Question

# The general solution of $\cot \theta +\tan \theta =2$ (a) $\theta =\dfrac{n\pi }{2}+{{\left( -1 \right)}^{n}}\dfrac{\pi }{8}$(b) $\theta =\dfrac{n\pi }{2}+{{\left( -1 \right)}^{n}}\dfrac{\pi }{4}$(c) $\theta =\dfrac{n\pi }{2}+{{\left( -1 \right)}^{n}}\dfrac{\pi }{6}$(d) $\theta =n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{8}$

Hint: The $\cot \theta$ in the left side of the equation can be written as $\dfrac{1}{\tan \theta }$ and further solving the equation gives us a quadratic equation
which can be solved to get $\tan \theta$.

Before proceeding with the question, we must know the general solution of an equation
involving $\tan \theta$. If we are given an equation,
$\tan \theta =m$
The general solution of this equation is given by the formula,
$\theta =n\pi +\left( Principal\text{ solution }of~{{\tan }^{-1}}m \right).............\left( 1 \right)$
In this question, we have to find the solution of the equation $\cot \theta +\tan \theta =2...........\left( 2 \right)$.
In trigonometry, we have a formula $\cot \theta =\dfrac{1}{\tan \theta }.............\left( 3 \right)$.
Substituting $\cot \theta =\dfrac{1}{\tan \theta }$ from equation $\left( 3 \right)$ in equation $\left( 2 \right)$, we get,
\begin{align} & \dfrac{1}{\tan \theta }+\tan \theta =2 \\ & \Rightarrow \dfrac{1+{{\tan }^{2}}\theta }{\tan \theta }=2 \\ & \Rightarrow 1+{{\tan }^{2}}\theta =2\tan \theta \\ & \Rightarrow {{\tan }^{2}}\theta -2\tan \theta +1=0 \\ \end{align}
The above equation is a quadratic equation which can be written as ${{\left( \tan \theta -1 \right)}^{2}}$since ${{\left( \tan \theta -1 \right)}^{2}}={{\tan }^{2}}\theta -2\tan \theta +1$. Substituting ${{\tan }^{2}}\theta -2\tan \theta +1={{\left( \tan \theta -1 \right)}^{2}}$, we get,
\begin{align} & {{\left( \tan \theta -1 \right)}^{2}}=0 \\ & \Rightarrow \left( \tan \theta -1 \right)=0 \\ & \Rightarrow \tan \theta =1 \\ \end{align}
Form equation $\left( 1 \right)$, we can find the solution of the above equation. Substituting

$m=1$ in equation $\left( 1 \right)$, we get,
\begin{align} & \theta =n\pi +\left( Principal\text{ solution }of~{{\tan }^{-1}}1 \right) \\ & \Rightarrow \theta =n\pi +\dfrac{\pi }{4} \\ \end{align}
Let us put different integral values of $n$ to obtain the solution set of $\theta$. Substituting
$n=-2,-1,0,1,2$, we get the solution set of $\theta$,
$\theta =\left\{ .....,\dfrac{-7\pi }{4},\dfrac{-3\pi }{4},\dfrac{\pi }{4},\dfrac{5\pi }{4},\dfrac{9\pi }{4},...... \right\}$
Now, let us find the solution set of all the options in the question. We will substitute different
integral values of $n$ in all the options.
Solution set of option (a) is $\left\{ .....,\dfrac{-7\pi }{8},\dfrac{-5\pi }{8},\dfrac{\pi }{8},\dfrac{3\pi }{8},\dfrac{9\pi }{8},...... \right\}$
Solution set of option (b) is $\left\{ .....,\dfrac{-3\pi }{4},\dfrac{\pi }{4},\dfrac{5\pi }{4},\dfrac{9\pi }{4},...... \right\}$
Solution set of option (c) is $\left\{ .....,\dfrac{-5\pi }{6},\dfrac{-2\pi }{3},\dfrac{\pi }{6},\dfrac{\pi }{3},\dfrac{2\pi }{3},...... \right\}$
Solution set of option (d) is $\left\{ .....,\dfrac{-15\pi }{8},\dfrac{-9\pi }{8},\dfrac{\pi }{8},\dfrac{7\pi }{8},\dfrac{17\pi }{8},...... \right\}$
We can clearly observe that the solution set of option (b) is matching to the solution set of
$\theta$.

Option (b) is the correct answer

Note: In such a type of question, sometimes it is possible that our general solution does not
match with any of the options given in the question. In that case we should make the solution set of our answer and then compare this solution set to the solution set of the
options to obtain the correct answer.