Question
Show that $\sec x+\tan x=\sqrt{\frac{1+\sin x}{1-\sin x}}$
✓
Answer
$\begin{aligned}
\sec x+\tan x & =\frac{1}{\cos x}+\frac{\sin x}{\cos x} \\
& =\frac{1+\sin x}{\cos x} \\
& =\sqrt{\frac{(1+\sin x)^2}{\cos ^2 x}} \\
& =\sqrt{\frac{(1+\sin x)(1+\sin x)}{1-\sin ^2 x}} \\
& =\sqrt{\frac{(1+\sin x)(1+\sin x)}{(1-\sin x)(1+\sin x)}} \\
& =\sqrt{\frac{1+\sin x}{1-\sin x}}
\end{aligned}
$
\sec x+\tan x & =\frac{1}{\cos x}+\frac{\sin x}{\cos x} \\
& =\frac{1+\sin x}{\cos x} \\
& =\sqrt{\frac{(1+\sin x)^2}{\cos ^2 x}} \\
& =\sqrt{\frac{(1+\sin x)(1+\sin x)}{1-\sin ^2 x}} \\
& =\sqrt{\frac{(1+\sin x)(1+\sin x)}{(1-\sin x)(1+\sin x)}} \\
& =\sqrt{\frac{1+\sin x}{1-\sin x}}
\end{aligned}
$
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