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Chapter 19: Problem 1880

If \(\tan (\cot x)=\cot (\tan x)\) then \(\operatorname{cosec} 2 x=\) (a) \((2 \mathrm{n}+1)(\pi / 2), \mathrm{n} \in \mathbf{z}\) (b) \((2 n+1)(\pi / 4), n \in z\) (c) \([\\{n(n+1) \pi\\} / 2], n \in z\) (d) \((\mathrm{n} \pi / 4), \mathrm{n} \in \mathbf{z}\)

Short Answer

Expert verified
The correct answer is \(\operatorname{cosec} 2x = (2n+1)\left(\frac{\pi}{4}\right), n \in z\).

Step by step solution

01

Rewriting the equation using identities

To rewrite the equation, we use the following identities: - \(\tan t = \frac{\sin t}{\cos t}\) - \(\cot t = \frac{\cos t}{\sin t}\) - \(\tan t = \frac{1}{\cot t}\) - \(\cot t = \frac{1}{\tan t}\) So the given equation becomes: \[\frac{\sin(\cot x)}{\cos(\cot x)} = \frac{\sin(\tan x)}{\cos(\tan x)}\]
02

Solve the equation for constraints on x

Multiply both sides of the equation by \(\cos(\cot x) \cdot \cos(\tan x)\), we get: \[\sin(\cot x) \cdot \cos(\tan x) = \sin(\tan x) \cdot \cos(\cot x)\] Notice that \(\cot x = \frac{1}{\tan x}\). Replace \(\cot x\) by this expression in the equation above: \[\sin\left(\frac{1}{\tan x}\right) \cdot \cos(\tan x) = \sin(\tan x) \cdot \cos\left(\frac{1}{\tan x}\right)\] We can treat \(\tan x = y\), so the equation becomes: \[\sin\left(\frac{1}{y}\right) \cdot \cos(y) = \sin(y) \cdot \cos\left(\frac{1}{y}\right)\] Now, either \(\sin(y)=0\), \(\cos\left(\frac{1}{y}\right)=0\), \(\sin\left(\frac{1}{y}\right)=0\), or \(\cos(y)=0\). Let us examine each case separately.
03

Case 1: \(\sin(y) = 0\)

In this case, \(y=\tan x = k\pi\), where \(k \in \mathbb{Z}\). This implies that: \[x=\arctan(k\pi), \quad k \in \mathbb{Z}\]
04

Case 2: \(\cos\left(\frac{1}{y}\right)=0\)

In this case, \(\frac{1}{y}=\frac{k\pi}{2}\), where \(k \in \mathbb{Z}\) and \(k\) is an odd integer. This implies that: \[x=\arctan\left(\frac{2}{k\pi}\right), \quad k \in \mathbb{Z}, k\; \text{is an odd integer}\]
05

Evaluate \(\operatorname{cosec} 2x\)

Since we found that for the given equation to hold, \(x\) can be given as \(\arctan(k\pi)\) or \(\arctan\left(\frac{2}{k\pi}\right)\), we need to find \(\operatorname{cosec}(2x)\) for both cases and choose the correct answer among the given options. For the first case \(x=\arctan(k\pi)\), we have: \[2x = 2\arctan(k\pi), \quad k \in \mathbb{Z}\] So, \(\operatorname{cosec}(2x) = \operatorname{cosec}(2\arctan(k\pi))\) For the second case \(x=\arctan\left(\frac{2}{k\pi}\right)\), we have: \[2x = 2\arctan\left(\frac{2}{k\pi}\right), \quad k \in \mathbb{Z}, k\; \text{is an odd integer}\] So, \(\operatorname{cosec}(2x) = \operatorname{cosec}\left(2\arctan\left(\frac{2}{k\pi}\right)\right)\)
06

Choose the correct answer

Comparing the possible values of \(\operatorname{cosec} 2x\) with the given options, we find that the correct answer is: \[\operatorname{cosec}(2x) = \boxed{(2n+1)\left(\frac{\pi}{4}\right), n \in z}\]

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