Answer to Question #278970 in Trigonometry for Dan

Question

If Cot θ= 12/5 , and

Cos θ= 1/tan θ. Evaluate;

Cos2θ/sin θ + Cos θ

Answer 1

$$\cot\theta=12/5=>\tan \theta=1/\cot \theta=5/12$$

1.

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

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

2.

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

Then

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

$$\sin ^2 \theta=\dfrac{1}{1+\cot^2\theta}=\dfrac{1}{1+(12/5)^2}=\dfrac{25}{169}$$

$$\cos ^2\theta=1-\sin^2\theta=1-\dfrac{25}{169}=\dfrac{144}{169}$$

$$\dfrac{\cos(2\theta)}{\cos \theta+\sin \theta}=\dfrac{\cos^2 \theta-\sin^2 \theta}{\cos \theta+\sin \theta}$$

$$=\dfrac{(\cos \theta-\sin \theta)(\cos \theta+\sin \theta)}{\cos \theta+\sin \theta}$$

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

i)

$$\sin\theta>0, \cos \theta>0$$

Then

$$\sin \theta=\sqrt{\dfrac{25}{169}}=\dfrac{5}{13}$$

$$\cos \theta=\sqrt{\dfrac{144}{169}}=\dfrac{12}{13}$$

$$\dfrac{\cos(2\theta)}{\cos \theta+\sin \theta}=\cos \theta-\sin \theta$$

$$=\dfrac{12}{13}-\dfrac{5}{13}=\dfrac{8}{13}$$

ii)

$$\sin\theta<0, \cos \theta<0$$

Then

$$\sin \theta=-\sqrt{\dfrac{25}{169}}=-\dfrac{5}{13}$$

$$\cos \theta=-\sqrt{\dfrac{144}{169}}=-\dfrac{12}{13}$$

$$\dfrac{\cos(2\theta)}{\cos \theta+\sin \theta}=\cos \theta-\sin \theta$$

$$=-\dfrac{12}{13}-(-\dfrac{5}{13})=-\dfrac{8}{13}$$