You need to use double angle identity for `tan 2a` , such that:

`tan 2a = (2tan a)/(1 - tan^2 a)`

You also need to use the following trigonometric identity, such that:

`tan a = (sin a)/(cos a)`

Substituting `(sin a)/(cos a)` for `tan a` yields:

`tan 2a = (2(sin a)/(cos a))/(1 - (sin^2 a)/(cos^2 a))`

Since the problem provides the information `cos a = 1 - sin a` , you need to substitute `1 - sin a` for cos a, such that:

`tan 2a = (2(sin a)/(1 - sin a )) / (1 - (sin^2 a)/(1 - sin a)^2)`

`tan 2a = (2(sin a)/(1 - sin a))/((1 - 2sin^2 a)/(1 - sin a)^2)`

Reducing duplicate terms yields:

`tan 2a = 2sin a(1 - sina)/(1 - 2sin^2 a)`

Substituting `cos a` for `1 - sin a` yields:

`tan 2a = (2sin a*cos a)/(cos 2a)`

`tan 2a = (sin 2a)/(cos2a) => tan 2a = tan 2a => 0 = 0`

**Hence, evaluating `tan 2a` , under the given conditions, yields **`tan 2a = (sin 2a)/(cos2a).`

We'll re-write the given constraint cosa = 1 - sina. We'll add sin a both sides and we'll have:

sin a + cos a = 1

We'll square raise the new relation sina + cosa = 1.

(sina + cosa)^2 = 1^2

(sina)^2 + (cosa)^2 + 2sina*cosa = 1 (1)

But, from the fundamental formula of trigonometry:

(sina)^2 + (cosa)^2 = 1

We'll substitute (sina)^2 + (cosa)^2 by 1:

The relation (1) will become:

1 + 2sina*cosa = 1

We'll eliminate like terms:

2sina*cosa = 0

But 2sina*cosa = sin (2a)

We'll write the formula for tan 2a:

tan 2a = sin 2a/cos 2a

tan 2a = 0/cos 2a

tan 2a = 0