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Maths Question 3 – JEE-MAIN 2025

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Let A={θ∈[0,2π]:1+10 Re (2cos⁡θ+isin⁡θcos⁡θ−3isin⁡θ)=0}. Then ∑θ∈Aθ2 is equal to

The problem requires finding the real part of a complex fraction and then solving a trigonometric equation.

Step 1: Simplify the Complex Expression✦ Active

First, we need to find the real part of the complex fraction 2cos⁡θ+isin⁡θcos⁡θ−3isin⁡θ. We multiply the numerator and denominator by the conjugate of the denominator.

Re(2cos⁡θ+isin⁡θcos⁡θ−3isin⁡θ)=Re((2cos⁡θ+isin⁡θ)(cos⁡θ+3isin⁡θ)(cos⁡θ−3isin⁡θ)(cos⁡θ+3isin⁡θ)) =Re(2cos2⁡θ+6isin⁡θcos⁡θ+isin⁡θcos⁡θ+3i2sin2⁡θcos2⁡θ−(3isin⁡θ)2) =Re(2cos2⁡θ−3sin2⁡θ+i(7sin⁡θcos⁡θ)cos2⁡θ+9sin2⁡θ)=2cos2⁡θ−3sin2⁡θcos2⁡θ+9sin2⁡θ
Step 2: Solve the Trigonometric Equation○ Expand

Substitute the real part back into the given equation 1+10 Re (2cos⁡θ+isin⁡θcos⁡θ−3isin⁡θ)=0 and solve for θ in the interval [0,2π].

1+10(2cos2⁡θ−3sin2⁡θcos2⁡θ+9sin2⁡θ)=0 10(2cos2⁡θ−3sin2⁡θ)=−(cos2⁡θ+9sin2⁡θ) 20cos2⁡θ−30sin2⁡θ=−cos2⁡θ−9sin2⁡θ 21cos2⁡θ=21sin2⁡θ⟹cos2⁡θ=sin2⁡θ⟹tan2⁡θ=1

This implies tan⁡θ=1 or tan⁡θ=−1. The solutions for θ∈[0,2π] are: For tan⁡θ=1: θ=π4,5π4 For tan⁡θ=−1: θ=3π4,7π4 Thus, the set A={π4,3π4,5π4,7π4}.

Step 3: Calculate the Sum of Squares○ Expand

Finally, we calculate the sum of the squares of the values of θ in set A.

∑θ∈Aθ2=(π4)2+(3π4)2+(5π4)2+(7π4)2 =π216(12+32+52+72) =π216(1+9+25+49) =π216(84)=21π24
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