# Prove the following identities. 1. `tan(x-45)+tan(x+45)=2tan2x` 2. `cos 2alpha=(cot alpha-tan alpha)/(cot alpha+tan alpha)` `` Thank you!

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1) `tan(x-45^o) + tan(x+45^o) = 2tan(2x)`

To prove, apply the sum-difference formula of tangent.

`(tanx - tan 45^o )/(1+tanxtan45^o) + (tan x + tan45^o)/(1-tanxtan45^o)=2tan(2x)`

Then, plug-in `tan45^o=1` .

`(tanx -1)/(1+tanx*1) + (tanx+1)/(1-tanx*1)=2tan(2x)`

`(tanx -1)/(1+tanx) + (tanx+1)/(1-tanx)=2tan(2x)`

Then, express the two fractions with same denominators. The LCD of the two fractions is `(1+tanx)(1-tanx)=1-tan^2x` . So, their equivalent fractions using the LCD as their denominators are:

`(tanx-1)/(1+tanx)*(1-tanx)/(1-tanx) +(tanx+1)/(1-tanx)*(1+tanx)/(1+tanx)=2tan(2x)`

`(-tan^2x +2tanx -1)/(1-tan^2x)+(tan^2x+2tanx+1)/(1-tan^2x)=2tan(2x)`

`(4tanx)/(1-tan^2x)=2tan(2x)`

Then, apply the double angle identity of tangent.

`2*2tanx/(1-tan^x)=2tan(2x)`

`2*tan(2x)=2tan(2x)`

`2tan(2x)=2tan(2x)`

Since left side simplifies to the same expression at the right side, **it proves that the given equation `tan(x-45^o) + tan(x+45^o) = 2tan(2x)` is an identity.**

2) `cos(2alpha)=(cotalpha-tanalpha)/(cotalpha+ tanalpha)`

To prove, express the tangent and cotangent in terms of cosine and sine.

`cos(2alpha) = (cosalpha/sinalpha-sinalpha/cosalpha)/(cosalpha/sinalpha+sinalpha/cosalpha)`

To simplify the complex fraction, multiply the top and bottom by the LCD of the fractions present. The LCD is `sinalphacosalpha` .

` cos(2alpha) = (cosalpha/sinalpha - sinalpha/cosalpha)/(cosalpha/sinalpha + sinalpha/cosalpha)*((sinalphacosalpha)/1)/((sinalphacosalpha)/1)`

`cos(2alpha)=(cos^2alpha - sin^2alpha)/(cos^2alpha+sin^2alpha)`

For the denominator, apply the Pythagorean Identity.

`cos(2alpha)=(cos^2alpha - sin^2alpha)/1`

`cos(2alpha)=cos^2alpha - sin^2alpha`

Then, apply the double angle identity of cosine.

`cos(2alpha)=cos(2alpha)`

**Therefore, the given equation is an identity.**

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Thank you! Helped SO much with my upcoming exam!