Ta có: $1+\frac{1}{\cos 2x}=\frac{1+\cos 2x}{\cos 2x}=\frac{2\cos^2 x}{\cos 2x}$ $=\frac{\cos^2 x.\sin 2x}{2\cos 2x.\sin x.\cos x}=\frac{\tan 2x}{\tan x}$ với $x \neq \frac{\pi}{2} (2)$Gọi $P_k=1+\frac{1}{\cos 2^k a}, k=0,1,2...,n$ ta có: $P_n=P_0.P_1....P_n=\frac{\tan a}{\tan \frac{a}{2} }.\frac{\tan 2a}{\tan a}.\frac{\tan 2^2a}{\tan 2^{n-1}a}=\frac{\tan 2^n a}{\tan \frac{a}{2} }$Vậy $P_n=\frac{\tan 2^n a}{\tan \frac{a}{2} }$
Ta có: $1+\frac{1}{\cos 2x}=\frac{1+\cos 2x}{\cos 2x}=\frac{2\cos^2 x}{\cos 2x}$ $=\frac{\cos^2 x.\sin 2x}{2\cos 2x.\sin x.\cos x}=\frac{\tan 2x}{\tan x}$ với $x \neq \frac{\pi}{2} (2)$Gọi $P_k=1+\frac{1}{\cos 2^k a}, k=0,1,2...,n$ ta có: $P_n=P_0.P_1....P_n=\frac{\tan a}{\tan \frac{a}{2} }.\frac{\tan 2a}{\tan a}.\frac{\tan 2^2a}{\tan 2^{n-1}a}=\frac{\tan 2^n a}{\tan \frac{a}{2} }$Vậy $P_n=\frac{\tan 2^n a}{\tan \frac{a}{2} }
Ta có: $1+\frac{1}{\cos 2x}=\frac{1+\cos 2x}{\cos 2x}=\frac{2\cos^2 x}{\cos 2x}$ $=\frac{\cos^2 x.\sin 2x}{2\cos 2x.\sin x.\cos x}=\frac{\tan 2x}{\tan x}$ với $x \neq \frac{\pi}{2} (2)$Gọi $P_k=1+\frac{1}{\cos 2^k a}, k=0,1,2...,n$ ta có: $P_n=P_0.P_1....P_n=\frac{\tan a}{\tan \frac{a}{2} }.\frac{\tan 2a}{\tan a}.\frac{\tan 2^2a}{\tan 2^{n-1}a}=\frac{\tan 2^n a}{\tan \frac{a}{2} }$Vậy $P_n=\frac{\tan 2^n a}{\tan \frac{a}{2} }
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