Question

How do you simplify \[(\csc {{30}^{\circ }})(\tan {{60}^{\circ }})\]?

Answer 1

Hint: We have been given with the trigonometric functions, which we have to simplify. There are two terms, we will take them separately and simplify them and then we will combine them later to solve further. We will write the given functions in terms of sine and cosine functions and then we will cancel the similar terms and hence we will obtain the simplified form.

Complete step by step solution:
According to the given question, we have the trigonometric functions which we have to simplify. We can see two functions in the given expression namely the cosecant function and the tangent function. We will take these functions separately and simplify it by writing these functions in terms of sine and cosine functions.
We will begin by writing the given expression first, we have,
\[(\csc {{30}^{\circ }})(\tan {{60}^{\circ }})\]----(1)
Taking \[(\csc {{30}^{\circ }})\], we get,
\[\csc {{30}^{\circ }}=\dfrac{1}{\sin {{30}^{\circ }}}\]-----(2)
As we know that sine and cosecant functions are inverse of each other, that is, \[\sin \theta =\dfrac{1}{\csc \theta }\]
Now, we will take the second function \[(\tan {{60}^{\circ }})\], we get,
\[\tan {{60}^{\circ }}=\dfrac{\sin {{60}^{\circ }}}{\cos {{60}^{\circ }}}\]----(3)
As we know that the ratio of sine and cosine function is tangent function, that is, \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\].
Substituting equation (2) and (3) in equation (!), we get,
\[(\csc {{30}^{\circ }})(\tan {{60}^{\circ }})\]
\[\Rightarrow \dfrac{1}{\sin {{30}^{\circ }}}\times \dfrac{\sin {{60}^{\circ }}}{\cos {{60}^{\circ }}}\]---(4)
We can use trigonometric formulas to solve the obtained expression. We will use the formula of \[\sin (2\theta )\].
We have, \[\sin (2\theta )=2\sin \theta \cos \theta \]
Using this formula for \[\sin {{60}^{\circ }}\] in equation (4), we get,
\[\sin {{60}^{\circ }}=\sin ({{2.30}^{\circ }})=2\sin {{30}^{\circ }}\cos {{30}^{\circ }}\]---(5)
Substituting this in equation (4), we get,
\[\Rightarrow \dfrac{1}{\sin {{30}^{\circ }}}\times \dfrac{2\sin {{30}^{\circ }}\cos {{30}^{\circ }}}{\cos {{60}^{\circ }}}\]
\[\sin {{30}^{\circ }}\] will get cancelled in the above equation and we have,
\[\Rightarrow 2\dfrac{\cos {{30}^{\circ }}}{\cos {{60}^{\circ }}}\]
We know the values of \[\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}\] and \[\cos {{60}^{\circ }}=\dfrac{1}{2}\], we will substitute these values and we will get,
\[\Rightarrow 2\dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}}\]
\[\Rightarrow 2\sqrt{3}\]

Therefore, the simplified form is \[2\dfrac{\cos {{30}^{\circ }}}{\cos {{60}^{\circ }}}\] or \[2\sqrt{3}\].

Note: The trigonometric formula used in the above solution must be done carefully as the angle to be taken in \[\sin (2\theta )\] gets very confusing at times, so must be done in order. \[\sin (2\theta )\] formula is a commonly used formula used in most trigonometry related questions. Also, \[\cos 2\theta \] is another important formula commonly used.