If tan theta + sec theta = e^x,then cos theta | vedclass

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(B) Given that $	an 	heta + \sec 	heta = e^x$ $(i)$.We know the identity $\sec^2 	heta - 	an^2 	heta = 1$,which can be written as $(\sec 	heta

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$\frac{2}{e^x + e^{-x}}$

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(B) Given that $	an 	heta + \sec 	heta = e^x$ $(i)$.We know the identity $\sec^2 	heta - 	an^2 	heta = 1$,which can be written as $(\sec 	heta - 	an 	heta)(\sec 	heta + 	an 	heta) = 1$.Substituting $(i)$ into this identity,we get $\sec 	heta - 	an 	heta = \frac{1}{e^x} = e^{-x}$ $(ii)$.Adding equations $(i)$ and $(ii)$:$(	an 	heta + \sec 	heta) + (\sec 	heta - 	an 	heta) = e^x + e^{-x}$.$2 \sec 	heta = e^x + e^{-x}$.Since $\cos 	heta = \frac{1}{\sec 	heta}$,we have $\cos 	heta = \frac{2}{e^x + e^{-x}}$.

If $\tan \theta + \sec \theta = e^x$,then $\cos \theta$ equals

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