How do you find tan θ if cos(-θ)= 12/13 and θ is in quadrant IV?

Question

#a^2 + b^2 = c^2#

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Answer 1 · 2018-05-19T01:35:05

#tantheta=-5/12#

.

#cos(-theta)=12/13#

#costheta=cos(-theta)#

#costheta=12/13#

#sin^2theta+cos^2 theta=1#

#sin^2theta=1-cos^2theta=1-144/169=(169-144)/169=25/169#

#sintheta=+-5/13#

In quadrant #IV#, cosine is positive and sine is negative. Therefore, we will only accept the negative value of #sintheta#:

#sintheta=-5/13#

#tantheta=sintheta/costheta=(-5/13)/(12/13)=-5/12#