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← IIT-JEE 2009 · Paper 2
MathematicsMultiple correct+4 / -1Question 28 of 57

IIT-JEE 2009 · Paper 2, Mathematics question 28

For 0<θ<π20<\theta<\dfrac\pi2, the solution(s) of
∑m=16cosec⁡(θ+(m−1)π4)cosec⁡(θ+mπ4)=42\sum_{m=1}^6\operatorname{cosec}\left(\theta+\frac{(m-1)\pi}4\right)\operatorname{cosec}\left(\theta+\frac{m\pi}4\right)=4\sqrt2
is(are)

Choose one or more options
Answer and worked solution

Official answer: C, D

Telescoping. For angles differing by π4\frac\pi4:
cosec⁡Acosec⁡B=sin⁡(B−A)sin⁡π4sin⁡Asin⁡B=2(cot⁡A−cot⁡B).\operatorname{cosec}A\operatorname{cosec}B=\frac{\sin(B-A)}{\sin\frac\pi4\sin A\sin B}=\sqrt2(\cot A-\cot B).

Sum. 2[cot⁡θ−cot⁡(θ+3π2)]=2(cot⁡θ+tan⁡θ)=22sin⁡2θ\sqrt2\left[\cot\theta-\cot\left(\theta+\frac{3\pi}2\right)\right]=\sqrt2(\cot\theta+\tan\theta)=\dfrac{2\sqrt2}{\sin2\theta}.

Solve. 22sin⁡2θ=42\dfrac{2\sqrt2}{\sin2\theta}=4\sqrt2 gives sin⁡2θ=12\sin2\theta=\frac12: θ=π12\theta=\frac\pi{12} or 5π12\frac{5\pi}{12}.

(C), (D)

Topics:Trigonometric equationsTrigonometric ratios and identities
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