Question

Find the value of the following:
\[\dfrac{\cos (0).\sin \dfrac{\pi }{2}.{{\cos }^{2}}\left( \dfrac{\pi }{6} \right).{{\sec }^{2}}\left( \dfrac{\pi }{4} \right)}{\tan \left( \dfrac{\pi }{3} \right)+\cot \left( \dfrac{\pi }{3} \right)}\]

Answer 1

Hint: In this question, first of all, recollect the value of \[\cos \left( {{0}^{o}} \right),\sin \dfrac{\pi }{2},\cos \dfrac{\pi }{6},\sec \dfrac{\pi }{4},\tan \dfrac{\pi }{3},\cot \dfrac{\pi }{3}\] by using a trigonometric table for general angles. Now, substitute these values in the given expression to get the required answer.

Complete step-by-step solution -
In this question, we have to find the value of the expression
\[\dfrac{\cos (0).\sin \dfrac{\pi }{2}.{{\cos }^{2}}\left( \dfrac{\pi }{6} \right).{{\sec }^{2}}\left( \dfrac{\pi }{4} \right)}{\tan \left( \dfrac{\pi }{3} \right)+\cot \left( \dfrac{\pi }{3} \right)}\]
Let us consider the expression given in the question,
\[E=\dfrac{\cos (0).\sin \dfrac{\pi }{2}.{{\cos }^{2}}\left( \dfrac{\pi }{6} \right).{{\sec }^{2}}\left( \dfrac{\pi }{4} \right)}{\tan \left( \dfrac{\pi }{3} \right)+\cot \left( \dfrac{\pi }{3} \right)}\]
We can also write the above expression as,
\[E=\dfrac{\cos (0).\sin \dfrac{\pi }{2}.{{\left[ \cos \left( \dfrac{\pi }{6} \right) \right]}^{2}}.{{\left[ \sec \left( \dfrac{\pi }{4} \right) \right]}^{2}}}{\tan \left( \dfrac{\pi }{3} \right)+\cot \left( \dfrac{\pi }{3} \right)}.....\left( i \right)\]
Now let us find the values of \[\cos \left( {{0}^{o}} \right),\sin \dfrac{\pi }{2},\cos \dfrac{\pi }{6},\sec \dfrac{\pi }{4},\tan \dfrac{\pi }{3},\cot \dfrac{\pi }{3}\] from the trigonometric table for general angles.

	\[\sin \theta \]	\[\cos \theta \]	\[\tan \theta \]	\[\operatorname{cosec}\theta \]	\[\sec \theta \]	\[\cot \theta \]
0	0	1	0	-	1	-
\[\dfrac{\pi }{6}\]	\[\dfrac{1}{2}\]	\[\dfrac{\sqrt{3}}{2}\]	\[\dfrac{1}{\sqrt{3}}\]	2	\[\dfrac{2}{\sqrt{3}}\]	\[\sqrt{3}\]
\[\dfrac{\pi }{4}\]	\[\dfrac{1}{\sqrt{2}}\]	\[\dfrac{1}{\sqrt{2}}\]	1	\[\sqrt{2}\]	\[\sqrt{2}\]	1
\[\dfrac{\pi }{3}\]	\[\dfrac{\sqrt{3}}{2}\]	\[\dfrac{1}{2}\]	\[\sqrt{3}\]	\[\dfrac{2}{\sqrt{3}}\]	2	\[\dfrac{1}{\sqrt{3}}\]
\[\dfrac{\pi }{2}\]	1	0	-	1	-	0

From the above table, we get the values of \[\cos \left( {{0}^{o}} \right),\sin \dfrac{\pi }{2},\cos \dfrac{\pi }{6},\sec \dfrac{\pi }{4},\tan \dfrac{\pi }{3},\cot \dfrac{\pi }{3}\] as follows:
\[\cos \left( 0 \right)=1\]
\[\sin \left( \dfrac{\pi }{2} \right)=1\]
\[\cos \left( \dfrac{\pi }{6} \right)=\dfrac{\sqrt{3}}{2}\]
\[\sec \left( \dfrac{\pi }{4} \right)=\sqrt{2}\]
\[\tan \left( \dfrac{\pi }{3} \right)=\sqrt{3}\]
\[\cot \left( \dfrac{\pi }{3} \right)=\dfrac{1}{\sqrt{3}}\]
By substituting these values in equation (i), we get,
\[E=\dfrac{\left( 1 \right)\left( 1 \right){{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}{{\left( \sqrt{2} \right)}^{2}}}{\sqrt{3}+\dfrac{1}{\sqrt{3}}}\]
By simplifying the above expression, we get,
\[E=\dfrac{\left( 1 \right).\left( 1 \right).\left( \dfrac{3}{4} \right).\left( 2 \right)}{\sqrt{3}+\dfrac{1}{\sqrt{3}}}\]
\[E=\dfrac{\dfrac{6}{4}}{\dfrac{{{\left( \sqrt{3} \right)}^{2}}+1}{\sqrt{3}}}\]
\[E=\dfrac{\dfrac{6}{4}}{\dfrac{3+1}{\sqrt{3}}}\]
\[E=\dfrac{\dfrac{6}{4}}{\dfrac{4}{\sqrt{3}}}\]
\[E=\dfrac{6}{4}\times \dfrac{\sqrt{3}}{4}\]
\[E=\dfrac{6\sqrt{3}}{16}\]
By simplifying the above fraction, we get,
\[E=\dfrac{3\sqrt{3}}{8}\]
Hence, we get the value of \[\dfrac{\cos (0).\sin \dfrac{\pi }{2}.{{\cos }^{2}}\left( \dfrac{\pi }{6} \right).{{\sec }^{2}}\left( \dfrac{\pi }{4} \right)}{\tan \left( \dfrac{\pi }{3} \right)+\cot \left( \dfrac{\pi }{3} \right)}\text{ as }\dfrac{3\sqrt{3}}{8}\].


Note: In these types of questions, first of all, it is very important to remember all the trigonometric ratios of general angles like \[{{0}^{o}},\dfrac{\pi }{6},\dfrac{\pi }{3},\dfrac{\pi }{4},\dfrac{\pi }{2},etc.\] and also substitute them properly. Students are advised to at least remember all the values for \[\sin \theta \text{ and }\cos \theta \] because by using these, we can easily find the values for other trigonometric ratios like \[\tan \theta =\dfrac{\sin \theta }{\cos \theta },\cot \theta =\dfrac{\cos \theta }{\sin \theta },\sec \theta =\dfrac{1}{\cos \theta },\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }\].