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Question: Prove that \(\dfrac{\cos (\theta )}{\sin ({{90}^{\circ }}+\theta )}+\dfrac{\sin (-\theta )}{\sin (...

Prove that
cos⁡(θ)sin⁡(90∘+θ)+sin⁡(−θ)sin⁡(180∘+θ)−tan⁡(90∘+θ)cot⁡(θ)=3\dfrac{\cos (\theta )}{\sin ({{90}^{\circ }}+\theta )}+\dfrac{\sin (-\theta )}{\sin ({{180}^{\circ }}+\theta )}-\dfrac{\tan ({{90}^{\circ }}+\theta )}{\cot (\theta )}=3

Explanation

Solution

Hint:In this case, we need to find out the trigonometric ratios by the addition of angles. Therefore, we can use the expressions for the sine and cosine of sum of an angle and a multiple of 90∘{{90}^{\circ }} to obtain the expressions in terms of only the angle θ\theta . Then, we can use the relations between the trigonometric ratios to find out the required answer.

Complete step-by-step answer:
We know that the sine and cosine of sum of an angle θ\theta and 90∘{{90}^{\circ }} is given by the following formulas:
sin⁡(90∘+θ)=cos⁡(θ).........(1.1)\sin ({{90}^{\circ }}+\theta )=\cos \left( \theta \right).........(1.1)
And
cos⁡(90∘+θ)=−sin⁡(θ).........(1.2)\cos ({{90}^{\circ }}+\theta )=-\sin \left( \theta \right).........(1.2)
Similarly, we have the sine of an angle and its sum with 180∘{{180}^{\circ }} as
sin⁡(180∘+θ)=−sin⁡(θ)...............(1.3)\sin ({{180}^{\circ }}+\theta )=-\sin \left( \theta \right)...............(1.3)
as the values of sin⁡(180∘)=0 andcos⁡(180∘)=−1\sin ({{180}^{\circ }})=0\text{ and}\cos ({{180}^{\circ }})=-1.
Also, we know that the sin function is an odd function, therefore sin⁡(−θ)=−sin⁡(θ)................(1.4)\sin (-\theta )=-\sin \left( \theta \right)................(1.4)
Also, we know that the tangent of an angle is the ratio of its sine and cosine and cot is the ratio of its cosine and sine, i.e.
tan⁡(a)=sin⁡(a)cos⁡(a) and cot(a)=1tan⁡(a)=cos⁡(a)sin⁡(a)..............(1.5)\tan (a)=\dfrac{\sin (a)}{\cos (a)}\text{ and cot(a)=}\dfrac{1}{\tan (a)}=\dfrac{\cos (a)}{\sin (a)}..............(1.5)
Therefore, from equations (1.1), (1.2) and (1.5), we get
tan⁡(90∘+θ)=sin⁡(90∘+θ)cos⁡(90∘+θ)=cos⁡(θ)−sin⁡(θ)=−cot⁡(θ).............(1.7)\tan ({{90}^{\circ }}+\theta )=\dfrac{\sin ({{90}^{\circ }}+\theta )}{\cos ({{90}^{\circ }}+\theta )}=\dfrac{\cos (\theta )}{-\sin (\theta )}=-\cot (\theta ).............(1.7)
Therefore, form equations (1.1), (1.3), (1.4) and (1.7), we obtain
cos⁡(θ)sin⁡(90∘+θ)+sin⁡(−θ)sin⁡(180∘+θ)−tan⁡(90∘+θ)cot⁡(θ)=cos⁡(θ)cos⁡(θ)+−sin⁡(θ)−sin⁡(θ)−−cot⁡(θ)cot⁡(θ)=1+1+1=3\dfrac{\cos (\theta )}{\sin ({{90}^{\circ }}+\theta )}+\dfrac{\sin (-\theta )}{\sin ({{180}^{\circ }}+\theta )}-\dfrac{\tan ({{90}^{\circ }}+\theta )}{\cot (\theta )}=\dfrac{\cos (\theta )}{\cos (\theta )}+\dfrac{-\sin (\theta )}{-\sin (\theta )}-\dfrac{-\cot (\theta )}{\cot (\theta )}=1+1+1=3
Which we wanted to prove to obtain the answer to the question.

Note: We should note that, in equation (1.7), we could also write tan⁡(90∘+θ)\tan ({{90}^{\circ }}+\theta ) in terms of sine and cosine. However, as the denominator is given in cot, we would have then expanded the denominator in terms of sine and cosine as well which would have resulted in more calculation steps and thus would have been more difficult to solve.