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.