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Question: If \(\cot x-\tan x=2\), the generalized solution is (here, n is integer): (a) \(x=\dfrac{n\pi }{2}...

If cotxtanx=2\cot x-\tan x=2, the generalized solution is (here, n is integer):
(a) x=nπ2+π8x=\dfrac{n\pi }{2}+\dfrac{\pi }{8}
(b) x=nπ4+π16x=\dfrac{n\pi }{4}+\dfrac{\pi }{16}
(c) x=2nπ+π2x=2n\pi +\dfrac{\pi }{2}
(d) x=nπ+π4x=n\pi +\dfrac{\pi }{4}

Explanation

Solution

The equation given in the above problem is as follows: cotxtanx=2\cot x-\tan x=2, writing cotx=1tanx\cot x=\dfrac{1}{\tan x} in this trigonometric equation and we will get the quadratic in tanx\tan x and then rearrange the equation and get the value of tanx\tan x. After that find the angle at which you are getting that value of tanx\tan x. Let us suppose the angle that you are getting is θ\theta which will look like tanx=tanθ\tan x=\tan \theta then the general solution of this equation is equal to: x=nπ+θx=n\pi +\theta .

Complete step by step solution:
The trigonometric equation given in the above problem is as follows:
cotxtanx=2\cot x-\tan x=2
Now, writing cotx=1tanx\cot x=\dfrac{1}{\tan x} in the above equation and we get,
1tanxtanx=2\dfrac{1}{\tan x}-\tan x=2
Taking tanx\tan x as L.C.M in the above equation we get,
1tan2xtanx=2\dfrac{1-{{\tan }^{2}}x}{\tan x}=2
On cross multiplying the above equation we get,
1tan2x=2tanx1-{{\tan }^{2}}x=2\tan x
Now, dividing (1tan2x)\left( 1-{{\tan }^{2}}x \right) on both the sides of the above equation we get,
1tan2x1tan2x=2tanx1tan2x\dfrac{1-{{\tan }^{2}}x}{1-{{\tan }^{2}}x}=\dfrac{2\tan x}{1-{{\tan }^{2}}x}
In the L.H.S of the above equation, (1tan2x)\left( 1-{{\tan }^{2}}x \right) will be cancelled out from the numerator and the denominator and we get,
1=2tanx1tan2x1=\dfrac{2\tan x}{1-{{\tan }^{2}}x} …………. (1)
We know that there is a double angle identity of tangent which is equal to:
tan2x=2tanx1tan2x\tan 2x=\dfrac{2\tan x}{1-{{\tan }^{2}}x}
Using the above relation in eq. (1) we get,
1=tan2x1=\tan 2x
Now, to find the general solution, in the above equation, both sides contain tangent term so in the L.H.S of the above equation, we can write 1 as tanπ4\tan \dfrac{\pi }{4} then the above equation will look like:
tanπ4=tan2x\tan \dfrac{\pi }{4}=\tan 2x
Rearranging the above equation we get,
tan2x=tanπ4\tan 2x=\tan \dfrac{\pi }{4}
We know that the general solution for tanx=tanθ\tan x=\tan \theta is equal to:
x=nπ+θx=n\pi +\theta
Now, using the above general solution we can write the general solution of the above equation.
2x=nπ+π42x=n\pi +\dfrac{\pi }{4}
Dividing 2 on both the sides of the above equation we get,
x=nπ2+π8x=\dfrac{n\pi }{2}+\dfrac{\pi }{8}
From the above, we got the general solution for the above equation as: x=nπ2+π8x=\dfrac{n\pi }{2}+\dfrac{\pi }{8}.

So, the correct answer is “Option A”.

Note: To solve the above problem, you should know how to write the general solution for tanx\tan x otherwise you could not solve this problem. Also, you should know the double angle identity of tangent. If you miss any of the two concepts then it will be very hard for you to solve this problem so make sure you have properly understood these concepts.