If tan θ + sec θ = eˣ, then cos θ equals(a) (eˣ + e⁻ˣ)/2(b) 2/(eˣ + e⁻ˣ)(c) (eˣ − e⁻ˣ)/2(d) (eˣ − e⁻ˣ)/(eˣ + e⁻ˣ)

Question

\[ \text{If } \tan\theta+\sec\theta=e^x, \]

\[ \text{then } \cos\theta \text{ equals} \]

(a) \(\dfrac{e^x+e^{-x}}{2}\)
(b) \(\dfrac{2}{e^x+e^{-x}}\)
(c) \(\dfrac{e^x-e^{-x}}{2}\)
(d) \(\dfrac{e^x-e^{-x}}{e^x+e^{-x}}\)

Using identity

\[ (\sec\theta+\tan\theta) (\sec\theta-\tan\theta)=1 \]

\[ \sec\theta-\tan\theta=e^{-x} \]

Adding,

\[ 2\sec\theta = e^x+e^{-x} \]

\[ \sec\theta = \frac{e^x+e^{-x}}{2} \]

\[ \cos\theta = \frac{2}{e^x+e^{-x}} \]

\[ \boxed{\frac{2}{e^x+e^{-x}}} \]

Correct Option: (b)

Spread the love