How do you solve \[\text{cosec}\left( x \right)+\cot \left( x \right)=1\] ?

Hint: The above mentioned problem is a simple example of trigonometric equations. Before solving this question we need to recall some of the basic equations and relations of trigonometry. We proceed by converting the given problem into simpler form having \[\sin x\] and \[\cos x\] as the main terms. The equations are described as:
\[\begin{align}
  & \sec \theta =\dfrac{1}{\cos \theta } \\
 & \text{cosec}\theta =\dfrac{1}{\sin \theta } \\
 & \cot \theta =\dfrac{1}{\tan \theta } \\
\end{align}\]
We also need to remember some of the basic defined relations, which are:
\[\begin{align}
  & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\
 & {{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1 \\
 & \text{cose}{{\text{c}}^{2}}\theta -{{\cot }^{2}}\theta =1 \\
\end{align}\]

Complete step-by-step solution:
We start solving the problem by converting the \[\text{cosec}\left( x \right)\] term into \[\sin x\] term and the \[\cot \left( x \right)\] term into \[\cos x\] term and proceeding by gradually applying the relations between \[\sin x\] and \[\cos x\] .
Now starting off with the solution, we can firstly notice the forms given are \[\cos ec\left( x \right)\] and \[\cot \left( x \right)\] . So we can replace these two terms with all the known terms, equations and relations we have. We can therefore modify the given problem as,
\[\dfrac{1}{\sin \left( x \right)}+\dfrac{\cos \left( x \right)}{\sin \left( x \right)}=1\] as we know from trigonometric relations that, \[\text{cosec}\left( x \right)=\dfrac{1}{\sin \left( x \right)}\] and \[\cot \left( x \right)=\dfrac{\cos \left( x \right)}{\sin \left( x \right)}\]
Now, proceeding with our intermediate equation, we evaluate that,
\[\dfrac{1+\cos \left( x \right)}{\sin \left( x \right)}=1\]
Now, as stated in the hint part in the basic defined relations, we have \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] . This equation is valid for any \[x\] or \[\dfrac{x}{2}\] . Thus we can also write this defined relation as \[{{\sin }^{2}}\left( \dfrac{x}{2} \right)+{{\cos }^{2}}\left( \dfrac{x}{2} \right)=1\] . In the intermediate equation we can thus replace \[1={{\sin }^{2}}\left( \dfrac{x}{2} \right)+{{\cos }^{2}}\left( \dfrac{x}{2} \right)\].
From the half angle formula, we also know that \[\sin \left( x \right)=2\sin \left( \dfrac{x}{2} \right)\cos \left( \dfrac{x}{2} \right)\] and \[\cos \left( x \right)={{\cos }^{2}}\left( \dfrac{x}{2} \right)-{{\sin }^{2}}\left( \dfrac{x}{2} \right)\]. Now, we apply these three above equations, in our original problem and thus evaluate it as,
\[\dfrac{\left( {{\sin }^{2}}\left( \dfrac{x}{2} \right)+{{\cos }^{2}}\left( \dfrac{x}{2} \right) \right)+\left( {{\cos }^{2}}\left( \dfrac{x}{2} \right)-{{\sin }^{2}}\left( \dfrac{x}{2} \right) \right)}{2\sin \left( \dfrac{x}{2} \right)\cos \left( \dfrac{x}{2} \right)}=1\]
Now, removing all the brackets we get,
\[\begin{align}
  & \dfrac{{{\sin }^{2}}\left( \dfrac{x}{2} \right)+{{\cos }^{2}}\left( \dfrac{x}{2} \right)+{{\cos }^{2}}\left( \dfrac{x}{2} \right)-{{\sin }^{2}}\left( \dfrac{x}{2} \right)}{2\sin \left( \dfrac{x}{2} \right)\cos \left( \dfrac{x}{2} \right)}=1 \\
 & \Rightarrow \dfrac{2{{\cos }^{2}}\left( \dfrac{x}{2} \right)}{2\sin \left( \dfrac{x}{2} \right)\cos \left( \dfrac{x}{2} \right)}=1 \\
 & \Rightarrow \dfrac{\cos \left( \dfrac{x}{2} \right)}{\sin \left( \dfrac{x}{2} \right)}=1 \\
 & \Rightarrow \cot \left( \dfrac{x}{2} \right)=1 \\
 & \Rightarrow \tan \left( \dfrac{x}{2} \right)=1 \\
\end{align}\]
Now, the general solution of the form \[\tan \left( x \right)=1\] is \[x=n\pi +\dfrac{\pi }{4}\] where \[n\]may be any integer.
Thus writing \[\tan \left( \dfrac{x}{2} \right)=1\] we write \[\dfrac{x}{2}=n\pi +\dfrac{\pi }{4}\] . Now solving for \[x\] we get,
\[x=2n\pi +\dfrac{\pi }{2}\] where \[n\] may be any integer.
This is thus the general solution to our given problem equation.

Note: We need to keep in mind all the trigonometric equations and general values. We should also carefully observe whether any range of values are given or not. Cross-checking of the answer can be done by putting the value in the original equation. Another way to solve the problem is by multiplying both sides of the given equation by \[\text{cosec}\left( x \right)-\cot \left( x \right)\] . Thus, the equation becomes
\[\Rightarrow \text{cosec}\left( x \right)-\cot \left( x \right)=1....\text{equation 1}\]
Adding the given equation and equation 1 , we get
\[\Rightarrow \text{2cosec}\left( x \right)=2\]
\[\Rightarrow \text{cosec}\left( x \right)=1....\text{equation}\left( a \right)\]
Similarly, subtracting equation 1 from the given equation, we get
\[\Rightarrow 2\cot \left( x \right)=2\]
\[\Rightarrow \cot \left( x \right)=1....\text{equation}\left( b \right)\]
Thus, we are left with \[x=2n\pi +\dfrac{\pi }{2}\] after evaluating $\text{equation}\left( a \right)$ and $\text{equation}\left( b \right)$ .