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

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Discussion Overview

The discussion centers around the expression y = tan inverse (cot x) + cot inverse (tan x), exploring its properties and implications, particularly in the context of calculus.

Discussion Character

  • Technical explanation, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant presents the expression and seeks clarification on its value.
  • Another participant questions the assertion that the answer is -2, prompting further exploration.
  • A later reply provides a detailed derivation, suggesting that y can be expressed as y = 2arctan(cot(x)), leading to the conclusion that dy/dx = -2 under certain conditions.
  • There is a note on the periodicity of the tangent and cotangent functions, indicating that multiple values for y may exist depending on k, an integer.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the value of y, with some supporting the derived conclusion of -2 while others question it. The discussion remains unresolved regarding the implications of the periodicity and the conditions under which the derivative holds.

Contextual Notes

The discussion involves assumptions about the periodicity of trigonometric functions and the conditions under which the derivative is calculated. There are unresolved mathematical steps related to the implications of the periodicity on the values of y.

rahulk1
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if y = tan inverse (cot x) + cot inverse (tan x)
 
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Then what?
 
Why -2

- - - Updated - - -

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

How answer is -2
 
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:

$$\d{y}{x}=-2$$
 

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