Question

Find the value of $2{{\sin }^{2}}30{{\tan }^{2}}60-3{{\cos }^{2}}60{{\sec }^{2}}30$

Answer 1

Hint: In this type of question we have to use the concept of trigonometry. We can find the value of a given expression by using the values of the trigonometric ratios of general angles. To solve the given expression we use the table of trigonometric ratios of general angles such as ${{0}^{\circ }},{{30}^{\circ }},{{45}^{\circ }},{{60}^{\circ }},{{90}^{\circ }}$ etc. From this table we can find the values of $\sin {{30}^{\circ }},\tan {{60}^{\circ }},\cos {{60}^{\circ }}\And \sec {{30}^{\circ }}$ and substitute them in the given expression. Then we have to simplify the expression by using the BODMAS rule to obtain the required answer. The table of trigonometric values is given as follows:

	$\sin \theta $	$\cos \theta $	$\tan \theta $	$\text{cosec}\theta $	$\sec \theta $	$\cot \theta $
$0$	$0$	$1$	$0$	$\infty$	$1$	$\infty$
$\dfrac{\pi }{6}$	$\dfrac{1}{2}$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{1}{\sqrt{3}}$	$2$	$\dfrac{2}{\sqrt{3}}$	$\sqrt{3}$
$\dfrac{\pi }{4}$	$\dfrac{1}{\sqrt{2}}$	$\dfrac{1}{\sqrt{2}}$	$1$	$\sqrt{2}$	$\sqrt{2}$	$1$
$\dfrac{\pi }{3}$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{1}{2}$	$\sqrt{3}$	$\dfrac{2}{\sqrt{3}}$	$2$	$\dfrac{1}{\sqrt{3}}$
$\dfrac{\pi }{2}$	$1$	$0$	$\infty$	$1$	$\infty$	$0$

Complete step by step answer:
Now, we have to find the value of the expression, $2{{\sin }^{2}}30{{\tan }^{2}}60-3{{\cos }^{2}}60{{\sec }^{2}}30$
To simplify the above expression, we have to substitute the values of $\sin {{30}^{\circ }},\tan {{60}^{\circ }},\cos {{60}^{\circ }}\And \sec {{30}^{\circ }}$.
Let us consider the given expression,
$\Rightarrow \text{E = }2{{\sin }^{2}}30{{\tan }^{2}}60-3{{\cos }^{2}}60{{\sec }^{2}}30\text{ }\cdots \cdots \cdots \left( i \right)$
Now, we know that, from the table of values of trigonometric ratios of general angles such as ${{0}^{\circ }},{{30}^{\circ }},{{45}^{\circ }},{{60}^{\circ }},{{90}^{\circ }}$ we have,
$\Rightarrow \sin {{30}^{\circ }}=\dfrac{1}{2},\tan {{60}^{\circ }}=\sqrt{3},\cos {{60}^{\circ }}=\dfrac{1}{2}\And \sec {{30}^{\circ }}=\dfrac{2}{\sqrt{3}}$
By substituting this values in above equation we can write,
$\begin{align}
  & \Rightarrow \text{E = }2{{\sin }^{2}}30{{\tan }^{2}}60-3{{\cos }^{2}}60{{\sec }^{2}}30\text{ } \\
 & \Rightarrow \text{E = }2{{\left( \dfrac{1}{2} \right)}^{2}}{{\left( \sqrt{3} \right)}^{2}}-3{{\left( \dfrac{1}{2} \right)}^{2}}{{\left( \dfrac{2}{\sqrt{3}} \right)}^{2}}\text{ } \\
\end{align}$
By simplifying we get,
$\begin{align}
  & \Rightarrow \text{E = }2\left( \dfrac{1}{4} \right)\left( 3 \right)-3\left( \dfrac{1}{4} \right)\left( \dfrac{4}{3} \right) \\
 & \Rightarrow \text{ E = }\left( \dfrac{3}{2} \right)-\left( 1 \right) \\
 & \Rightarrow \text{ E = }\dfrac{1}{2} \\
\end{align}$
Hence, we get, the value of the expression $2{{\sin }^{2}}30{{\tan }^{2}}60-3{{\cos }^{2}}60{{\sec }^{2}}30$ is equal to $\dfrac{1}{2}$.

Note: In this type of question students have to take care during the calculation and have to follow the BODMAS rule. Also, students have to remember at least the values of $\sin \theta \And \cos \theta $ at different angles, so that from these values and by using the relations $\tan \theta =\dfrac{\sin \theta }{\cos \theta },\cot \theta =\dfrac{\cos \theta }{\sin \theta },\cos ec\theta =\dfrac{1}{\sin \theta },\sec \theta =\dfrac{1}{\cos \theta }$, they can find all the other trigonometric ratios which are required in the expressions.