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Question Number 115345 by bemath last updated on 25/Sep/20

$$\sec \theta (\sec \theta \left(\sin {}^2\theta \right)+2\sqrt{3}\sin \theta )=1$$$$hastherootsare{\theta }_{1}and{\theta }_2.Findthe$$$$valueof\tan {\theta }_1\times \tan {\theta }_2.$$

Answered by bobhans last updated on 25/Sep/20

$$\iff \frac{\sin {}^2\theta }{\cos \theta }+2\sqrt{3}\sin \theta =\cos \theta$$$$\iff 2\sqrt{3}\sin \theta .\cos \theta =\cos 2\theta$$$$\Rightarrow \sqrt{3}\sin 2\theta =\cos 2\theta \to \tan 2\theta =\frac{1}{\sqrt{3}}$$$$\to \frac{2\tan \theta }{1-\tan {}^2\theta }=\frac{1}{\sqrt{3}}\Rightarrow \tan {}^2\theta +2\sqrt{3}\tan \theta -1=0$$$$justapply{Vieta}^{\prime }sruleweget$$$$\tan {\theta }_1\times \tan {\theta }_2=-1$$

Answered by Dwaipayan Shikari last updated on 25/Sep/20

$$\frac{{\sin }^2\theta }{{\cos }^2\theta }+\frac{2\sqrt{3}\sin \theta }{\cos \theta }=1$$$${\tan }^2\theta +2\sqrt{3}\tan \theta -1=0$$$$\tan \theta =\frac{-2\sqrt{3}\pm \sqrt{12+4}}{2}=2-\sqrt{3}\mathrm{or}-(2+\sqrt{3})$$$$\tan {\theta }_1=2-\sqrt{3}$$$$\tan {\theta }_2=-(2+\sqrt{3})$$$$\tan {\theta }_1.\tan {\theta }_2=-1$$$$$$$$$$$$$$

Answered by MJS_new last updated on 25/Sep/20

$$\frac{1}{c}(\frac{s^2}{c}+2\sqrt{3}s)=1$$$$t^2+2\sqrt{3}t-1=0$$$$\Rightarrow t_1\times t_2=-1$$

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