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tan(-x)/cot(-x) = ?
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Remember cot x = 1/tan x
1
no
Tan x * cot x =1 ; since they are inverse of each other
\[\dfrac{\tan(-x)}{\cot(-x)} = \dfrac{\tan(-x)}{\dfrac{1}{\tan(-x)}} = \tan^2(-x)\]
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Since \(\tan(-x) \ -tan(x)\), \(\tan^2(-x) = --tan^2(x) = \boxed{\tan^2(x)}\)
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