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

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Question

(b) $	an 	heta + \sec 	heta = {e^x}$&hellip;..$(i)$

$	herefore \,\,\,\sec 	heta - 	an 	heta = {e^{ - x}}$&hellip;..$(ii)$

From $(i)$ and

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) $	an 	heta + \sec 	heta = {e^x}$&hellip;..$(i)$

$	herefore \,\,\,\sec 	heta - 	an 	heta = {e^{ - x}}$&hellip;..$(ii)$

From $(i)$ and $(ii),$

$\,2\sec 	heta = {e^x} + {e^{ - x}}\,$

$\Rightarrow \,\cos 	heta = \frac{2}{{{e^x} + {e^{ - x}}}}.$

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

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