If #sin theta = -5/13# and #cos theta = -12/13#, how do you find #cot theta#?

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Answer 1 · 2018-04-01T06:27:35

#cottheta=12/5#

We know that:

#tantheta=sintheta/costheta#

and

#cottheta=1/tantheta#

So we can plug in our values:

#color(white)=cot(theta)#

#=1/tan(theta)#

#=1/(sintheta/costheta)#

#=1/(sintheta/costheta)color(blue)(*costheta/costheta)#

#=1/(sintheta/color(red)cancelcolor(black)costheta)color(blue)(*costheta/color(red)cancelcolor(blue)costheta)#

#=1/sinthetacolor(blue)(*costheta)#

#=costheta/sintheta#

Now plug in our values:

#=(-12/13)/(-5/13)#

#=(-12/13)/(-5/13)color(blue)(*13/13)#

#=(-12/color(red)cancelcolor(black)13)/(-5/color(red)cancelcolor(black)13)color(blue)(*color(red)cancelcolor(blue)13/color(red)cancelcolor(blue)13)#

#=(-12)/(-5)color(blue)(*1/1)#

#=(-12)/(-5)#

#=(color(red)cancelcolor(black)-12)/(color(red)cancelcolor(black)-5)#

#=12/5#

That's the solution. Hope this helped!