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Question Number 14077 by Tinkutara last updated on 27/May/17

Answered by ajfour last updated on 28/May/17

$$let\frac{\pi }{4}\cot \theta =\alpha and\frac{\pi }{4}\tan \theta =\beta$$$$then\sin \alpha =\cos \beta =f\left(say\right)$$$$\beta =2p\pi \pm {\cos }^{-1}f;p\in Z....\left(a\right)$$$$\alpha =m\pi +{(-1)}^m{\sin }^{-1}f;m\in Z..\left(b\right)$$$$\alpha +\beta =\frac{\pi }{4}(\cot \theta +\tan \theta )$$$$\Rightarrow \alpha +\beta =\frac{\pi }{2\sin 2\theta }......\left(i\right)$$$$\alpha -\beta =\frac{\pi }{4}(\cot \theta -\tan \theta )$$$$\Rightarrow \alpha -\beta =\frac{\pi }{2\tan 2\theta }.......\left(ii\right)$$$$comparing\left(b\right)+\left(a\right)with\left(i\right):$$$$\frac{\pi }{2\sin 2\theta }=(2p+m)\pi \pm {\cos }^{-1}f$$$$+{(-1)}^m{\sin }^{-1}f...\left(I\right)$$$$comparing\left(b\right)-\left(a\right)with\left(ii\right):$$$$\frac{\pi }{2\tan 2\theta }=(m-2p)\pi +{(-1)}^m{\sin }^{-1}f$$$$\mp {\cos }^{-1}f...\left(II\right)$$$$\mathit{f}istobeeliminated,fourcases$$$$arisefrom\left(I\right)\mathrm{\&}\left(II\right):$$$$$$$$Ineqn\left(I\right)letmis{}^{\prime }{even}^{\prime }and$$$$letusfirstuse+{\cos }^{-1}f..then$$$$\frac{\pi }{2\sin 2\theta }=2n\pi +\frac{\pi }{2}$$$$\Rightarrow \frac{1}{\sin 2\theta }=4n+1$$$$\Rightarrow \sin 2\theta =\frac{1}{4n+1}$$$$\Rightarrow 2\theta =k\pi +{(-1)}^k{\sin }^{-1}\left(\frac{1}{4n+1}\right)$$$$or\mathit{\theta }=\frac{\mathit{k}\mathit{\pi }}{2}+\frac{{(-1)}^{\mathit{n}}}{2}{\sin }^{-1}\left(\frac{1}{4\mathit{n}+1}\right)$$$$Inean\left(II\right)letmis{}^{\prime }{even}^{\prime }and$$$$letususefirst+{\cos }^{-1}f...then$$$$\frac{\pi }{2\tan 2\theta }=-2n\pi +\frac{\pi }{2}$$$$\frac{1}{\tan 2\theta }=-4n+1$$$$\Rightarrow \tan 2\theta =\frac{-1}{4n-1}$$$$2\theta =k\pi +{\tan }^{-1}\left(\frac{-1}{4n-1}\right)$$$$or\mathit{\theta }=\frac{\mathit{k}\mathit{\pi }}{2}+\frac{1}{2}{\tan }^{-1}\left(\frac{-1}{4\mathit{n}-1}\right)$$$$insuchawaytwomore$$$$solutionsarepossible;(arenot$$$$inyourbook,asyouindicated).$$

Answered by Tinkutara last updated on 27/Jul/17

$$\cos \left(\frac{\pi }{4}\tan \theta \right)=\cos (\frac{\pi }{2}-\frac{\pi }{4}\cot \theta )$$$$\frac{\pi }{4}\tan \theta =2n\pi \pm (\frac{\pi }{2}-\frac{\pi }{4}\cot \theta )$$$$\mathrm{Taking}\mathrm{positive}\mathrm{sign},$$$$\frac{1}{4}(\tan \theta +\cot \theta )=2n+\frac{1}{2}$$$$\frac{1+{\tan }^2\theta }{2\tan \theta }=4n+1=\mathrm{cosec}2\theta$$$$2\theta =n\pi +{(-1)}^n{\mathrm{cosec}}^{-1}(4n+1)$$$$\theta =\frac{n\pi }{2}+\frac{{(-1)}^n}{2}{\sin }^{-1}\left(\frac{1}{4n+1}\right)$$$$\mathrm{Now}\mathrm{taking}\mathrm{negative}\mathrm{sign},$$$$\frac{1}{4}(\tan \theta -\cot \theta )=2n-\frac{1}{2}$$$$\frac{{\tan }^2\theta -1}{2\tan \theta }=4n-1=-\cot 2\theta$$$$2\theta =n\pi +{\cot }^{-1}[-(4n-1)]$$$$\theta =\frac{n\pi }{2}+\frac{1}{2}{\tan }^{-1}\left(\frac{-1}{4n-1}\right)$$

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