If a trigonometric equation is given as $\tan A+\cot A=2$ , then find the value of ${{\tan }^{2}}A+{{\cot }^{2}}A$

Hint: To find the value of the term ${{\tan }^{2}}A+{{\cot }^{2}}A$, we will have to first find the value of $\tan A$ and $\cot A$ and as the tan and cot functions are ratios of sine and cosine functions, we will first convert tan and cot functions in terms of sine and cosine functions and then solve.

Complete step-by-step solution -
In the question, we are given that $\tan A+\cot A=2$ . So, the first thing we are going to do is to assign this equation as equation (i). So, we will get
$\tan A+\cot A=2................\left( i \right)$
Now, in the next step we are going to convert the tan and cot functions into sine and cosine functions respectively. To do this, we are going to use the following identities:
$\begin{align}
  & \tan A=\dfrac{\sin A}{\cos A}................\left( ii \right) \\
 & \cot A=\dfrac{\cos A}{\sin A}................\left( iii \right) \\
\end{align}$
Now, we will put the values of $\tan A$ and $\cot A$ from equation (ii) and (iii) respectively to the equation (i). After substituting these values, we will get:
$\dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A}=2$
Taking LCM on the left-hand side, we get:
$\begin{align}
  & \dfrac{\left( \sin A \right)\left( \sin A \right)+\left( \cos A \right)\left( \cos A \right)}{\sin A\cos A}=2 \\
 & \Rightarrow \dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{\sin A\cos A}=2................\left( iv \right) \\
\end{align}$
Here, we are going to use another identity
$\Rightarrow {{\sin }^{2}}A+{{\cos }^{2}}A=1..............\left( v \right)$
Now we will put the value of the term ${{\sin }^{2}}A+{{\cos }^{2}}A$ as 1 from equation (v) in equation (iv). After doing this, we will get:
$\Rightarrow \dfrac{1}{\sin A\cos A}=2$
After cross multiplication, we get
$\Rightarrow 2\sin A\cos A=1..............\left( vi \right)$
Here, as we know that $\sin 2A=2\sin A\cos A$ , we will put his value in equation (vi)
$\begin{align}
  & \Rightarrow \sin 2A=1................\left( vii \right) \\
 & \Rightarrow \sin 2A=\sin \dfrac{\pi }{2} \\
\end{align}$
Here, we have used the principle solution for equation (vii). Therefore, we get,
$\begin{align}
  & \Rightarrow 2A=\dfrac{\pi }{2} \\
 & \Rightarrow A=\dfrac{\pi }{4}radian \\
\end{align}$
Now, we will evaluate the term given in question we have to find the value of
${{\tan }^{2}}A+{{\cot }^{2}}A$
We have found the value of A as $\dfrac{\pi }{4}$ . So, we will solve the above equation by substituting this value. Thus
$\begin{align}
  & {{\left( \tan A \right)}^{2}}+{{\left( \cot A \right)}^{2}}=\dfrac{{{\left( \sin A \right)}^{2}}}{{{\left( \cos A \right)}^{2}}}+\dfrac{{{\left( \cos A \right)}^{2}}}{{{\left( \sin A \right)}^{2}}} \\
 & =\dfrac{{{\left( \sin \dfrac{\pi }{4} \right)}^{2}}}{{{\left( \cos \dfrac{\pi }{4} \right)}^{2}}}+\dfrac{{{\left( \cos \dfrac{\pi }{4} \right)}^{2}}}{{{\left( \sin \dfrac{\pi }{4} \right)}^{2}}}
\end{align}$
The value of $\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$ and the value of $\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$ . Thus, we get
$\begin{align}
  & {{\left( \tan A \right)}^{2}}+{{\left( \cot A \right)}^{2}}=\dfrac{{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}}{{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}}+\dfrac{{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}}{{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}} \\
 & =\dfrac{\left( \dfrac{1}{2} \right)}{\left( \dfrac{1}{2} \right)}+\dfrac{\left( \dfrac{1}{2} \right)}{\left( \dfrac{1}{2} \right)} \\
 & =1+1 \\
 & =2
\end{align}$
Hence the value of \[{{\tan }^{2}}A+{{\cot }^{2}}A=2\].

Note: We can solve this question by without actually finding the angle A. This can be done by:
$\begin{align}
  & \left( \tan A+\cot A \right)=2\Rightarrow {{\left( \tan A+\cot A \right)}^{2}}={{2}^{2}} \\
 & \Rightarrow {{\tan }^{2}}A+{{\cot }^{2}}A+2=4\Rightarrow {{\tan }^{2}}A+{{\cot }^{2}}A=2 \\
\end{align}$