Is tan(90-theta) the same as tan(theta)???
No it's not.
sin ( 90° - θ ) = cos θ
cos ( 90° - θ ) = sin θ
tan ( 90° - θ ) = sin ( 90° - θ ) / cos ( 90° - θ ) = cos θ / sin θ = cot θ
tan ( 90° - θ ) = cot θ
okay thanks!
also is there any way to simplify (cos^4(theta) - sin^4(theta)) / (cos^2(theta)*sin(theta))
a^4-b^4 = (a^2-b^2)(a^2+b^2)
cos^4 -sin^4 = (cos^2-sin^2)(cos^2 + sin^2) but cos^2+sin^2 = 1
so you really have
(cos^2-sin^2)/(cos^2 sin)
well
I suppose that is
cos (2 theta)/(cos^2 theta sin theta)
or maybe
(1 - tan^2)/sin
but
(1-tan^2) = 2 tan/tan 2theta
so
2 tan theta/ (sin theta tan 2 theta
(2 /cos theta)/tan 2 Theta
2/ [ cos (theta) tan (2 theta)]
or just
so you really have
(cos^2-sin^2)/(cos^2 sin)
= (1 - tan^2 theta)/sin theta
To determine if tan(90-theta) is the same as tan(theta), we can apply the trigonometric identity:
tan(90-theta) = cot(theta)
In trigonometry, we know that the tangent (tan) of an angle is equal to the reciprocal of the cotangent (cot) of the same angle:
tan(theta) = 1/cot(theta)
Therefore, tan(90-theta) is not the same as tan(theta), but it is equal to cot(theta).
draw it on x and y axis system
if tan T = a/b
then
tan (90-b) = b/a