equation $(ii)$ from equation $(i)$:$(\sec 	heta + 	an 	heta) - (\sec 	heta - 	an 	heta) = p - \frac{1}{p}$$2 	an 	heta = \frac{p^2 - 1}{p}$$	an 	heta = \frac{p^2 - 1}{2p}$.

If $\sec \theta + \tan \theta = p,$ then $\tan \theta$ is equal to

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