Evaluate the expression $\dfrac{1}{\sqrt{3}}\sec 60-\operatorname{cosec}60$.
Answer
596.4k+ views
Hint: This question is based on trigonometric ratios. To solve such kind of problems, our approach will be to put the trigonometric ratio values in the problem. So, we will use results like $\operatorname{cosec}{{60}^{\circ }}=\dfrac{2}{\sqrt{3}}$ and $\sec {{60}^{\circ }}=2$. We will find the values and use them to get the desired answer.
Complete step-by-step solution:
We have been asked to find the value of the given expression, $\dfrac{1}{\sqrt{3}}\sec 60-\operatorname{cosec}60$. We can see that this is based on trigonometric ratios. Now, standardly speaking, we know the value of certain trigonometric ratios, namely, $\sin ,\cos ,\tan ,\cot ,\operatorname{cosec},\sec $ at certain angles only, like ${{0}^{\circ }},{{30}^{\circ }},{{45}^{\circ }},{{60}^{\circ }},{{90}^{\circ }}$. We know that angles such as these lie in the first quadrant. For angles which are lying in other quadrants, we use the trigonometric identities to solve them and convert into the first quadrant angles and then solve the equation.
Here, the equation given to us is,
$\dfrac{1}{\sqrt{3}}\sec {{60}^{\circ }}-\operatorname{cosec}{{60}^{\circ }}$
Now, since both the given angles are ${{60}^{\circ }}$, lying in the first quadrant and that too, they are among the known angles, we can directly put the value of those ratios and then solve the equation. We know that,
$\operatorname{cosec}{{60}^{\circ }}=\dfrac{2}{\sqrt{3}}$
And $\sec {{60}^{\circ }}=2$
So, we will substitute these values in the given equation. Doing so, we get,
$\begin{align}
& \dfrac{1}{\sqrt{3}}\sec {{60}^{\circ }}-\operatorname{cosec}{{60}^{\circ }} \\
& \Rightarrow \dfrac{1}{\sqrt{3}}\times 2-\dfrac{2}{\sqrt{3}} \\
& \Rightarrow \dfrac{2}{\sqrt{3}}-\dfrac{2}{\sqrt{3}} \\
& \Rightarrow 0 \\
\end{align}$
Hence, we get the value of the given expression as 0.
Note: While solving such questions, we have to make sure to solve angles by making them fall in the first quadrant and that too they should be among the angle values that we are aware of. We should be careful while substituting the values of the angles, otherwise you will end up with a wrong answer. If students do not know the values of sec and cosec functions, they can convert in terms of sin and cos to simplify.
Complete step-by-step solution:
We have been asked to find the value of the given expression, $\dfrac{1}{\sqrt{3}}\sec 60-\operatorname{cosec}60$. We can see that this is based on trigonometric ratios. Now, standardly speaking, we know the value of certain trigonometric ratios, namely, $\sin ,\cos ,\tan ,\cot ,\operatorname{cosec},\sec $ at certain angles only, like ${{0}^{\circ }},{{30}^{\circ }},{{45}^{\circ }},{{60}^{\circ }},{{90}^{\circ }}$. We know that angles such as these lie in the first quadrant. For angles which are lying in other quadrants, we use the trigonometric identities to solve them and convert into the first quadrant angles and then solve the equation.
Here, the equation given to us is,
$\dfrac{1}{\sqrt{3}}\sec {{60}^{\circ }}-\operatorname{cosec}{{60}^{\circ }}$
Now, since both the given angles are ${{60}^{\circ }}$, lying in the first quadrant and that too, they are among the known angles, we can directly put the value of those ratios and then solve the equation. We know that,
$\operatorname{cosec}{{60}^{\circ }}=\dfrac{2}{\sqrt{3}}$
And $\sec {{60}^{\circ }}=2$
So, we will substitute these values in the given equation. Doing so, we get,
$\begin{align}
& \dfrac{1}{\sqrt{3}}\sec {{60}^{\circ }}-\operatorname{cosec}{{60}^{\circ }} \\
& \Rightarrow \dfrac{1}{\sqrt{3}}\times 2-\dfrac{2}{\sqrt{3}} \\
& \Rightarrow \dfrac{2}{\sqrt{3}}-\dfrac{2}{\sqrt{3}} \\
& \Rightarrow 0 \\
\end{align}$
Hence, we get the value of the given expression as 0.
Note: While solving such questions, we have to make sure to solve angles by making them fall in the first quadrant and that too they should be among the angle values that we are aware of. We should be careful while substituting the values of the angles, otherwise you will end up with a wrong answer. If students do not know the values of sec and cosec functions, they can convert in terms of sin and cos to simplify.
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