Answer

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

Complete step-by-step answer:

We have ${\tan ^2}x = 1$

We do ${\tan ^2}x - 1 = 0$

As we know ${{\text{A}}^2} - {{\text{B}}^2} = ({\text{A + B}})({\text{A - B}})$

On applying the same to the given equation we get,

${\tan ^2}x - {1^2} = (\tan x - 1)(\tan x + 1) = 0$

Now, either $\tan x = 1$ or $\tan x = - 1$

Then we can say $\tan x = \pm 1$

With the help of trigonometric values we get to know that ,

$x = \dfrac{\pi }{4}$

We know on adding or subtracting $\pi $ with $\,\dfrac{\pi }{4}$.

We got the same value, that is $ \pm {\text{1 }}$.

Therefore the general solution of the equation is

$x = n\pi \pm \dfrac{\pi }{4}$

So, the correct option is ${\text{C}}$.

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.

Complete step-by-step answer:

We have ${\tan ^2}x = 1$

We do ${\tan ^2}x - 1 = 0$

As we know ${{\text{A}}^2} - {{\text{B}}^2} = ({\text{A + B}})({\text{A - B}})$

On applying the same to the given equation we get,

${\tan ^2}x - {1^2} = (\tan x - 1)(\tan x + 1) = 0$

Now, either $\tan x = 1$ or $\tan x = - 1$

Then we can say $\tan x = \pm 1$

With the help of trigonometric values we get to know that ,

$x = \dfrac{\pi }{4}$

We know on adding or subtracting $\pi $ with $\,\dfrac{\pi }{4}$.

We got the same value, that is $ \pm {\text{1 }}$.

Therefore the general solution of the equation is

$x = n\pi \pm \dfrac{\pi }{4}$

So, the correct option is ${\text{C}}$.

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.

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