A little Problem in Trigonometry (Multiple Angle)

Question

If $\tan^2 \theta = 1 + 2\tan^2 \phi$, show that $\cos 2\phi = 1 + 2\cos2\theta$.

What I have done..

$$\implies \tan^2 \theta = 1 + 2\tan^2 \phi\\ \implies 1 + \tan^2 \theta = 2 + 2\tan^2 \phi\\ \implies 1 + \tan^2 \theta = 2(1 + \tan^2 \phi)\\ \implies \sec^2 \theta = 2(\sec^2 \phi)$$

Answer 1

Your steps are fine you just need to do the following (I'm putting your steps first): \begin{align*} \tan^{2}\theta & = 1+2 \tan^{2}\phi\\ 1+\tan^{2}\theta & = 2+2 \tan^{2}\phi\\ \sec^2 \theta & = 2 \sec^2 \phi\\ \text{Do the following to continue:}\\ \frac{1}{\cos^2 \theta} & = \frac{2}{\cos^2 \phi}\\ \cos^2 \phi & = 2 \cos^2 \theta\\ \text{using the half angle formula}\\ \frac{1+\cos 2 \phi}{2} & = 1+\cos2 \theta\\ 1+\cos 2 \phi & = 2(1+\cos2 \theta)\\ \cos 2\phi & = 1 +2\cos 2\theta. \end{align*}

Answer 2

Apply Componendo & Dividendo on $$\frac{\tan^2\theta}1=\frac{1+2\tan^2\phi}1$$

$$\frac{1-\tan^2\theta}{1+\tan^2\theta}=\frac{1-(1+2\tan^2\phi)}{1+(1+2\tan^2\phi)}$$

$$\implies\cos2\theta=-\frac{\tan^2\phi}{1+\tan^2\phi}=-\sin^2\phi$$ $$\implies\cos2\theta=-\frac{1-\cos2\phi}2$$