how to prove $\tan A+\sec A=\frac{1}{(\sec A-\tan A)}$ ?
I already tried:
$$\begin{align}
\sin A/\cos A+1/\cos A&=1/(\sec A-\tan A)\\
\sin A+1/\cos A&=1/(\sec A-\tan A)\\
\end{align}$$
[Math] how to prove $\tan A+\sec A=\frac{1}{(\sec A-\tan A)}$
trigonometry
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Best Answer
$\mathbf{HINT:}$$$ \begin{align} (\tan A+\sec A)(\sec A-\tan A)&=\require{cancel}\cancel{\tan A\sec A}-\tan^2 A+\sec^2 A\,\, \require{cancel}\cancel{\ -\tan A \sec A} \\ &=\sec^2 A -\tan^2 A\equiv1 \end{align}$$