Question
Prove the following trigonometric identities.
$\frac{1+\cos\text{A}}{\sin^2\text{A}}=\frac{1}{1-\cos\theta}$
$\frac{1+\cos\text{A}}{\sin^2\text{A}}=\frac{1}{1-\cos\theta}$
✓
Answer
We know that $\sin^2\text{A}+\cos^2\text{A}=1$
$\sin^2\text{A}=1-\cos^2\text{A}$
$\Rightarrow\ \sin^2\text{A}=(1-\cos\text{A})(1+\cos\text{A})$
$\Rightarrow\ \text{L.H.S}=\frac{(1+\cos\text{A})}{(1-\cos\text{A})(1+\cos\text{A})}=\frac{1}{1-\cos\text{A}}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
Hence proved.
$\sin^2\text{A}=1-\cos^2\text{A}$
$\Rightarrow\ \sin^2\text{A}=(1-\cos\text{A})(1+\cos\text{A})$
$\Rightarrow\ \text{L.H.S}=\frac{(1+\cos\text{A})}{(1-\cos\text{A})(1+\cos\text{A})}=\frac{1}{1-\cos\text{A}}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
Hence proved.
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