Question #278970

If Cot θ= 12/5 , and

Cos θ= 1/tan θ. Evaluate;

Cos2θ/sin θ + Cos θ


1
Expert's answer
2021-12-13T14:21:27-0500

cotθ=12/5=>tanθ=1/cotθ=5/12\cot\theta=12/5=>\tan \theta=1/\cot \theta=5/12

1.

If cosθ=1/tanθ,\cos θ= 1/\tan θ, then cosθ=1/tanθ=12/5.\cos θ= 1/\tan θ=12/5.

There is no solution, because 1cosθ1-1\leq \cos \theta\leq 1 for θR.\theta \in \R.


2.

Suppose that the condition cosθ=1/tanθ\cos θ= 1/\tan θ is False.

Then


1+cot2θ=1sin2θ1+\cot^2\theta=\dfrac{1}{\sin^2\theta}

sin2θ=11+cot2θ=11+(12/5)2=25169\sin ^2 \theta=\dfrac{1}{1+\cot^2\theta}=\dfrac{1}{1+(12/5)^2}=\dfrac{25}{169}

cos2θ=1sin2θ=125169=144169\cos ^2\theta=1-\sin^2\theta=1-\dfrac{25}{169}=\dfrac{144}{169}


cos(2θ)cosθ+sinθ=cos2θsin2θcosθ+sinθ\dfrac{\cos(2\theta)}{\cos \theta+\sin \theta}=\dfrac{\cos^2 \theta-\sin^2 \theta}{\cos \theta+\sin \theta}

=(cosθsinθ)(cosθ+sinθ)cosθ+sinθ=\dfrac{(\cos \theta-\sin \theta)(\cos \theta+\sin \theta)}{\cos \theta+\sin \theta}

=cosθsinθ,cosθsinθ=\cos \theta-\sin \theta,\cos \theta\not=-\sin \theta

i)


sinθ>0,cosθ>0\sin\theta>0, \cos \theta>0

Then


sinθ=25169=513\sin \theta=\sqrt{\dfrac{25}{169}}=\dfrac{5}{13}

cosθ=144169=1213\cos \theta=\sqrt{\dfrac{144}{169}}=\dfrac{12}{13}

cos(2θ)cosθ+sinθ=cosθsinθ\dfrac{\cos(2\theta)}{\cos \theta+\sin \theta}=\cos \theta-\sin \theta

=1213513=813=\dfrac{12}{13}-\dfrac{5}{13}=\dfrac{8}{13}

ii)


sinθ<0,cosθ<0\sin\theta<0, \cos \theta<0

Then


sinθ=25169=513\sin \theta=-\sqrt{\dfrac{25}{169}}=-\dfrac{5}{13}

cosθ=144169=1213\cos \theta=-\sqrt{\dfrac{144}{169}}=-\dfrac{12}{13}

cos(2θ)cosθ+sinθ=cosθsinθ\dfrac{\cos(2\theta)}{\cos \theta+\sin \theta}=\cos \theta-\sin \theta

=1213(513)=813=-\dfrac{12}{13}-(-\dfrac{5}{13})=-\dfrac{8}{13}




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