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Mathematics

If A = 30°, then prove that :

cos 2 A=cos2 A−sin2 A=1−tan2 A1+tan2 A\text{cos 2 A} = \text{cos}^2 \text{ A} - \text{sin}^2 \text{ A} = \dfrac{1 - \text{tan}^2 \text{ A}}{1 + \text{tan}^2 \text{ A}}

Trigonometric Identities

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Answer

cos 2 A=cos2 A−sin2 A=1−tan2 A1+tan2 A\text{cos 2 A} = \text{cos}^2 \text{ A} - \text{sin}^2 \text{ A} = \dfrac{1 - \text{tan}^2 \text{ A}}{1 + \text{tan}^2 \text{ A}}

1st term=cos 2 A=cos (2 x 30°)=cos 60°=12\text{1st term} = \text{cos 2 A} = \text{cos (2 x 30°)} = \text{cos 60°} = \dfrac{1}{2}

2nd term=cos2 A−sin2 A=cos2 30°−sin2 30°=(32)2−(12)2=34−14=3−14=24=12\text{2nd term} = \text{cos}^2 \text{ A} - \text{sin}^2 \text{ A}\\[1em] = \text{cos}^2 \text{ 30°} - \text{sin}^2 \text{ 30°}\\[1em] = \Big(\dfrac{\sqrt3}{2}\Big)^2 - \Big(\dfrac{1}{2}\Big)^2\\[1em] = \dfrac{3}{4} - \dfrac{1}{4}\\[1em] = \dfrac{3 - 1}{4}\\[1em] = \dfrac{2}{4}\\[1em] = \dfrac{1}{2}

3rd term=1−tan2 A1+tan2 A=1−tan2 30°1+tan2 30°=1−(13)21+(13)2=33−1333+13=3−133+13=2343=2343=24=12\text{3rd term} = \dfrac{1 - \text{tan}^2 \text{ A}}{1 + \text{tan}^2 \text{ A}}\\[1em] = \dfrac{1 - \text{tan}^2 \text{ 30°}}{1 + \text{tan}^2 \text{ 30°}}\\[1em] = \dfrac{1 - \Big(\dfrac{1}{\sqrt3}\Big)^2}{1 + \Big(\dfrac{1}{\sqrt3}\Big)^2}\\[1em] = \dfrac{\dfrac{3}{3} - \dfrac{1}{3}}{\dfrac{3}{3} + \dfrac{1}{3}}\\[1em] = \dfrac{\dfrac{3 - 1}{3}}{\dfrac{3 + 1}{3}}\\[1em] = \dfrac{\dfrac{2}{3}}{\dfrac{4}{3}}\\[1em] = \dfrac{\dfrac{2}{\cancel{3}}}{\dfrac{4}{\cancel{3}}}\\[1em] = \dfrac{2}{4}\\[1em] = \dfrac{1}{2}

∴ 1st term = 2nd term = 3rd term

Hence, cos 2 A=cos2 A−sin2 A=1−tan2 A1+tan2 A\text{cos 2 A} = \text{cos}^2 \text{ A} - \text{sin}^2 \text{ A} = \dfrac{1 - \text{tan}^2 \text{ A}}{1 + \text{tan}^2 \text{ A}}

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