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

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

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

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

JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Easy

1 of 1

Application +4 | -1

\tan \theta+\sec \theta=e^x

, then

\cos \theta

equals

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because by using the identity

\sec^2 \theta - \tan^2 \theta = 1

, we can find

\cos \theta

Practice All Chapter MCQs →

\tan \alpha=\frac{p}{q}

\alpha=6 \beta

\alpha

\frac{1}{2}[p \operatorname{cosec} 2 \beta-q \sec 2 \beta]

\cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ}

\triangle A B C

C

{\sin ^6}\,\,\theta + {\cos ^6}\,\,\theta + 3{\sin ^2}\,\,\theta \,\,{\cos ^2}\,\,\theta

\sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}+\sin ^2 \frac{C}{2}=

\sin ^2(A / 2)+\sin ^2(B / 2)-\sin ^2(C / 2)=