Question
Prove the following trigonometric identities.
$\frac{1}{1+\sin\text{A}}+\frac{1}{1-\sin\text{A}}=2\sec^2\text{A}$
$\frac{1}{1+\sin\text{A}}+\frac{1}{1-\sin\text{A}}=2\sec^2\text{A}$
✓
Answer
$\text{L.H.S}=\frac{1-\sin\text{A}+1+\sin\text{A}}{(1+\sin\text{A})(1-\sin\text{A})}$
$\Rightarrow\ \frac{2}{1-\sin^2\text{A}} \big[\because (1+\sin\text{A})(1-\sin\text{A})=1-\sin^2\text{A}\big]$
$\Rightarrow\ \frac{2}{\cos^2\text{A}}\Rightarrow 2\sec^2\text{A} \big[\because 1-\sin\text{A}=\cos\text{A}\big]$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
Hence proved
$\Rightarrow\ \frac{2}{1-\sin^2\text{A}} \big[\because (1+\sin\text{A})(1-\sin\text{A})=1-\sin^2\text{A}\big]$
$\Rightarrow\ \frac{2}{\cos^2\text{A}}\Rightarrow 2\sec^2\text{A} \big[\because 1-\sin\text{A}=\cos\text{A}\big]$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
Hence proved
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