    Question

# For 0<θ<π2, the solution (s) of ∑6m=1cosec(θ+(m−1)π4)cosec(θ+mπ4)=4√2 is/are

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Solution

## The correct options are C D For 0<θ<π2 ∑6m=1cosec(θ+(m−1)π4)cosec(θ+mπ4)=4√2 ⇒∑6m=11sin(θ+(m−1)π4)sin(θ+mπ4)=4√2 ⇒∑6m=1sin[θ+mπ4−(θ+(m−1)π4)]sin(θ+(m−1)π4)sin(θ+mπ4)=4 ⇒∑6m=1[cot(θ+(m−1)π4)−cot(θ+mπ4)]=4 ⇒cot(θ)−cot(θ+π4)+cot(θ+π4) −cot(θ+2π4)+...+cot(θ+5π4)−cot(θ+6π4)=4 ⇒cotθ−cot(3π2+θ)=4 ⇒cotθ+tanθ=4 ⇒tan2θ−4tanθ+1=0 ⇒(tanθ−2)2−3=0 ⇒(tanθ−2+√3)(tanθ−2−√3)=0 ⇒tanθ=2−√3ortanθ=2+√3 ⇒θ=π12;θ=5π12 ∵θ∈(0,θ2)  Suggest Corrections  0      Similar questions
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