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Question Number 69953 by Shamim last updated on 29/Sep/19

$$\mathrm{if}\tan \theta +\sec \theta =\sqrt{3}\mathrm{than}\mathrm{find}\mathrm{out}\mathrm{the}$$$$\mathrm{value}\mathrm{of}\underset{}{}\theta \mathrm{where}{0}^{\mathrm{o}}⩽\theta ⩽2\pi .$$

Commented by Shamim last updated on 29/Sep/19

$$\mathrm{ans}\mathrm{ki}\mathrm{only}\mathrm{x}=\frac{\pi }{6}???$$

Commented by mathmax by abdo last updated on 29/Sep/19

$$tan\theta +\frac{1}{cos\theta }=\sqrt{3}\Rightarrow cos\theta tan\theta +1=\sqrt{3}cos\theta \Rightarrow$$$$sin\theta +1=\sqrt{3}cos\theta \Rightarrow \sqrt{3}cos\theta -sin\theta =1(with\theta \ne \frac{\pi }{2}+k\pi )$$$$2(\frac{\sqrt{3}}{2}cos\theta -\frac{1}{2}sin\theta )=1\Rightarrow cos\theta cos\left(\frac{\pi }{6}\right)-sin\theta sin\left(\frac{\pi }{6}\right)=\frac{1}{2}\Rightarrow$$$$cos(\theta +\frac{\pi }{6})=\frac{1}{2}\Rightarrow cos(\theta +\frac{\pi }{6})=cos\left(\frac{\pi }{3}\right)\Rightarrow \theta +\frac{\pi }{6}=\frac{\pi }{3}+2k\pi or$$$$\theta +\frac{\pi }{6}=-\frac{\pi }{3}+2k\pi \Rightarrow \theta =\frac{\pi }{6}+2k\pi or\theta =-\frac{\pi }{2}+2k\pi (k\in Z)$$$$solutionson[0,2\pi ]$$$$0⩽\frac{\pi }{6}+2k\pi ⩽2\pi \Rightarrow 0⩽\frac{1}{6}+2k⩽2\Rightarrow -\frac{1}{6}⩽2k⩽\frac{11}{6}\Rightarrow$$$$-\frac{1}{12}⩽k⩽\frac{11}{12}\Rightarrow k=0\Rightarrow x=\frac{\pi }{6}$$$$c_2\to 0⩽-\frac{\pi }{2}+2k\pi ⩽2\pi \Rightarrow 0⩽-\frac{1}{2}+2k⩽2\Rightarrow \frac{1}{2}⩽2k⩽\frac{5}{2}\Rightarrow$$$$\frac{1}{4}⩽k⩽\frac{5}{4}\Rightarrow k=1\Rightarrow x=\frac{3\pi }{2}butthisisnotso<ution!$$$$$$$$$$

Commented by Prithwish sen last updated on 29/Sep/19

$$\tan \theta +\sec \theta =\sqrt{3}....\left(\mathrm{i}\right)$$$$\Rightarrow \sec \theta -\tan \theta =\frac{1}{\sqrt{3}}...\left(\mathrm{ii}\right)$$$$\mathrm{Adding}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)$$$$2\sec \theta =\frac{4}{\sqrt{3}}$$$$\cos \theta =\frac{\sqrt{3}}{2}=\cos \frac{\pi }{6}$$$$\Rightarrow \mathit{\theta }=\mathbf{n}\mathit{\pi }\pm \frac{\mathit{\pi }}{6}$$

Commented by mathmax by abdo last updated on 29/Sep/19

$$yes.$$

Commented by Shamim last updated on 29/Sep/19

$$\mathrm{okk}.$$$$\mathrm{all}\mathrm{kind}\mathrm{of}\mathrm{like}\mathrm{this}\mathrm{math}\mathrm{maintain}$$$$\mathrm{same}\mathrm{rule}???$$

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