Question

Prove: sec2θ=csc^2θ/(2 cot^2θ-csc^2 θ)

Answer 1

`\bf \textit{Double Angle Identities} \\ \quad \\ sin(2\theta)=2sin(\theta)cos(\theta) \\ \quad \\ cos(2\theta)= \begin{cases} cos^2(\theta)-sin^2(\theta)\\ 1-2sin^2(\theta)\\ \boxed{2cos^2(\theta)-1} \end{cases} \\ \quad \\\\ tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\\\\ -------------------------------\\\\`


`\bf sec(\theta)=\cfrac{csc^2(\theta)}{2cot^2(\theta)-csc^2(\theta)}\\\\ -------------------------------\\\\ \textit{doing the right-hand-side} \\\\\\ \cfrac{(1)/(sin^2(\theta))}{(2cos^2(\theta))/(sin^2(\theta))-(1)/(sin^2(\theta))}\implies \cfrac{(1)/(sin^2(\theta))}{(2cos^2(\theta)-1)/(sin^2(\theta))}\implies \cfrac{1}{sin^2(\theta)}\cdot \cfrac{sin^2(\theta)}{2cos^2(\theta)-1} \\\\\\ \cfrac{1}{2cos^2(\theta)-1}\implies \cfrac{1}{cos(2\theta)}\implies sec(2\theta)`