Question
Prove the following identity :
secA(1 + sinA)(secA - tanA) = 1
secA(1 + sinA)(secA - tanA) = 1
✓
Answer
$
\begin{aligned}
& \text { LHS }=\sec \mathrm{A}(1+\sin \mathrm{A})(\sec \mathrm{A}-\tan \mathrm{A}) \\
& =\frac{1}{\cos A}(1+\sin A)\left(\frac{1}{\cos A}-\frac{\sin A}{\cos A}\right) \\
& =\frac{(1+\sin A)}{\cos A}\left(\frac{1-\sin A}{\cos A}\right)=\frac{1-\sin ^2 A}{\cos ^2 A} \\
& =\left(\frac{\cos ^2 A}{\cos ^2 A}\right)=1=\text { RHS }
\end{aligned}
$
\begin{aligned}
& \text { LHS }=\sec \mathrm{A}(1+\sin \mathrm{A})(\sec \mathrm{A}-\tan \mathrm{A}) \\
& =\frac{1}{\cos A}(1+\sin A)\left(\frac{1}{\cos A}-\frac{\sin A}{\cos A}\right) \\
& =\frac{(1+\sin A)}{\cos A}\left(\frac{1-\sin A}{\cos A}\right)=\frac{1-\sin ^2 A}{\cos ^2 A} \\
& =\left(\frac{\cos ^2 A}{\cos ^2 A}\right)=1=\text { RHS }
\end{aligned}
$
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