Question
Evaluate the following:
If A = 30°, verify that:
$\cos2\text{A}=\frac{1-\tan^2\text{A}}{1+\tan^2\text{A}}$
If A = 30°, verify that:
$\cos2\text{A}=\frac{1-\tan^2\text{A}}{1+\tan^2\text{A}}$
✓
Answer
$\text{A}=30^\circ$
$\Rightarrow2\text{A}=2\times30^\circ=60^\circ$
$\cos2\text{A}=\cos60^\circ=\frac12$
$\frac{1-\tan^2\text{A}}{1+\tan^2\text{A}}=\frac{1-\tan^230^\circ}{1+\tan^230^\circ}$
$=\frac{1-\Big(\frac{1}{\sqrt{3}}\Big)^2}{1+\Big(\frac{1}{\sqrt{3}}\Big)^2}=\frac{\Big(1-\frac{1}{{3}}\Big)}{\Big(1+\frac13\Big)}$
$=\Big(\frac{2}{3}\Big)\times\frac34=\frac12$
$\therefore\ \cos2\text{A}=\frac{1-\tan^2\text{A}}{1+\tan^2\text{A}}$
$\Rightarrow2\text{A}=2\times30^\circ=60^\circ$
$\cos2\text{A}=\cos60^\circ=\frac12$
$\frac{1-\tan^2\text{A}}{1+\tan^2\text{A}}=\frac{1-\tan^230^\circ}{1+\tan^230^\circ}$
$=\frac{1-\Big(\frac{1}{\sqrt{3}}\Big)^2}{1+\Big(\frac{1}{\sqrt{3}}\Big)^2}=\frac{\Big(1-\frac{1}{{3}}\Big)}{\Big(1+\frac13\Big)}$
$=\Big(\frac{2}{3}\Big)\times\frac34=\frac12$
$\therefore\ \cos2\text{A}=\frac{1-\tan^2\text{A}}{1+\tan^2\text{A}}$
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