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

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(B) We are given that $	an 	heta + \sec 	heta = e^x$  --- $(i)$We know the trigonometric identity $\sec^2 	heta - 	an^2 	heta = 1$.This can be f

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

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(B) We are given that $	an 	heta + \sec 	heta = e^x$  --- $(i)$We know the trigonometric identity $\sec^2 	heta - 	an^2 	heta = 1$.This can be factored as $(\sec 	heta - 	an 	heta)(\sec 	heta + 	an 	heta) = 1$.Substituting the value from $(i)$,we get $(\sec 	heta - 	an 	heta) e^x = 1$,which implies $\sec 	heta - 	an 	heta = e^{-x}$ --- $(ii)$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 $\sec 	heta = \frac{e^x + e^{-x}}{2}$.Therefore,$\cos 	heta = \frac{2}{e^x + e^{-x}}$.

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

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