Differentiation (Part-38) | S N De

 

$\,\,1(xii)\,\,\,\cot \theta-\tan\theta=2 \\ \Rightarrow \frac{\cos\theta}{\sin\theta}-\frac{\sin\theta}{\cos\theta}=2 \\ \Rightarrow \frac{\cos^2\theta-\sin^2\theta}{\sin\theta\cos\theta}=2 \\ \Rightarrow \cos2\theta=2\sin\theta\cos\theta \\ \Rightarrow \cos2\theta=\sin2\theta \\ \Rightarrow \tan2\theta=1 \\ \Rightarrow \tan2\theta=\tan \frac{\pi}{4} \\ \Rightarrow 2\theta=n\pi+\pi/4 ,\quad n\,\,\text{being any integer.} \\ \Rightarrow \theta=\frac{n\pi}{2}+\pi/8\,\,\,\text{(ans.)}$

$\,\,1(xiii)\,\,\,\tan^2x+\cot^2x=2 \\ \Rightarrow \tan^2x-2\tan x\cot x+\cot^2x=0\,\,\,[*] \\ \Rightarrow (\tan x-\cot x)^2=0\\ \Rightarrow \tan x=\cot x \\ \Rightarrow \tan x=\frac{1}{\tan x}\\ \Rightarrow \tan^2x=1 \\ \Rightarrow \tan x= \pm 1 =\tan\left(\pm\frac{\pi}{4}\right) \\ \therefore x= n\pi \pm \pi/4,\,\,\,n \in\mathbb Z. $

Note[*] : $\,\,\tan x\cot x=\tan x \times \frac{1}{\tan x}=1$

$\,\,1(xiv)\,\,\,4\sin4\theta+1=\sqrt5 \\ \Rightarrow \sin4\theta=\frac{\sqrt5-1}{4}=\sin\frac{\pi}{10} \\ \Rightarrow 4\theta=n\pi+(-1)^n \pi/10 \\ \Rightarrow \theta=\frac{n\pi}{4}+(-1)^n\frac{\pi}{40},\,\,\,n\in\mathbb Z.$

$\,\,1(xv)\,\,\,4\sin4\theta-1=\sqrt5 \\ \Rightarrow \sin4\theta=\frac{\sqrt5+1}{4}=\sin\frac{3\pi}{10}\\ \Rightarrow 4\theta=n\pi+(-1)^n\frac{3\pi}{10} \\ \Rightarrow \theta=\frac{n\pi}{4}+(-1)^n\frac{3\pi}{40},\,\,\,n\in\mathbb Z.$

$\,\,1(xvi)\,\,\,4\cos^2x+\sqrt3=2(\sqrt3+1) \\ \Rightarrow 2\cos^2x=\frac{2(\sqrt3+1)-\sqrt3}{2} \\ \Rightarrow 1+\cos2x=\frac{\sqrt3+2}{2} \\ \Rightarrow \cos2x=\frac{\sqrt3+2}{2}-1 =\frac{\sqrt3}{2} \\ \Rightarrow \cos2x=\cos\left(\pm \frac{\pi}{6}\right) \\ \Rightarrow 2x=2n\pi \pm \pi/6 \\ \Rightarrow x= n\pi\pm\pi/12,\,\,\,\,n \in \mathbb Z.$

ADDITIONAL PROBLEMS : (S N De  Bengali Version)

$\,\,1(xvii)\,\, \sec^3\theta-2\tan^2\theta=2 \\ \Rightarrow \sec^3\theta=2(1+\tan^2\theta)\\ \Rightarrow \sec^3\theta=2\sec^2\theta \\ \Rightarrow \sec^2\theta(\sec\theta-2)=0  \\ \Rightarrow \sec\theta=2\,\,\,[\sec\theta \neq 0]\\ \therefore \cos\theta=\frac 12=\cos(\pm\pi/3) \\ \Rightarrow \theta=2n\pi \pm \pi/3\, ,\,\,\,n\in\mathbb Z.$

$\,\,\,1(xviii)\,\,\,\sec\theta+\csc\theta=2\sqrt2 \\ \Rightarrow \frac{1}{\cos\theta}+\frac{1}{\sin\theta}=2\sqrt2 \\ \Rightarrow \frac{\sin\theta+\cos\theta}{\sin\theta\cos\theta}=2\sqrt2 \\ \Rightarrow \frac{1}{\sqrt2}\left(\sin\theta+\cos\theta\right)=2\sin\theta\cos\theta \\ \Rightarrow \frac{1}{\sqrt2}\cos\theta+\frac{1}{\sqrt2}\sin\theta=\sin2\theta \\ \Rightarrow \cos\theta\cos\frac{\pi}{4}+\sin\theta\sin\frac{\pi}{4}=\sin2\theta \\ \Rightarrow \cos\left(\theta-\pi/4\right)=\cos(\pi/2-2\theta) \\ \therefore \theta-\pi/4=2n\pi \pm (\pi/2-2\theta) \\ \Rightarrow \theta=\frac{2n\pi}{3}+\frac{\pi}{4}, \\ \text{or,}\,\,\,\theta=-2n\pi+\frac{\pi}{4},\,\,n \in\mathbb Z.$

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