Question

It is given that if $\sin \theta =\dfrac{\sqrt{3}}{2}$ , find the value of all other T-ratios of \[\theta \] .

Answer 1

Hint: In this question, we are given the value of \[\sin \theta \] , hence, we can use the following relations which involve \[\sin \theta \] and hence, we can get the values of all other T-ratios.
\[\begin{align}
  & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\
 & \text{sec }\!\!\theta\!\!\text{ }=\dfrac{1}{\cos \theta } \\
 & \tan \theta =\dfrac{\sin \theta }{\cos \theta } \\
 & \text{cosec }\!\!\theta\!\!\text{ }=\dfrac{1}{\sin \theta } \\
 & \cot \theta =\dfrac{1}{\tan \theta } \\
\end{align}\]

Complete step by step answer:
Now, by using these above relations, we can get the required results and values.

Now, from the given question we have
\[\sin \theta =\dfrac{\sqrt{3}}{2}\ \ \ \ \ ...(a)\]
Now, by using the trigonometric identity which gives the relation between the function that are mentioned in the hint, we get the following substitute the value from the question and as well as from equation (a)
\[\begin{align}
  & \Rightarrow {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\
 & \Rightarrow {{\cos }^{2}}\theta =1-{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}} \\
 & \Rightarrow {{\cos }^{2}}\theta =1-\dfrac{3}{4} \\
 & \Rightarrow {{\cos }^{2}}\theta =\dfrac{1}{4} \\
 & \Rightarrow \cos \theta =\sqrt{\dfrac{1}{4}}=\dfrac{1}{2} \\
\end{align}\]
For finding the value of sec function, we could again use the relations given in the hint as follows
\[\begin{align}
  & \Rightarrow \left( \dfrac{1}{\cos \theta } \right)=\sec \theta \\
 & \Rightarrow \sec \theta =\left( \dfrac{1}{\dfrac{1}{2}} \right)=2 \\
\end{align}\]
Now, again from the hint, we know that we can get the tan function as follows
\[\begin{align}
  & \Rightarrow \tan \theta =\dfrac{\sin \theta }{\cos \theta } \\
 & \Rightarrow \tan \theta =\dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}} \\
 & \Rightarrow \tan \theta =\sqrt{3} \\
\end{align}\]
Now, again referring to the relations, we get the value of $\cot $ function as follows
\[\begin{align}
  & \Rightarrow \tan \theta =\dfrac{1}{\cot \theta } \\
 & \Rightarrow \cot \theta =\dfrac{1}{\tan \theta } \\
 & \Rightarrow \cot \theta =\dfrac{1}{\sqrt{3}} \\
\end{align}\]
Now, for finding the value of cosec function, we can write the following
\[\Rightarrow \text{cosec }\!\!\theta\!\!\text{ }=\dfrac{1}{\sin \theta }\]
\[\begin{align}
  & \Rightarrow \text{cosec }\!\!\theta\!\!\text{ }=\dfrac{1}{\dfrac{\sqrt{3}}{2}} \\
 & \Rightarrow \text{cosec }\!\!\theta\!\!\text{ }=\dfrac{2}{\sqrt{3}} \\
\end{align}\]
Hence, these are the value of all the T-ratios of \[\theta \] . using the value of sin function that was given in the hint.

So, the correct answer is “Option A”.

Note: The other way of solving the above problem is that:
We have given $\sin \theta =\dfrac{\sqrt{3}}{2}$. And we know that, $\sin \theta =\dfrac{\sqrt{3}}{2}$ when $\theta ={{60}^{\circ }}$ so to find the other trigonometric ratios, we just have to put $\theta ={{60}^{\circ }}$ in those trigonometric ratios.
The trigonometric ratios we have already discussed above:
$\cos \theta =\cos {{60}^{\circ }}$
And we know that $\cos {{60}^{\circ }}=\dfrac{1}{2}$ so the value of $\cos \theta $ is equal to:
$\cos \theta =\dfrac{1}{2}$
$\tan \theta =\tan {{60}^{\circ }}$
We know that the value of $\tan {{60}^{\circ }}=\sqrt{3}$ so substituting this value in the above equation we get,
$\tan \theta =\sqrt{3}$
We know that, $\cot \theta $ is the reciprocal of $\tan \theta $ so:
$\cot \theta =\dfrac{1}{\tan \theta }=\dfrac{1}{\sqrt{3}}$
According to trigonometric ratios, $\sec \theta \And \text{cosec }\!\!\theta\!\!\text{ }$ is the reciprocal of $\cos \theta \And \sin \theta $ so we are going to write the values of $\sec \theta \And \text{cosec }\!\!\theta\!\!\text{ }$ as follows:
$\sec \theta =\dfrac{1}{\cos \theta }=\dfrac{1}{\cos {{60}^{\circ }}}=2$
$\text{cosec }\!\!\theta\!\!\text{ }=\dfrac{1}{\sin \theta }=\dfrac{1}{\sin {{60}^{\circ }}}=\dfrac{2}{\sqrt{3}}$