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Verify the identity. cot (x - pi/2) = - tan x
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\(\sin(x-\frac{\pi}{2})=-\cos(x)\) and \(\cos(x-\frac{\pi}{2})= \sin(x)\)
what am i suppose to do with those?
\[\cot (x-\frac {\pi}{2})=\frac{\cos (x-\frac{\pi}{2})}{\sin (x-\frac{\pi}{2})}=\frac{\cos x \cos \frac{\pi}{2}+\sin x \sin \frac{\pi}{2}}{\sin x \cos \frac{\pi}{2}-\cos x \sin \frac{\pi}{2}}=\]\[\frac{\cos x(0)+\sin x(1)}{\sin x(0)-\cos x(1)}=\frac{\sin x}{-\cos x}=- \tan x\]
thats all?
\[\cot(x-\frac{\pi}{2})=\dfrac{\cos(x-\frac{\pi}{2})}{\sin(x-\frac{\pi}{2})}=\dfrac{-\sin(x-\frac{\pi}{2})}{\cos(x-\frac{\pi}{2})}=-\tan(x)\]
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