Answer
418.5k+ views
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:
$\sec \theta =\dfrac{1}{\cos \theta }=\dfrac{1}{\cos {{60}^{\circ }}}=2$
$\text{cosec }\!\!\theta\!\!\text{ }=\dfrac{1}{\sin \theta }=\dfrac{1}{\sin {{60}^{\circ }}}=\dfrac{2}{\sqrt{3}}$
\[\begin{align}
& {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\
& \text{sec }\!\!\theta\!\!\text{ }=\dfrac{1}{\cos \theta } \\
& \tan \theta =\dfrac{\sin \theta }{\cos \theta } \\
& \text{cosec }\!\!\theta\!\!\text{ }=\dfrac{1}{\sin \theta } \\
& \cot \theta =\dfrac{1}{\tan \theta } \\
\end{align}\]
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:
$\sec \theta =\dfrac{1}{\cos \theta }=\dfrac{1}{\cos {{60}^{\circ }}}=2$
$\text{cosec }\!\!\theta\!\!\text{ }=\dfrac{1}{\sin \theta }=\dfrac{1}{\sin {{60}^{\circ }}}=\dfrac{2}{\sqrt{3}}$
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)