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$$\sec(2x)=\frac{\sec^2x+\sec^4x}{2+\sec^2x-\sec^4x}$$

I have no idea how to verify this. I've tried changing it into cosine but it doesn't work.

user108452
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  • I've edited your post to include MathJax. Please verify that it's correct and says what you intend. –  Nov 13 '13 at 03:24
  • Change each $\sec(x)$ by $\frac{2}{z+z^{-1}}$ and $\sec(2x)$ by $\frac{2}{z^2+z^{-2}}$. Clear denominators and check that the polynomials in both sides are equal. If they are the identity is true. – OR. Nov 13 '13 at 03:51

2 Answers2

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Note $$\dfrac{\sec^2{x}+\sec^4{x}}{2+\sec^2{x}-\sec^4{x}}=\dfrac{\sec^2{x}(1+\sec^2{x})}{(1+\sec^2{x})(2-\sec^2{x})}=\dfrac{\sec^2{x}}{2-\sec^2{x}}=\dfrac{1}{2\cos^2{x}-1}=\sec{(2x)}$$

math110
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You can also use Wolfram Alpha http://www.wolframalpha.com to verify it 

Simplify[(Sec[x]^2+Sec[x]^4)/(2+Sec[x]^2-Sec[x]^4)]
Andrey Sokolov
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  • The question was how to verify an identity. Surely, using wolfram alpha is a valid way to verify it. I hasten to add that it is probably more reliable too than doing it by hand. – Andrey Sokolov Nov 13 '13 at 04:25
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    Wolfram|Alpha would be an invalid way of verifying the identity if it had the identity automatically programmed into it (as it does many trig identities), because then it would be assuming the conclusion. After looking through the documentation for Mathematica (and by extension Wolfram|Alpha), it does not assume the identity, and is therefore a valid way of verifying it. – Juan Sebastian Lozano Nov 13 '13 at 21:37