Question
If $\theta=30^\circ,$ verify that.
$\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}$
$\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}$
✓
Answer
$\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}$
Substitutc $\theta=30^\circ$
$\text{L.H.S}=\cos2\theta,\ \text{R.H.S}=\frac{1-\tan^2\theta}{1+\tan^2\theta}$
$=\cos2(30^\circ)=\frac{1-\tan^230^\circ}{1+\tan^230^\circ}$
$=\cos60^\circ=\frac{1}{2}=\frac{1-\Big(\frac{1}{\sqrt{3}}\Big)^2}{1+\Big(\frac{1}{\sqrt{3}}\Big)^2}=\frac{1-\frac{1}{3}}{1+\frac{1}{3}}=\frac{\frac{2}{3}}{\frac{4}{3}}=\frac{1}{2}$
Substitutc $\theta=30^\circ$
$\text{L.H.S}=\cos2\theta,\ \text{R.H.S}=\frac{1-\tan^2\theta}{1+\tan^2\theta}$
$=\cos2(30^\circ)=\frac{1-\tan^230^\circ}{1+\tan^230^\circ}$
$=\cos60^\circ=\frac{1}{2}=\frac{1-\Big(\frac{1}{\sqrt{3}}\Big)^2}{1+\Big(\frac{1}{\sqrt{3}}\Big)^2}=\frac{1-\frac{1}{3}}{1+\frac{1}{3}}=\frac{\frac{2}{3}}{\frac{4}{3}}=\frac{1}{2}$
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