The number of solutions of the given equation | vedclass

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(B) Given equation: $\sec 	heta + 	an 	heta = \sqrt{3}$ ... $(i)$We know that $\sec^2 	heta - 	an^2 	heta = 1$,which can be written as $(\sec

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$1$

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(B) Given equation: $\sec 	heta + 	an 	heta = \sqrt{3}$ ... $(i)$We know that $\sec^2 	heta - 	an^2 	heta = 1$,which can be written as $(\sec 	heta - 	an 	heta)(\sec 	heta + 	an 	heta) = 1$.Substituting $(i)$ into this,we get $\sec 	heta - 	an 	heta = \frac{1}{\sqrt{3}}$ ... $(ii)$Adding $(i)$ and $(ii)$ gives $2 \sec 	heta = \sqrt{3} + \frac{1}{\sqrt{3}} = \frac{4}{\sqrt{3}}$,so $\sec 	heta = \frac{2}{\sqrt{3}}$,which implies $\cos 	heta = \frac{\sqrt{3}}{2}$.Subtracting $(ii)$ from $(i)$ gives $2 	an 	heta = \sqrt{3} - \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}}$,so $	an 	heta = \frac{1}{\sqrt{3}}$.Since $\cos 	heta = \frac{\sqrt{3}}{2}$ and $	an 	heta = \frac{1}{\sqrt{3}}$,$	heta$ must be in the first quadrant.For $0 Thus,there is only $1$ solution.

The number of solutions of the given equation $\tan \theta + \sec \theta = \sqrt{3}$,where $0 < \theta < 2\pi$ is

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