(i)[4]
Prove that $\left(\frac{1}{\cos x} - \tan x\right)^2 = \frac{1 - \sin x}{1 + \sin x}$.
(ii)[3]
Hence solve $\left( \frac{1}{\cos 2x} - \tan 2x \right)^2 = \frac{1}{3}$ for $0 \leq x \leq \pi$.
Mathematics 9709 · AS & A Level · Trigonometry
Mathematics 9709 · AS & A Level · Trigonometry
Prove that $\left(\frac{1}{\cos x} - \tan x\right)^2 = \frac{1 - \sin x}{1 + \sin x}$.
Hence solve $\left( \frac{1}{\cos 2x} - \tan 2x \right)^2 = \frac{1}{3}$ for $0 \leq x \leq \pi$.