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Question: If \(\tan\theta + \sec\theta = e^{x},\) then \(\cos\theta\) equals...

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

A

(ex+e−x)2\frac{(e^{x} + e^{- x})}{2}

B

2(ex+e−x)\frac{2}{(e^{x} + e^{- x})}

C

(ex−e−x)2\frac{(e^{x} - e^{- x})}{2}

D

(ex−e−x)(ex+e−x)\frac{(e^{x} - e^{- x})}{(e^{x} + e^{- x})}

Answer

2(ex+e−x)\frac{2}{(e^{x} + e^{- x})}

Explanation

Solution

tan⁡θ+sec⁡θ=ex\tan\theta + \sec\theta = e^{x} …..(i)

∴sec⁡θ−tan⁡θ=e−x\therefore\sec\theta - \tan\theta = e^{- x} …..(ii)

From (i) and (ii), 2sec⁡θ=ex+e−x⇒cos⁡θ=2ex+e−x.2\sec\theta = e^{x} + e^{- x} \Rightarrow \cos\theta = \frac{2}{e^{x} + e^{- x}}.