Question

Evaluate the following ${{\tan }^{2}}{{30}^{\circ }}+{{\tan }^{2}}{{60}^{\circ }}+{{\tan }^{2}}{{45}^{\circ }}$.

Answer 1

Hint:In order to solve this question, we should know a few values of trigonometric ratios where, trigonometric ratios are the ratios of two of the three sides of a right angled triangle, like $\tan \theta =\dfrac{perpendicular}{base}$. We should know trigonometric standard angles i.e $\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}},\tan {{45}^{\circ }}=1$ and $\tan {{60}^{\circ }}=\sqrt{3}$ to solve this question.By using these we can find the answer.

Complete step-by-step answer:
In this question, we are asked to evaluate an expression, that is, ${{\tan }^{2}}{{30}^{\circ }}+{{\tan }^{2}}{{60}^{\circ }}+{{\tan }^{2}}{{45}^{\circ }}$. To solve this question, we should know about the trigonometric ratios, which are the ratios of two of the three sides of a right angled triangle, like we know that, $\tan \theta =\dfrac{perpendicular}{base}$. In the given expression, we will put the value of the tan angles, that are, $\tan {{30}^{\circ }},\tan {{45}^{\circ }}$ and $\tan {{60}^{\circ }}$, that are, $\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}},\tan {{45}^{\circ }}=1$ and $\tan {{60}^{\circ }}=\sqrt{3}$. Therefore, we can write the given expression as,
${{\tan }^{2}}{{30}^{\circ }}+{{\tan }^{2}}{{60}^{\circ }}+{{\tan }^{2}}{{45}^{\circ }}$
${{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}+{{\left( \sqrt{3} \right)}^{2}}+{{\left( 1 \right)}^{2}}$
Now, we know that ${{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}=\dfrac{1}{3},{{\left( \sqrt{3} \right)}^{2}}=3$ and ${{\left( 1 \right)}^{2}}=1$. So, we can write the expression as follows,
$\begin{align}
  & \dfrac{1}{3}+3+1 \\
 & \Rightarrow \dfrac{1}{3}+4 \\
\end{align}$
Now, we will take the LCM of the above values. So, we will get,
$\begin{align}
  & \dfrac{1+4\times 3}{3} \\
 & \Rightarrow \dfrac{1+12}{3} \\
 & \Rightarrow \dfrac{13}{3} \\
\end{align}$
Hence, we get the value of the given expression, ${{\tan }^{2}}{{30}^{\circ }}+{{\tan }^{2}}{{60}^{\circ }}+{{\tan }^{2}}{{45}^{\circ }}$ as $\dfrac{13}{3}$.

Note: While solving this question, the students can think of applying the identity $1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta $ from which we will get ${{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1$ which is also a correct method to solve the question, but the number of terms will increase in the expression and the solution will become more complex and hence, the chances of calculation mistakes will be very high. So, it is better to remember the trigonometric standard angle values of $\tan {{30}^{\circ }},\tan {{45}^{\circ }},\tan {{60}^{\circ }}$, that are, $\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}},\tan {{45}^{\circ }}=1$ and $\tan {{60}^{\circ }}=\sqrt{3}$ so that the students can directly substitute these values and solve the question.