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:

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

Then we should observe that:

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

And so we may write:

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

or:

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

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

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

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

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

or:

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

Thus:

\(\displaystyle \d{y}{x}=-2\)