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

What I have done..

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

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  • $\begingroup$ On the left for example if you want tan squared theta, put it between dollar signs. "dollar" \tan ^2 \theta "dollar" becomes $\tan^2 \theta$ by the interpreter here. Note how you need initial backslashes before tan and theta, to tell interpreter tan is a function and theta a variable. (Looks like Sanath already fixed...) $\endgroup$ – coffeemath Jun 1 '14 at 23:11
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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*}

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  • $\begingroup$ I think this is a homework question; so please do not post the full answer. Rather, give a hint. $\endgroup$ – user122283 Jun 1 '14 at 23:17
  • $\begingroup$ @SanathDevalapurkar I am feeling completely helpless on the part $\cos ^2 phi\ = 2 cos ^2 theta\ . $\endgroup$ – Swetank Jun 1 '14 at 23:26
  • $\begingroup$ recall $\cos \theta = 1/\sec \theta$. $\endgroup$ – Anurag A Jun 1 '14 at 23:27
  • $\begingroup$ @SanathDevalapurkar even if that is true Swetank had done most of it and any hint that I give would be just reiterating what I posted. $\endgroup$ – Anurag A Jun 1 '14 at 23:29
  • $\begingroup$ @Anurag still not :( sec^2 theta to 2cos^2 theta? $\endgroup$ – Swetank Jun 1 '14 at 23:35
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Apply Componendo & Dividendo on $$\frac{\tan^2\theta}1=\frac{1+2\tan^2\phi}1$$

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

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

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  • $\begingroup$ @Swetank, How about this? $\endgroup$ – lab bhattacharjee Jun 2 '14 at 3:42

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