Question
Prove the following trigonometric identities.
$\sec\text{A}(1-\sin\text{A})(\sec\text{A}+\tan\text{A})=1$
$\sec\text{A}(1-\sin\text{A})(\sec\text{A}+\tan\text{A})=1$
✓
Answer
$\text{L.H.S.}=\sec\text{A}(1-\sin\text{A})(\sec\text{A}+\tan\text{A})$
$=\frac{1}{\cos\text{A}}(1-\sin\text{A})\Big(\frac{1}{\cos\text{A}}+\frac{\sin\text{A}}{\cos\text{A}}\Big)$
$=\Big(\frac{1-\sin\text{A}}{\cos\text{A}}\Big)\Big(\frac{1+\sin\text{A}}{\cos\text{A}}\Big)$
$=\frac{1-\sin^2\text{A}}{\cos^2\text{A}}=\frac{\cos^2\text{A}}{\cos^2\text{A}}$
$=1=\text{R.H.S.}$
$=\frac{1}{\cos\text{A}}(1-\sin\text{A})\Big(\frac{1}{\cos\text{A}}+\frac{\sin\text{A}}{\cos\text{A}}\Big)$
$=\Big(\frac{1-\sin\text{A}}{\cos\text{A}}\Big)\Big(\frac{1+\sin\text{A}}{\cos\text{A}}\Big)$
$=\frac{1-\sin^2\text{A}}{\cos^2\text{A}}=\frac{\cos^2\text{A}}{\cos^2\text{A}}$
$=1=\text{R.H.S.}$
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