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Verify the identity. cot ( theta - pi / 2 ) = - tan theta
\[\cot(x-\pi/2)=\tan(\pi/2-(x-\pi/2))=\tan(-x)=-\tan(x)\]
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)
For any CO - trigonometric functions m(x) and co-m(x), this property is always true.\[m(x)=co -m(\pi/2-x)=co -m(x-\pi/2).....or ..vise..versa----if ..m(x)..is ..even\]\[m(x)=co -m(\pi/2-x)=-co -m(x-\pi/2).....or ..vise..versa----if ..m(x)..is ..odd\]
The first one is for even function and the second one for odd.
Alright I get it thank you.!