Could someone please show me how to solve this

Prove
tan(-theta) = -tan(theta)

take right triangle a,b,c

tan T = a/c first quadrant
tan-T = -a/c fourth quadrant
but
-a/c = -(a/c)

you can also do it by recalling some properties of sin x and cos x:

sin (-x) = -sin x
cos (-x) = cos x
since tan(theta) = (sin(theta))/(cos(theta)),,
tan(-theta) = (sin(-theta))/(cos(-theta))
applying their properties we have:
-(sin(theta))/(cos(theta)) which is equal to -tan(theta)

so there,, =)

To prove the identity tan(-θ) = -tan(θ), we will use the definition of the tangent function and some trigonometric identities.

The tangent function is defined as the ratio of the sine and cosine functions:

tan(θ) = sin(θ) / cos(θ)

We can also express the sine and cosine functions in terms of their periodicity:

sin(-θ) = -sin(θ) (odd function)
cos(-θ) = cos(θ) (even function)

Now, let's substitute these expressions into the tangent equation:

tan(-θ) = sin(-θ) / cos(-θ)

Replacing the sine and cosine functions:

tan(-θ) = -sin(θ) / cos(θ)

Since -sin(θ) is equal to sin(-θ) and -cos(θ) is equal to cos(-θ), we can further simplify:

tan(-θ) = -sin(θ) / cos(θ)

We end up with the same expression as tan(θ) but with a negative sign:

tan(-θ) = -tan(θ)

Therefore, we have proven that tan(-θ) = -tan(θ), which means that the tangent of the negative angle is equal to the negative of the tangent of the positive angle.