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I'm having some trouble proving trig identities, this time with the double-angle formula. I want to prove that:

$$ \frac{1 + \tan^2 A}{1 - \tan^2 A}=\sec 2A $$

I know that:

$$ 1 - \tan^2 A = \frac{2 \tan A}{\tan 2A} $$

But I don't know how to get the numerator of the LHS. I also know that:

$$ \sec 2A = \frac{1}{2\sin A \cos A} $$

But I can't marry them together. Can anybody point me in the right direction?

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  • $\begingroup$ Did you mean $\dfrac{1+\tan^2 A}{1-\tan^2 A}=\sec 2A$ $\endgroup$
    – Hawk
    Feb 9, 2014 at 15:03

1 Answer 1

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First of all, $\displaystyle2\sin A\cos A=\sin2A$

If $\displaystyle\sec2A=\frac1{2\sin A\cos A}$

we need $\displaystyle\sec2A=\frac1{\sin2A}\iff1=\frac{\sin2A}{\cos2A}$

$\displaystyle\implies\tan2A=1\iff2A=n\pi+\frac\pi4$ where $n$ is any integer

Now, $$\frac{1+\tan^2A}{1-\tan^2A}=\frac{1+\dfrac{\sin^2A}{\cos^2A}}{1-\dfrac{\sin^2A}{\cos^2A}}=\frac{\cos^2A+\sin^2A}{\cos^2A-\sin^2A}=\frac1{\cos2A}$$

Reference : Double-Angle Formulas

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  • $\begingroup$ I liked the previous formatting of your answer more...if you don't mind $\endgroup$
    – Hawk
    Feb 9, 2014 at 15:05
  • $\begingroup$ @Hawk, please find the edited version $\endgroup$ Feb 9, 2014 at 15:07
  • $\begingroup$ How did you get from $\frac{1+\tan^2A}{1-\tan^2A}$ to $\frac{\cos^2A+\sin^2A}{\cos^2A-\sin^2A}$? $\endgroup$
    – hohner
    Feb 9, 2014 at 15:08
  • $\begingroup$ Yes, this is fine...Your solution is good as always $\endgroup$
    – Hawk
    Feb 9, 2014 at 15:09
  • $\begingroup$ @hohner, please find the edited version $\endgroup$ Feb 9, 2014 at 15:11

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