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QuestionRewrite cos⁡(90∘−θ)cot⁡θ\cos \left(90^{\circ}-\theta\right) \cot \theta using one of the six trigonometric functions of angle θ\theta. cos⁡(90∘−θ)cot⁡θ=\cos \left(90^{\circ}-\theta\right) \cot \theta=

Studdy Solution

STEP 1

Assumptions1. The expression is cos⁡(90∘−θ)cot⁡θ\cos \left(90^{\circ}-\theta\right) \cot \theta. . We need to rewrite this expression as one of the six trigonometric functions of acute angle θ\theta.

STEP 2

We start by using the co-function identity for cosine, which states that cos⁡(90∘−θ)=sin⁡θ\cos \left(90^{\circ}-\theta\right) = \sin \theta.
cos⁡(90∘−θ)cot⁡θ=sin⁡θcot⁡θ\cos \left(90^{\circ}-\theta\right) \cot \theta = \sin \theta \cot \theta

STEP 3

Next, we use the fact that cot⁡θ=1tan⁡θ\cot \theta = \frac{1}{\tan \theta}.
sin⁡θcot⁡θ=sin⁡θ⋅1tan⁡θ\sin \theta \cot \theta = \sin \theta \cdot \frac{1}{\tan \theta}

STEP 4

Now, we use the identity tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta}.
sin⁡θ⋅1tan⁡θ=sin⁡θ⋅1sin⁡θcos⁡θ\sin \theta \cdot \frac{1}{\tan \theta} = \sin \theta \cdot \frac{1}{\frac{\sin \theta}{\cos \theta}}

STEP 5

implify the expression.
sin⁡θ⋅1sin⁡θcos⁡θ=sin⁡θ⋅cos⁡θsin⁡θ\sin \theta \cdot \frac{1}{\frac{\sin \theta}{\cos \theta}} = \sin \theta \cdot \frac{\cos \theta}{\sin \theta}

STEP 6

The sin⁡θ\sin \theta terms cancel out, leaving us with cos⁡θ\cos \theta.
sin⁡θ⋅cos⁡θsin⁡θ=cos⁡θ\sin \theta \cdot \frac{\cos \theta}{\sin \theta} = \cos \thetaSo, the expression cos⁡(90∘−θ)cot⁡θ\cos \left(90^{\circ}-\theta\right) \cot \theta can be rewritten as cos⁡θ\cos \theta.

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