Question

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

Answer 1

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 ${{90}^{\circ }}$ to obtain the expressions in terms of only the angle $\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 $\theta $ and ${{90}^{\circ }}$ is given by the following formulas:
$\sin ({{90}^{\circ }}+\theta )=\cos \left( \theta \right).........(1.1)$
And
$\cos ({{90}^{\circ }}+\theta )=-\sin \left( \theta \right).........(1.2)$
Similarly, we have the sine of an angle and its sum with ${{180}^{\circ }}$ as
$\sin ({{180}^{\circ }}+\theta )=-\sin \left( \theta \right)...............(1.3)$
as the values of $\sin ({{180}^{\circ }})=0\text{ and}\cos ({{180}^{\circ }})=-1$.
Also, we know that the sin function is an odd function, therefore $\sin (-\theta )=-\sin \left( \theta \right)................(1.4)$
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.
$\tan (a)=\dfrac{\sin (a)}{\cos (a)}\text{ and cot(a)=}\dfrac{1}{\tan (a)}=\dfrac{\cos (a)}{\sin (a)}..............(1.5)$
Therefore, from equations (1.1), (1.2) and (1.5), we get
$\tan ({{90}^{\circ }}+\theta )=\dfrac{\sin ({{90}^{\circ }}+\theta )}{\cos ({{90}^{\circ }}+\theta )}=\dfrac{\cos (\theta )}{-\sin (\theta )}=-\cot (\theta ).............(1.7)$
Therefore, form equations (1.1), (1.3), (1.4) and (1.7), we obtain
\[\dfrac{\cos (\theta )}{\sin ({{90}^{\circ }}+\theta )}+\dfrac{\sin (-\theta )}{\sin ({{180}^{\circ }}+\theta )}-\dfrac{\tan ({{90}^{\circ }}+\theta )}{\cot (\theta )}=\dfrac{\cos (\theta )}{\cos (\theta )}+\dfrac{-\sin (\theta )}{-\sin (\theta )}-\dfrac{-\cot (\theta )}{\cot (\theta )}=1+1+1=3\]
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 $\tan ({{90}^{\circ }}+\theta )$ 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.