Question

Assertion \((A):\) If \(y=\tan ^{-1}(\cot x)+\cot ^{-1}(\tan x), \pi / 2

Short answer

Question: Show that the derivative of the function \(y = \tan^{-1}(\cot x) + \cot^{-1}(\tan x)\) equals -2 for the interval \(\pi/2 < x < \pi\). Answer: In the interval \(\pi/2 < x < \pi\), the derivative of the function \(y = \tan^{-1}(\cot x) + \cot^{-1}(\tan x)\) is indeed equal to -2.

Step-by-step solution

Find the derivative using the Chain Rule and Inverse Trigonometric Functions derivatives

The function \(y\) is a sum of two inverse trigonometric functions. To find the derivative \(\frac{dy}{dx}\), we will use the Chain Rule for derivatives and the known formulas for derivatives of inverse trigonometric and simple alternatives. Recall the following formulas: 1. \(\frac{d}{dx}(\tan^{-1}u) = \frac{1}{1+u^2}\cdot\frac{du}{dx}\) 2. \(\frac{d}{dx}(\cot^{-1}u) = -\frac{1}{1+u^2}\cdot\frac{du}{dx}\) 3. \(\frac{d}{dx}(\tan x) = \sec^2 x\) 4. \(\frac{d}{dx}(\cot x) = -\csc^2 x\)

Find the derivative of \(y\) using the formulas from Step 1

Using the formulas from Step 1, we have: \(\frac{dy}{dx} = \left[\frac{1}{1+(\cot x)^2}\cdot(-\csc^2 x)\right] + \left[-\frac{1}{1+(\tan x)^2}\cdot(\sec^2 x)\right]\) Since \(\cot x = \frac{1}{\tan x}\) and \(\csc^2 x = \frac{1}{\sin^2 x}\), we can simplify the expression: \(\frac{dy}{dx} = \left[\frac{-1}{\sin^2 x(1+(\frac{1}{\tan x})^2)}\right]+\left[-\frac{(\sec^2 x)}{1+(\tan x)^2}\right]\)

Verify the reason \((R)\)

Now, recall that we are given the reason \((R)\): If \(\pi/2 < x < \pi\), then \(y = 2\pi - 2x\) Taking the derivative with respect to x of the equation above, we get: \(\frac{dy}{dx}= -2\)

Check if the condition \(\pi / 2

Now we want to check whether the derivative we found in Step 2 equals -2 when the condition \(\pi/2 < x < \pi\) is satisfied. Let's recall what we have: \(\frac{dy}{dx} = \left[\frac{-1}{\sin^2 x(1+(\frac{1}{\tan x})^2)}\right] + \left[-\frac{(\sec^2 x)}{1+(\tan x)^2}\right]\) Since \(\pi/2 < x < \pi\), we are in the second quadrant, where both \(\sin^2 x\) and \(\sec^2 x\) are positive. This means that the expression above is transformed into: \(\frac{dy}{dx} = -\frac{1}{(1+\cot^2 x)} - \frac{1}{(1+\tan^2 x)}\) As we're in the second quadrant, there is no problem, as both \(\tan^2 x\) and \(\cot^2 x\) are always positive in this interval. So the expression above becomes: \(\frac{dy}{dx} = \frac{-(\tan^2 x+\cot^2 x+2\tan x \cot x)}{1+\tan^2 x+\cot^2 x}=-2\) So, in the interval \(\pi/2 < x < \pi\), Assertion \((A)\) holds, and the derivative equals -2 as required.