Asked by maath
Is tan(90-theta) the same as tan(theta)???
Answers
Answered by
Bosnian
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
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
Answered by
Damon
draw it on x and y axis system
if tan T = a/b
then
tan (90-b) = b/a
if tan T = a/b
then
tan (90-b) = b/a
Answered by
maath
okay thanks!
also is there any way to simplify (cos^4(theta) - sin^4(theta)) / (cos^2(theta)*sin(theta))
also is there any way to simplify (cos^4(theta) - sin^4(theta)) / (cos^2(theta)*sin(theta))
Answered by
Damon
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)]
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)]
Answered by
Damon
or just
so you really have
(cos^2-sin^2)/(cos^2 sin)
= (1 - tan^2 theta)/sin theta
so you really have
(cos^2-sin^2)/(cos^2 sin)
= (1 - tan^2 theta)/sin theta
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.