For 0<θ<π2 , the solution of 6∑m=1cosec(θ+(m−1)π4)cosec(θ+mπ4)=4√2 is/are
We have,
6∑m=1cosec[θ+(m−1)π4]cosec(θ+mπ4)=4√2
6∑m=11sin[θ+(m−1)π4]sin(θ+mπ4)=4√2
6∑m=11sin[θ+(m−1)π4]sin(θ+mπ4)=4√2
1sinθsin(θ+π4)+1sin(θ+π4)sin(θ+π2)+.....+1sin(θ+5π4)sin(θ+3π2)=4√2
1sinπ4⎡⎢ ⎢ ⎢ ⎢⎣sin(θ+π4−θ)sinθsin(θ+π4)+sin(θ+π2−(θ+π4))sin(θ+π4)sin(θ+π2)+.....+sin(θ+3π2−(θ+5π4))sin(θ+5π4)sin(θ+3π2)⎤⎥ ⎥ ⎥ ⎥⎦=4√2
√2⎡⎢ ⎢ ⎢ ⎢⎣sin(θ+π4)cosθ−cos(θ+π4)sinθsinθsin(θ+π4)+sin(θ+π2)cos(θ+π4)−cos(θ+π2)sin(θ+π4)sin(θ+π4)sin(θ+π2)+.....⎤⎥ ⎥ ⎥ ⎥⎦=4√2
[cotθ−cot(θ+π4)+cot(θ+π4)−cot(θ+π2)+.....+cot(θ+5π4)−cot(θ+3π2)]=4
cotθ−cot(θ+π2)=4
cotθ+tanθ=4
tan2θ−4tanθ+1=0
tanθ=4±√16−4×1×12
tanθ=4±√16−42
tanθ=4±√122
tanθ=4±2√32
tanθ=2±√3
θ=π12 or 5π12
Hence, the value of θ is π12 or 5π12