Question

Verify the identity.
cot ( theta - pi / 2 ) = - tan theta

Answer 1

\[\cot(x-\pi/2)=\tan(\pi/2-(x-\pi/2))=\tan(-x)=-\tan(x)\]

Answer 2

cot ( theta - pi / 2 )= cos ( theta - pi / 2 ) / sin ( theta - pi / 2 )

cos ( theta - pi / 2 )= cos(theta) cos (pi/2) + sin(theta) sin (pi/2)
= cos(theta) (0) + sin(theta) (1)
= sin(theta)

sin ( theta - pi / 2 )= sin(theta) cos(pi/2) - cos(theta) sin(pi/2)
= sin(theta) (0) - cos(theta) (1)
= -cos(theta)
so you get,
cot ( theta - pi / 2 )= sin(theta) / -cos(theta)
cot ( theta - pi / 2 )= -tan(theta)

Answer 3

For any CO - trigonometric functions m(x) and co-m(x), this property is always true.\[m(x)=co -m(\pi/2-x)=co -m(x-\pi/2).....or ..vise..versa----if ..m(x)..is ..even\]\[m(x)=co -m(\pi/2-x)=-co -m(x-\pi/2).....or ..vise..versa----if ..m(x)..is ..odd\]

Answer 4

got it now..??

Answer 5

The first one is for even function and the second one for odd.

Answer 6

@gjhfdfg ????

Answer 7

I think so.?

Answer 8

Alright I get it thank you.!