A little Problem in Trigonometry (Multiple Angle) If $\tan^2 \theta = 1 + 2\tan^2 \phi$, show that $\cos 2\phi = 1 + 2\cos2\theta$.
What I have done..
$$\implies \tan^2 \theta = 1 + 2\tan^2 \phi\\
\implies 1 + \tan^2 \theta = 2 + 2\tan^2 \phi\\
\implies 1 + \tan^2 \theta = 2(1 + \tan^2 \phi)\\
\implies \sec^2 \theta = 2(\sec^2 \phi)$$

Apply Componendo & Dividendo on $$\frac{\tan^2\theta}1=\frac{1+2\tan^2\phi}1$$
$$\frac{1-\tan^2\theta}{1+\tan^2\theta}=\frac{1-(1+2\tan^2\phi)}{1+(1+2\tan^2\phi)}$$
$$\implies\cos2\theta=-\frac{\tan^2\phi}{1+\tan^2\phi}=-\sin^2\phi$$
$$\implies\cos2\theta=-\frac{1-\cos2\phi}2$$
