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

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(B) We are given that $\sec 	heta + 	an 	heta = p$ ........... $(i)$We know the identity $\sec^2 	heta - 	an^2 	heta = 1$,which can be factored

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

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(B) We are given that $\sec 	heta + 	an 	heta = p$ ........... $(i)$We know the identity $\sec^2 	heta - 	an^2 	heta = 1$,which can be factored as $(\sec 	heta - 	an 	heta)(\sec 	heta + 	an 	heta) = 1$.Substituting the value from $(i)$,we get $(\sec 	heta - 	an 	heta) \cdot p = 1$,which implies $\sec 	heta - 	an 	heta = \frac{1}{p}$ ........... $(ii)$To find $	an 	heta$,subtract 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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