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