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Eccentric-Angle Sum for Equilateral Triangle | JEE

JEE Maths question with a full step-by-step solution.

Question
If the circumcentre of an equilateral triangle inscribed in x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1, with vertices having eccentric angles α,β,γ\alpha,\beta,\gamma, is (x1,y1)(x_1,y_1), then ∑cos⁡αcos⁡β+∑sin⁡αsin⁡β\sum\cos\alpha\cos\beta+\sum\sin\alpha\sin\beta is:
A9x12a2+9y12b2+32\dfrac{9x_1^2}{a^2}+\dfrac{9y_1^2}{b^2}+\dfrac{3}{2}
B9x12−9y12+a2b29x_1^2-9y_1^2+a^2b^2
C9x122a2+9y122b2−32\dfrac{9x_1^2}{2a^2}+\dfrac{9y_1^2}{2b^2}-\dfrac{3}{2}correct
D9x12a2+9y12b2+3\dfrac{9x_1^2}{a^2}+\dfrac{9y_1^2}{b^2}+3
Solution
Step 1: For an equilateral triangle the circumcentre coincides with the centroid, so
x1=a3∑cos⁡α,y1=b3∑sin⁡α  ⇒  ∑cos⁡α=3x1a, ∑sin⁡α=3y1b.x_1=\frac{a}{3}\sum\cos\alpha,\quad y_1=\frac{b}{3}\sum\sin\alpha \;\Rightarrow\; \sum\cos\alpha=\frac{3x_1}{a},\ \sum\sin\alpha=\frac{3y_1}{b}.
 Step 2: Use the identities
∑cos⁡αcos⁡β=(∑cos⁡α)2−∑cos⁡2α2,∑sin⁡αsin⁡β=(∑sin⁡α)2−∑sin⁡2α2.\sum\cos\alpha\cos\beta=\frac{(\sum\cos\alpha)^2-\sum\cos^2\alpha}{2},\qquad \sum\sin\alpha\sin\beta=\frac{(\sum\sin\alpha)^2-\sum\sin^2\alpha}{2}.
 Step 3: Add and use ∑(cos⁡2α+sin⁡2α)=3\sum(\cos^2\alpha+\sin^2\alpha)=3:
∑cos⁡αcos⁡β+∑sin⁡αsin⁡β=(∑cos⁡α)2+(∑sin⁡α)2−32.\sum\cos\alpha\cos\beta+\sum\sin\alpha\sin\beta=\frac{(\sum\cos\alpha)^2+(\sum\sin\alpha)^2-3}{2}.
 Step 4: Substitute the sums.
=12(9x12a2+9y12b2−3)=9x122a2+9y122b2−32.=\frac{1}{2}\left(\frac{9x_1^2}{a^2}+\frac{9y_1^2}{b^2}-3\right)=\frac{9x_1^2}{2a^2}+\frac{9y_1^2}{2b^2}-\frac{3}{2}.
 Correct answer: (3)
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