MHB If y = tan inverse (cot x) + cot inverse (tan x)

[rahulk1]

if y = tan inverse (cot x) + cot inverse (tan x)

 

[Evgeny.Makarov]

Then what?

 

[rahulk1]

Why -2

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y = tan inverse (cot x) + cot inverse (tan x)

How answer is -2

 

[MarkFL]

This is a calculus question...please don't continue to post calculus questions in other forums.

If given:

$$y=\arctan\left(\cot(x)\right)+\arccot\left(\tan(x)\right)$$

Then we should observe that:

$$\arccot\left(\tan(x)\right)=\arctan\left(\cot(x)\right)$$

And so we may write:

$$y=2\arctan\left(\cot(x)\right)$$

or:

$$\frac{y}{2}=\arctan\left(\cot(x)\right)$$

Now, we may take the tangent of both sides to get:

$$\tan\left(\frac{y}{2}\right)=\cot(x)$$

This implies (because of the relationship between co-functions and the periodicity of the tangent/cotangent functions):

$$\frac{y}{2}=\frac{\pi}{2}(2k+1)-x$$ where $$k\in\mathbb{Z}$$

or:

$$y=\pi(2k+1)-2x$$

Thus:

$$\d{y}{x}=-2$$