Certainly! Let's solve this step-by-step.
Given:
[tex]\[
\operatorname{cosec} \theta = 2
\][/tex]
Since [tex]\(\operatorname{cosec} \theta\)[/tex] is the reciprocal of [tex]\(\sin \theta\)[/tex], we have:
[tex]\[
\sin \theta = \frac{1}{\operatorname{cosec} \theta} = \frac{1}{2}
\][/tex]
Now we know [tex]\(\sin \theta = \frac{1}{2}\)[/tex].
Using the Pythagorean identity for sine and cosine:
[tex]\[
\sin^2 \theta + \cos^2 \theta = 1
\][/tex]
Substitute [tex]\(\sin \theta = \frac{1}{2}\)[/tex]:
[tex]\[
\left(\frac{1}{2}\right)^2 + \cos^2 \theta = 1
\][/tex]
[tex]\[
\frac{1}{4} + \cos^2 \theta = 1
\][/tex]
[tex]\[
\cos^2 \theta = 1 - \frac{1}{4}
\][/tex]
[tex]\[
\cos^2 \theta = \frac{3}{4}
\][/tex]
[tex]\[
\cos \theta = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}
\][/tex]
Next, we need to find [tex]\(\cot \theta\)[/tex]. The cotangent function is given by the reciprocal of the tangent function:
[tex]\[
\cot \theta = \frac{\cos \theta}{\sin \theta}
\][/tex]
Substitute [tex]\(\cos \theta = \frac{\sqrt{3}}{2}\)[/tex] and [tex]\(\sin \theta = \frac{1}{2}\)[/tex]:
[tex]\[
\cot \theta = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \frac{\sqrt{3}}{2} \div \frac{1}{2} = \sqrt{3}
\][/tex]
Then, we need to find:
[tex]\[
\frac{\sin \theta}{\cos \theta}
\][/tex]
Substitute [tex]\(\sin \theta = \frac{1}{2}\)[/tex] and [tex]\(\cos \theta = \frac{\sqrt{3}}{2}\)[/tex]:
[tex]\[
\frac{\sin \theta}{\cos \theta} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{2} \div \frac{\sqrt{3}}{2} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}
\][/tex]
Finally, we can find the required expression:
[tex]\[
\cot \theta + \frac{\sin \theta}{\cos \theta} = \sqrt{3} + \frac{\sqrt{3}}{3}
\][/tex]
To combine these terms into a single fraction:
[tex]\[
\sqrt{3} + \frac{\sqrt{3}}{3} = \frac{3\sqrt{3}}{3} + \frac{\sqrt{3}}{3} = \frac{4\sqrt{3}}{3}
\][/tex]
Thus, the required expression [tex]\(\cot \theta + \frac{\sin \theta}{\cos \theta}\)[/tex] is:
[tex]\[
2.309401076758503
\][/tex]
