JEE maths#13

Geometry Level 4

If cot ⁡ ( θ − α ) \cot ( \theta -\alpha) , 3 cot ⁡ θ 3 \cot \theta , and cot ⁡ ( θ + α ) \cot ( \theta + \alpha) are in an arithmetic progression and θ \theta is not a integral multiple of π 2 \dfrac \pi 2 , then find the value of 2 sin ⁡ 2 θ sin ⁡ 2 α \dfrac{2 \sin^{2} \theta}{ \sin^{2} \alpha} .

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The answer is 3.

Rahil Sehgal by Rahil Sehgal , 20, India

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1 solution

If a a , b b and c c are in an arithmetic progression, then a + c = 2 b a+c = 2b . Therefore,

cot ⁡ ( θ − α ) + cot ⁡ ( θ + α ) = 2 ⋅ 3 cot ⁡ θ cot ⁡ α cot ⁡ θ + 1 cot ⁡ α − cot ⁡ θ + cot ⁡ α cot ⁡ θ − 1 cot ⁡ α + cot ⁡ θ = 6 cot ⁡ θ ( cot ⁡ α + cot ⁡ θ ) ( cot ⁡ α cot ⁡ θ + 1 ) + ( cot ⁡ α − cot ⁡ θ ) ( cot ⁡ α cot ⁡ θ − 1 ) ( cot ⁡ α − cot ⁡ θ ) ( cot ⁡ α + cot ⁡ θ ) = 6 cot ⁡ θ 2 cot ⁡ 2 α cot ⁡ θ + 2 cot ⁡ θ cot ⁡ 2 α − cot ⁡ 2 θ = 6 cot ⁡ θ cot ⁡ 2 α + 1 = 3 cot ⁡ 2 α − 3 cot ⁡ 2 θ 3 cot ⁡ 2 θ = 2 cot ⁡ 2 α − 1 3 cos ⁡ 2 θ sin ⁡ 2 θ = 2 cos ⁡ 2 α sin ⁡ 2 α − 1 3 ( 1 − sin ⁡ 2 θ ) sin ⁡ 2 θ = 2 ( 1 − sin ⁡ 2 α ) sin ⁡ 2 α − 1 3 sin ⁡ 2 θ − 3 = 2 sin ⁡ 2 α − 2 − 1 3 sin ⁡ 2 θ = 2 sin ⁡ 2 α ⟹ 2 sin ⁡ 2 θ sin ⁡ 2 α = 3 \begin{aligned} \cot (\theta - \alpha) + \cot (\theta + \alpha) & = 2\cdot 3 \cot \theta \\ \frac {\cot \alpha \cot \theta + 1}{\cot \alpha - \cot \theta} + \frac {\cot \alpha \cot \theta - 1}{\cot \alpha + \cot \theta} & = 6 \cot \theta \\ \frac {(\cot \alpha + \cot \theta)(\cot \alpha \cot \theta + 1) + (\cot \alpha - \cot \theta)(\cot \alpha \cot \theta - 1)}{(\cot \alpha - \cot \theta)(\cot \alpha + \cot \theta)} & = 6 \cot \theta \\ \frac {2 \cot^2 \alpha \cot \theta + 2 \cot \theta}{\cot^2 \alpha - \cot^2 \theta} & = 6 \cot \theta \\ \cot^2 \alpha + 1 & = 3 \cot^2 \alpha - 3 \cot^2 \theta \\ 3 \cot^2 \theta & = 2 \cot^2 \alpha - 1 \\ \frac {3\cos^2 \theta}{\sin^2 \theta} & = \frac {2\cos^2 \alpha}{\sin^2 \alpha} - 1 \\ \frac {3 (1 - \sin^2 \theta)}{\sin^2 \theta} & = \frac {2(1- \sin^2 \alpha)}{\sin^2 \alpha} - 1 \\ \frac 3{\sin^2 \theta} - 3 & = \frac 2{\sin^2 \alpha} - 2 - 1 \\ \frac 3{\sin^2 \theta} & = \frac 2{\sin^2 \alpha} \\ \implies \frac {2\sin^2 \theta}{\sin^2 \alpha} & = \boxed{3} \end{aligned}

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