To simplify: `(tan 2x + sec 2x)^2`

We know that `tan x = (sin x)/(cos x) ` and that `sec x = 1/(cos x)`

`therefore (tan 2x + sec 2x)^2`

`= ((sin2x)/(cos 2x) + 1/(cos 2x))^2`

We have an LCD of cos 2x

`therefore = ((sin2x + 1)/ (cos 2x))^2`

At this stage we can choose any one of various options to solve. To keep it simple and easy to follow:

We know that `sin^2 x+cos ^2x = 1 ` so substitute it for the +1

`therefore = ((sin 2x+sin^2x +cos^2 x)/ (cos 2x))^2`

we also know that `sin 2x = 2sinx.cos x ` and the most appropriate choice in this particular question for cos 2x is `cos 2x = cos^2 x- sin^2 x`

`therefore = ((2sinx.cosx + sin^2 x+cos^2 x)/(cos ^2x-sin^2x))^2`

Now rearrange the numerator and you will recognize a trinomial:

`therefore = ((sin^2 x+2 sinx.cosx + cos^2)/ (cos^2x-sin^2x))^2`

and factorize. Note the denominator is a difference of squares:

`therefore = ((sinx+cosx)^2/((cosx-sinx)(cosx+sin x)))^2`

cross-cancelling we will be left with:

`=((sinx+cosx)/(cos x-sinx))^2`

and expanding the square `= ((sin^2x + 2sinx.cos x + cos^2x)/(cos ^2x-2sinxcosx+sin^2 x))`

Remember `sin^2x+cos^2 = 1` so substitute it into the numerator and denominator:

`therefore = (1+2sinxcosx)/(1-2sinxcosx)`

`therefore = (1+sin2x)/(1-sin2x)`

**Ans: ** `(tan2x+sec2x)^2 = ((1+sin2x)/(1-sin2x))`

Please note that there is more than one way to simplify this particular question. This is a fairly comprehensive answer to ensure that you understand the steps.