Question

If we have a trigonometric expression as $\tan \theta +\cot \theta =2$, then find the value of ${{\tan }^{2}}\theta +{{\cot }^{2}}\theta $.

Answer 1

Hint: We first define the identity formula of a square of two numbers. We also consider the relation of $\tan \theta $ and $\cot \theta $ to find the value of $\tan \theta .\cot \theta $. We take both sides’ square of the given and then replace the value of $\tan \theta .\cot \theta $. Finally, we find a solution to the problem.

Complete step-by-step solution:
We know the identity formula of ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
We also know the trigonometric form
$\begin{align}
  & \tan \theta =\dfrac{1}{\cot \theta } \\
 & \Rightarrow \tan \theta .\cot \theta =1 \\
\end{align}$
We use these two forms in the squaring off $\tan \theta +\cot \theta =2$.
Taking both sides square of $\tan \theta +\cot \theta =2$ we get ${{\left( \tan \theta +\cot \theta \right)}^{2}}={{2}^{2}}=4$.
We apply the identity formula and get ${{\tan }^{2}}\theta +{{\cot }^{2}}\theta +2\tan \theta \cot \theta =4$.
Now we place the value of $\tan \theta .\cot \theta =1$ in the equation ${{\tan }^{2}}\theta +{{\cot }^{2}}\theta +2\tan \theta \cot \theta =4$.
$\begin{align}
  & {{\tan }^{2}}\theta +{{\cot }^{2}}\theta +2\times 1=4 \\
 & \Rightarrow {{\tan }^{2}}\theta +{{\cot }^{2}}\theta +2=4 \\
 & \Rightarrow {{\tan }^{2}}\theta +{{\cot }^{2}}\theta =4-2=2 \\
\end{align}$
So, the value of ${{\tan }^{2}}\theta +{{\cot }^{2}}\theta $ is 2.

Note: We can also convert $\cot \theta $ into $\tan \theta $ from the very start in the equation of $\tan \theta +\cot \theta =2$. Then we don’t need to use the value of $\tan \theta .\cot \theta $ as we have already considered the relation between them as $\tan \theta =\dfrac{1}{\cot \theta }$. We then just need to square both sides to get the value of the problem.