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Find the value of $\dfrac{\sin \left( -{{660}^{\circ }} \right)\tan \left( {{1050}^{\circ }} \right)\sec \left( -{{420}^{\circ }} \right)}{\cos \left( {{225}^{\circ }} \right)\csc \left( {{315}^{\circ }} \right)\cos \left( {{510}^{\circ }} \right)}$?
(a) $\dfrac{\sqrt{3}}{4}$
(b) $\dfrac{\sqrt{3}}{2}$
(c) $\dfrac{2}{\sqrt{3}}$
(d) $\dfrac{4}{\sqrt{3}}$

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Last updated date: 13th Jul 2024
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Answer
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Hint: Assume the given expression as E and use the relations $\sin \left( -x \right)=-\sin x$ and $\sec \left( -x \right)=\sec x$ to make the angles of all the trigonometric ratios positive. Now, convert all the angles from degrees into radians by multiplying then with $\dfrac{\pi }{{{180}^{\circ }}}$. Use the sign convention of the given trigonometric functions in certain quadrants and simplify the expression by substituting the values of trigonometric ratios of special angles to get the answer.

Complete step by step answer:
Here we have been provided with the expression $\dfrac{\sin \left( -{{660}^{\circ }} \right)\tan \left( {{1050}^{\circ }} \right)\sec \left( -{{420}^{\circ }} \right)}{\cos \left( {{225}^{\circ }} \right)\csc \left( {{315}^{\circ }} \right)\cos \left( {{510}^{\circ }} \right)}$ and we have to find its value. Let us assume the given expression as E, so we have,
$\Rightarrow E=\dfrac{\sin \left( -{{660}^{\circ }} \right)\tan \left( {{1050}^{\circ }} \right)\sec \left( -{{420}^{\circ }} \right)}{\cos \left( {{225}^{\circ }} \right)\csc \left( {{315}^{\circ }} \right)\cos \left( {{510}^{\circ }} \right)}$
Using the formulas $\sin \left( -x \right)=-\sin x$ and $\sec \left( -x \right)=\sec x$ to make the angles positive we get,
$\Rightarrow E=-\left[ \dfrac{\sin \left( {{660}^{\circ }} \right)\tan \left( {{1050}^{\circ }} \right)\sec \left( {{420}^{\circ }} \right)}{\cos \left( {{225}^{\circ }} \right)\csc \left( {{315}^{\circ }} \right)\cos \left( {{510}^{\circ }} \right)} \right]$
Converting all the angles from degrees into radians we by multiplying them with $\dfrac{\pi }{{{180}^{\circ }}}$ we get,
\[\Rightarrow E=-\left[ \dfrac{\sin \left( \dfrac{11\pi }{3} \right)\tan \left( \dfrac{35\pi }{6} \right)\sec \left( \dfrac{7\pi }{3} \right)}{\cos \left( \dfrac{5\pi }{4} \right)\csc \left( \dfrac{7\pi }{4} \right)\cos \left( \dfrac{17\pi }{6} \right)} \right]\]
Here we can write the angles of different trigonometric function present above as the sum or difference of two angles, so we get,
\[\Rightarrow E=-\left[ \dfrac{\sin \left( 4\pi -\dfrac{\pi }{3} \right)\tan \left( 6\pi -\dfrac{\pi }{6} \right)\sec \left( 2\pi +\dfrac{\pi }{3} \right)}{\cos \left( \pi +\dfrac{\pi }{4} \right)\csc \left( 2\pi -\dfrac{\pi }{4} \right)\cos \left( 3\pi -\dfrac{\pi }{6} \right)} \right]\]
We know that in the first quadrant all the trigonometric functions are positive, in the second quadrant only sine and cosecant function is positive, in the third quadrant only tangent and cotangent function is positive and in the fourth quadrant only cosine and secant function is positive. Therefore the above expression can be simplified as: -
\[\begin{align}
  & \Rightarrow E=-\left[ \dfrac{\left( -\sin \left( \dfrac{\pi }{3} \right) \right)\left( -\tan \left( \dfrac{\pi }{6} \right) \right)\sec \left( \dfrac{\pi }{3} \right)}{\left( -\cos \left( \dfrac{\pi }{4} \right) \right)\left( -\csc \left( \dfrac{\pi }{4} \right) \right)\left( -\cos \left( \dfrac{\pi }{6} \right) \right)} \right] \\
 & \Rightarrow E=\left[ \dfrac{\sin \left( \dfrac{\pi }{3} \right)\tan \left( \dfrac{\pi }{6} \right)\sec \left( \dfrac{\pi }{3} \right)}{\cos \left( \dfrac{\pi }{4} \right)\csc \left( \dfrac{\pi }{4} \right)\cos \left( \dfrac{\pi }{6} \right)} \right] \\
\end{align}\]
Using and substituting the values $\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$, $\tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}}$, $\sec \dfrac{\pi }{3}=2$, $\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$, $\csc \dfrac{\pi }{4}=\sqrt{2}$ and $\cos \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2}$ in the above expression we get,
\[\begin{align}
  & \Rightarrow E=\left[ \dfrac{\dfrac{\sqrt{3}}{2}\times \dfrac{1}{\sqrt{3}}\times 2}{\dfrac{1}{\sqrt{2}}\times \sqrt{2}\times \dfrac{\sqrt{3}}{2}} \right] \\
 & \therefore E=\dfrac{2}{\sqrt{3}} \\
\end{align}\]

So, the correct answer is “Option c”.

Note: Note that the only reason to convert the given angles from degrees into radians is that in higher classes we use the notation of radian and all the formulas are written in terms of $\pi $. You must remember the signs of all the trigonometric functions in different quadrants otherwise you may make sign mistakes while solving. Also, remember the values of all the functions for certain special angles like $\dfrac{\pi }{6}$, $\dfrac{\pi }{4}$, $\dfrac{\pi }{3}$, $\dfrac{\pi }{2}$ etc.