If A = 30°, verify that:
Question:

If A = 30°, verify that:

(i) $\sin 2 A=\frac{2 \tan A}{1+\tan ^{2} A}$

(ii) $\cos 2 A=\frac{1-\tan ^{2} A}{1+\tan ^{2} A}$

(iii) $\tan 2 A=\frac{2 \tan A}{1-\tan ^{2} A}$

Solution:

A = 30o

$\Rightarrow 2 A=2 \times 30^{\circ}=60^{\circ}$

(i) $\sin 2 A=\sin 60^{\circ}=\frac{\sqrt{3}}{2}$

$\frac{2 \tan A}{1+\tan ^{2} A}=\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=\frac{2 \times\left(\frac{1}{\sqrt{3}}\right)}{1+\left(\frac{1}{\sqrt{3}}\right)^{2}}=\frac{\left(\frac{2}{\sqrt{3}}\right)}{1+\frac{1}{3}}=\frac{\left(\frac{2}{\sqrt{3}}\right)}{\frac{4}{3}}=\left(\frac{2}{\sqrt{3}}\right) \times \frac{3}{4}=\frac{\sqrt{3}}{2}$

$\therefore \sin 2 A=\frac{2 \tan A}{1+\tan ^{2} A}$

(ii) $\cos 2 \mathrm{~A}=\cos 60^{\circ}=\frac{1}{2}$

$\frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\frac{1-\tan ^{2} 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=\frac{1-\left(\frac{1}{\sqrt{3}}\right)^{2}}{1+\left(\frac{1}{\sqrt{3}}\right)^{2}}=\frac{\left(1-\frac{1}{3}\right)}{\left(1+\frac{1}{3}\right)}=\frac{\left(\frac{2}{3}\right)}{\frac{4}{3}}=\left(\frac{2}{3}\right) \times \frac{3}{4}=\frac{1}{2}$

$\therefore \cos 2 A=\frac{1-\tan ^{2} A}{1+\tan ^{2} A}$

(iii) $\tan 2 \mathrm{~A}=\tan 60^{\circ}=\sqrt{3}$

$\frac{2 \tan A}{1-\tan ^{2} A}=\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}=\frac{2 \times\left(\frac{1}{\sqrt{3}}\right)}{1-\left(\frac{1}{\sqrt{3}}\right)^{2}}=\frac{\left(\frac{2}{\sqrt{3}}\right)}{1-\frac{1}{3}}=\frac{\left(\frac{2}{\sqrt{3}}\right)}{\frac{2}{3}}=\left(\frac{2}{\sqrt{3}}\right) \times \frac{3}{2}=\sqrt{3}$

$\therefore \tan 2 A=\frac{2 \tan A}{1-\tan ^{2} A}$

 

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