# Use the double angle formulas to verify the identity. `tan^2x = (1-cos2x)/(1+cos2x)`

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The double angle formula we can use here to verify the equation given is the *double angle formula for the cosine function*:

`cos2x = cos^2(x) - sin^2(x)`

Substituting `sin^2(x) = 1-cos^2(x)` (from the unit circle formula) gives

`cos2x = 2cos^2(x) - 1` (1)

Now, because the cosine in terms of the tangent is given by

`cos(x) = pm 1/(sqrt(1+tan^2(x))` this``

`cos^2(x) = 1/(1+tan^2(x))`

Substituting this into the result in equation (1) gives

`cos2x = 2cos^2(x) - 1 = 2/(1+tan^2(x)) - (1+tan^2(x))/(1+tan^2(x))`

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

The identity we are required to verify is

`tan^2(x) = (1-cos2x)/(1+cos2x)`

Using the double angle formula for cos2x as described above, we have that

`(1-cos2x)/(1+cos2x) = (1-(1-tan^2(x))/(1+tan^2(x)))/(1+(1-tan^2(x))/(1+tan^2(x)))` `= (1+tan^2(x) - 1 + tan^2(x))/(1+tan^2(x) + 1 - tan^2(x))` `= (2tan^2(x))/2`

`= tan^2(x)` **as required**