Question
If $\theta=30^\circ,$ verify that.
$\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}$
$\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}$
✓
Answer
Given:
$\theta=30^\circ\ \dots(1)$
To verify:
$\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}\ \dots(2)$
$\sin2\theta=\sin2\times30$
$=\sin60$
$=\frac{\sqrt{3}}{2}$
Now consider right hand side
$\frac{2\tan\theta}{1+\tan^2\theta}=\frac{2\tan30}{1+\tan^230}$
$=\frac{2\times\frac{1}{\sqrt{3}}}{1+\Big(\frac{1}{\sqrt{3}}\Big)^2}$
$=\frac{\sqrt{3}}{2}$
Hence it is verified that,
$\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}$
$\theta=30^\circ\ \dots(1)$
To verify:
$\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}\ \dots(2)$
$\sin2\theta=\sin2\times30$
$=\sin60$
$=\frac{\sqrt{3}}{2}$
Now consider right hand side
$\frac{2\tan\theta}{1+\tan^2\theta}=\frac{2\tan30}{1+\tan^230}$
$=\frac{2\times\frac{1}{\sqrt{3}}}{1+\Big(\frac{1}{\sqrt{3}}\Big)^2}$
$=\frac{\sqrt{3}}{2}$
Hence it is verified that,
$\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}$
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