If $\tan A+\sec A=e^x$, find $\cos A$ If $\tan A+\sec A=e^x$, find $\cos A$
Here is what I've tried:
\begin{align}&\frac{\sin A}{\cos A}+\frac{1}{\cos A}&=e^x\\
\implies&\frac{1+\sin A}{\cos A}&=e^x\end{align}
Now, I've squared both sides, but, in vain. How should I continue? Am I proceeding wrong? Any help is much appreciated.
 A: HINT:
As $\sec A+\tan A=e^x$
Now  $\sec A-\tan A=\dfrac1{\sec A+\tan A}=?$
Can you solve for $\sec A?$
Finally $\cos A\cdot\sec A=?$
A: HINT
shorthand instead of latex
$$ \frac{\sqrt{1-c^2}}{c} +  \frac{1}{c} = e $$
Simplify with algebra after squaring, make it a quadratic in $c= \cos A.$ Then one root is $ c= \cos A=0,$ and the other
$$ c\,=  \frac{2e}{e^2+1}$$
$$ c= \cos A =  \frac{1}{\cosh x }$$
A: \begin{align}
\frac{1+\sin A}{\cos A}&=\frac{(\cos \frac A2+\sin \frac A2)^2}{\cos^2 \frac A2-\sin^2 \frac A 2}\\
&=\frac{\cos \frac A2+\sin \frac A2}{\cos \frac A2-\sin \frac A 2}\\
\end{align}
implying that $$(1-e^x)\cos \frac A2 =-(1+e^x)\sin \frac A2$$
the rest shall be doable for you.
A: $$\tan A+\sec A=e^x\implies\tan A+\frac{1}{\cos A}=e^x\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\implies\frac{1}{\cos A}=e^x-\tan A\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\implies\cos A=\frac{1}{e^x-\tan A}.$$
A: Given 
tanA+secA=e^x.   ------------(I)
Calculate.   ( secA-tanA) using tan^2A+1=sec^2A
secA-tanA=1/e^x.  ------------(ii)
Add.  (i)&(ii)
2secA=e^x+1/e^x
And then calculate cosA
