\hspace{-16}\bf{P=\left(1+\frac{1}{\cos 1^0}\right).\left(1+\frac{1}{\cos 2^0}\right).\left(1+\frac{1}{\cos 4^0}\right)...\left(1+\frac{1}{\cos 1024^0}\right)}
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1 Answers
9cos2θ+1cosθ=2cosθ
P=2cos212(Πr=010cos2rθ+1cos2r-1)sec1024°
P=211cos12°(Πr=-19cos2r)sec1024°
P=211cot12°tan1024°
i think i did the calculations right basically this is the basis of the sum
