Question

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

Answer 1

Hint: 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.