I need to prove that:1+tan⁡xtan⁡2x=sec⁡2x

Question

King Osborn · Accepted Answer

I always like to rewrite things in terms of

\cos

and

\sin

when trying to verify identities. I tried to write out the steps in as much detail as possible.

1 + \tan (x) \tan (2 x) = 1 + \frac{\sin (x)}{\cos (x)} \frac{\sin (2 x)}{\cos (2 x)}

= 1 + \frac{\sin (x) 2 \sin (x) \cos (x)}{\cos (x) \cos (2 x)}

using the double angle formula for

\sin (2 x)

= 1 + \frac{2 \sin^{2} (x)}{\cos (2 x)}

cancelling

\cos (x)

= \frac{\cos (2 x) + 2 \sin^{2} (x)}{\cos (2 x)}

getting a common denominator

= \frac{\cos^{2} (x) - \sin^{2} (x) + 2 \sin^{2} (x)}{\cos (2 x)}

using the identity for

\cos (2 x)

= \frac{\cos^{2} (x) + \sin^{2} (x)}{\cos (2 x)}

= \frac{1}{\cos (2 x)}

= \sec (2 x)

Brody Buckley · Accepted Answer

Using the double angle formulas for sin⁡ and cos⁡, we gettan⁡(2x)=sin⁡(2x)cos⁡(2x)=2sin⁡(x)cos⁡(x)cos2(x)−sin2(x)Multiplying the top and bottom by 1cos2(x), we gettan⁡(2x)=2sin⁡(x)cos⁡(x)cos2(x)−sin2(x)=2sin⁡(x)cos⁡(x)1−sin2(x)cos2(x)=2tan⁡(x)1−tan2(x)It is a common trick in mathematics to multiply an expression by AA, where A is some clever choice of expresssion. Here we used A=1cos2(x). This obviously won&#039;t change the value, because AA=1 and multiplying by 1 doesn&#039;t do anything, but it might collapse some things in what you started with to a form you know how to deal with.

I need to prove that: 1+\tan x\tan2x=\sec2x

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