Question

What does this expression ‘\[\tan \theta +\cot \theta \]’ equals to?
\[\begin{align}
  & A.\ \sec \theta \cdot \cos ec\theta \\
 & B.\ \sin \theta \cdot \cos \theta \\
 & C.\ \tan \theta \cdot \cot \theta \\
 & D.\ \sin \theta +\cos \theta \\
\end{align}\] \[\]

Answer 1

HINT: -
One should know this before that
\[{{\sec }^{2}}\theta ={{\tan }^{2}}\theta +1\ \ \ \ \ \ \ \ \ \ \ \ \,...\left( a \right)\]
And this as well
\[\tan \theta =\dfrac{1}{\cot \theta }\ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(b)\]
And these too
\[\sec \theta =\dfrac{1}{\cos \theta }\ and\ \cos ec\theta =\dfrac{1}{\sin \theta }\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(c)\]

Complete step-by-step answer:

We can see that the relations and equations of trigonometry can be used to get a solution to this question. We can see that we have chosen only these equations and relations of trigonometry because the question consists of these trigonometric functions that are tan \[\theta \] , cot \[\theta \] , sec \[\theta \] and cosec \[\theta \].
Therefore, using equations (a),(b) and (c), we get:-
     \[\begin{align}
  & \tan \theta +\dfrac{1}{\tan \theta } \\
 & \dfrac{{{\tan }^{2}}\theta +1}{\tan \theta } \\
 & \dfrac{{{\sec }^{2}}\theta }{\tan \theta } \\
 & \dfrac{{{\sec }^{2}}\theta \cdot \cos ec\theta }{\sec \theta } \\
 & \sec \theta \cdot \cos ec\theta \\
\end{align}\]
One could also think of that the options that have been provided with the question, all of them do not comprise of sin \[\theta \] and cos \[\theta \]terms so , one could get an idea about which relations and equations could be used to get to an answer.


NOTE: -
There are many ways to solve a trigonometry question and the mentioned solution is just one of them. Different solutions can vary from each other. Some of them might be similar as well which is mostly the case.
We can also see that we have not used sin \[\theta \] and cos \[\theta \] because they are nowhere in the question. However, one can use them as well as there are many ways to get to a solution in trigonometric questions.