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cot ( x - pi/2)= -tan x Can anybody help verify???? Medal... @phi @mathstudent55 @mathmale
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do you know the formula for the tangent of the difference of two angles ?
\[\cot \left( x-\frac{ \pi }{ 2} \right)=\cot \left\{ -\left( \frac{ \pi }{ 2 } -x\right) \right\}=-\cot \left( \frac{ \pi }{ 2 }-x \right)=-\tan x\]
Is it cot(pi/2-theta)=tan
I didnt know you were able to just multiply by -1
Is that all you have to do?
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no i have not multiplied by -1 if you do not follow i do otherwise. \[\cot \left( -\theta \right) =-\cot \theta \]
ohhhh I get it so you're basically just manipulating the negative sign
Thank you
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