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Verify the identity. cot (x minus pi divided by two) = -tan x
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\[\cot(x-\pi/2)=-tanx\]
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is it your Q??
Yes
here is trigonometric formula for cot Cot(A-B) = (CotACotB+1)/(CotB-CotA)
\[\cot(x-\pi/2)=\frac{ cotx*\cot (\pi/2) +1}{ \cot (\pi/2)- cotx }\]
cot(pi/2)=0 standard value =(0+1)/(0-cotx)
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\[=\frac{ 1 }{ -cotx }=-tanx\]
as cotx=1/tanx trigonometric value
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