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Mathematics

If A = 30°, prove that :

(i) sin 2A = 2tan⁡A1+tan⁡2A\dfrac{2 \tan A}{1 + \tan^2A}

(ii) cos 2A = (1−tan⁡2A1+tan⁡2A)\Big(\dfrac{1 - \tan^2A}{1 + \tan^2A}\Big)

Trigonometrical Ratios

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Answer

(i) Left Hand Side:

sin 2A = sin 2(30°) = sin 60°

= 32\dfrac{\sqrt{3}}{2}

Right Hand Side:

2tan⁡A1+tan⁡2A=2×131+13=2343=23×34=32.\dfrac{2 \tan A}{1 + \tan^2A} = \dfrac{2\times \dfrac{1}{\sqrt{3}}}{1 + \dfrac{1}{3}}\\[1em] = \dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{4}{3}}\\[1em] = \dfrac{2}{\sqrt{3}}\times \dfrac{3}{4}\\[1em] = \dfrac{\sqrt{3}}{2}.

Hence, proved that sin 2A = 2tan⁡A1+tan⁡2A\dfrac{2 \tan A}{1 + \tan^2A}

(ii) Left Hand Side :

cos 2A = cos 2(30°) = cos 60°

= 12\dfrac{1}{2}

Right Hand Side :

1−tan⁡2A1+tan⁡2A=1−tan⁡230°1+tan⁡230°=1−(13)21+(13)2=1−131+13=2343=12.\dfrac{1 - \tan^2A}{1 + \tan^2A} \\[1em] = \dfrac{1 - \tan^230°}{1 + \tan^230°} \\[1em] =\dfrac{1 - \Big(\dfrac{1}{\sqrt{3}}\Big)^2}{1 + \Big(\dfrac{1}{\sqrt{3}}\Big)^2}\\[1em] = \dfrac{1 - \dfrac{1}{3}}{1 + \dfrac{1}{3}} \\[1em] = \dfrac{\dfrac{2}{3}}{\dfrac{4}{3}}\\[1em] = \dfrac{1}{2}.

Hence, proved that cos 2A = (1−tan⁡2A1+tan⁡2A)\Big(\dfrac{1 - \tan^2A}{1 + \tan^2A}\Big).

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