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Question: The general solution of \({\tan ^2}x = 1\) is: \( {\text{A}}{\text{. }}\,\,n\pi + \dfrac{\pi }...

The general solution of tan2x=1{\tan ^2}x = 1 is:
A. nπ+π4 B.nππ4 C.nπ±π4 D.2nπ±π4  {\text{A}}{\text{. }}\,\,n\pi + \dfrac{\pi }{4} \\\ {\text{B}}{\text{.}}\,\,\,n\pi - \dfrac{\pi }{4} \\\ {\text{C}}{\text{.}}\,\,\,n\pi \pm \dfrac{\pi }{4} \\\ {\text{D}}{\text{.}}\,\,\,\,{\text{2}}n\pi \pm \dfrac{\pi }{4} \\\

Explanation

Solution

Hint: Here, we have to use the formula A2B2=(A + B)(A - B){{\text{A}}^2} - {{\text{B}}^2} = ({\text{A + B}})({\text{A - B}}) to find the general solution of the given equation.

Complete step-by-step answer:

We have tan2x=1{\tan ^2}x = 1
We do tan2x1=0{\tan ^2}x - 1 = 0
As we know A2B2=(A + B)(A - B){{\text{A}}^2} - {{\text{B}}^2} = ({\text{A + B}})({\text{A - B}})
On applying the same to the given equation we get,
tan2x12=(tanx1)(tanx+1)=0{\tan ^2}x - {1^2} = (\tan x - 1)(\tan x + 1) = 0
Now, either tanx=1\tan x = 1 or tanx=1\tan x = - 1
Then we can say tanx=±1\tan x = \pm 1
With the help of trigonometric values we get to know that ,
x=π4x = \dfrac{\pi }{4}
We know on adding or subtracting π\pi with π4\,\dfrac{\pi }{4}.
We got the same value, that is ± \pm {\text{1 }}.
Therefore the general solution of the equation is
x=nπ±π4x = n\pi \pm \dfrac{\pi }{4}
So, the correct option is C{\text{C}}.

Note: Whenever we are struck with these types of problems of finding general solutions always try to find the basic angle from the value obtained by algebraic operations. And then use the quadrant rule in trigonometry to get the general solution.