Question

How do you solve $\tan x + \cot x - 2 = 0$?

Answer 1

To solve $\tan x + \cot x - 2 = 0$ we will transform cotangent value to tangent value and then we will form a quadratic equation in tangent. We will assume $\tan x = t$ and we will solve the quadratic equation in $t$ using the factorization method. Then we will substitute back $t$ as $\tan x$. At last, we will take the inverse trigonometric function on both sides to find the result.

Complete step by step answer:
Given, $\tan x + \cot x - 2 = 0$.
Substituting $\cot x = \dfrac{1}{{\tan x}}$ in the above equation, we get
$\Rightarrow \tan x + \dfrac{1}{{\tan x}} - 2 = 0$
Taking the LCM on the left hand side of the equation, we get
$\Rightarrow \dfrac{{{{\tan }^2}x - 2\tan x + 1}}{{\tan x}} = 0$
Cross multiplying the value from the denominator of left hand side of the equation to right hand side of the equation, we get
$\Rightarrow {\tan ^2}x - 2\tan x + 1 = 0$
We can see that this is a quadratic equation in $\tan x$.
Now, putting $\tan x = t$, we get
$\Rightarrow {\operatorname{t} ^2} - 2\operatorname{t} + 1 = 0$
Now splitting the coefficient of $\operatorname{t}$ in a way that its product is equal to the product of coefficient of other two terms and sum is equal to coefficient to $\operatorname{t}$.
$\Rightarrow {t^2} - t - t + 1 = 0$
Taking common, we get
$\Rightarrow t\left( {t - 1} \right) - 1\left( {t - 1} \right) = 0$
$\Rightarrow \left( {t - 1} \right)\left( {t - 1} \right) = 0$
On solving we get
$\Rightarrow t = 1$
Now, substituting $t = \tan x$, we get
$\Rightarrow \tan x = 1$
Taking inverse trigonometric function on both the sides, we get
$\Rightarrow {\tan ^{ - 1}}\left( {\tan x} \right) = {\tan ^{ - 1}}\left( 1 \right)$
As ${\tan ^{ - 1}}\left( {\tan x} \right) = x$, we get
$\Rightarrow x = {\tan ^{ - 1}}\left( 1 \right)$
Therefore, the solution of $\tan x + \cot x - 2 = 0$ is ${\tan ^{ - 1}}\left( 1 \right)$.

Note:
If we further solve the result ${\tan ^{ - 1}}\left( 1 \right)$, then we will get $\dfrac{\pi }{4}$ as the principal solution. A function that repeats its values after every particular interval is called the periodic function. As we know, $\tan x$ is a periodic function and the period is $\pi$. The general solution is given by $\dfrac{\pi }{4} + n\pi$. So, $\dfrac{\pi }{4} + \pi$, $\dfrac{\pi }{4} + 2\pi$, $\dfrac{\pi }{4} + 3\pi$, etc. are also the solution.