The number of distinct solutions of sec theta | vedclass

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Question

$\sec 	heta+	an 	heta=\sqrt{3}$

$\Rightarrow \quad 1+\sin 	heta=\sqrt{3} \cos 	heta(\cos 	heta 
eq 0)$

$\Rightarrow \quad \sqrt{3} \cos \

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Vedclass · Answer

$\sec 	heta+	an 	heta=\sqrt{3}$

$\Rightarrow \quad 1+\sin 	heta=\sqrt{3} \cos 	heta(\cos 	heta 
eq 0)$

$\Rightarrow \quad \sqrt{3} \cos 	heta-\sin 	heta=1$

$\Rightarrow \quad \frac{\sqrt{3}}{2} \cos 	heta-\frac{1}{2} \sin 	heta=\frac{1}{2}$

$\Rightarrow \quad \cos 	heta \cos \frac{\pi}{6}-\sin 	heta \sin \frac{\pi}{6}=\cos \frac{\pi}{3}$

$\Rightarrow \quad \cos \left(	heta+\frac{\pi}{6}ight)=\cos \frac{\pi}{3}$

$\Rightarrow \quad 	heta+\frac{\pi}{6}=2 n \pi \pm \frac{\pi}{3}$             $\left( {n \in Z} ight)$

$\Rightarrow \quad 	heta=2 n \pi-\frac{\pi}{6} \pm \frac{\pi}{3}$              $\left( {n \in Z} ight)$

$=2 n \pi+\frac{\pi}{6} 	ext { or } 2 n \pi-\frac{\pi}{2}$

But we reject the values $	heta=2 \mathrm{n} \pi-\frac{\pi}{2}$

$\because \quad \cos 	heta=0$ for this values of $	heta$

$	herefore $ $	heta=2 \mathrm{n} \pi+\frac{\pi}{6}, \mathrm{n} \in \mathrm{I}$

Thus, $	heta=\frac{\pi}{6}$ is the only value of $	heta \in[0,2, \pi]$

The number of distinct solutions of $\sec \theta \,\, + \,\,\tan \theta \, = \,\sqrt 3 \,,\,0\,\, \leqslant \,\,\theta \,\, \leqslant \,\,2\pi$

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