The number of solutions of the given equation | vedclass

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Question

(c) $\sec 	heta + 	an 	heta = \sqrt 3 $....$(i)$

Also we have ${\sec ^2}	heta - {	an ^2}	heta = 1$...$(ii)$

$ \Rightarrow $ $\sec 	heta -

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(c) $\sec 	heta + 	an 	heta = \sqrt 3 $....$(i)$

Also we have ${\sec ^2}	heta - {	an ^2}	heta = 1$...$(ii)$

$ \Rightarrow $ $\sec 	heta - 	an 	heta = \frac{1}{{\sqrt 3 }}$...$(iii)$

Now $(i)$ and $(iii)$ gives,

$	an 	heta = \frac{1}{2}\left( {\sqrt 3 - \frac{1}{{\sqrt 3 }}} ight) = \frac{1}{{\sqrt 3 }} = 	an \left( {\frac{\pi }{6}} ight)$

$ \Rightarrow $ $	heta = n\pi + \frac{\pi }{6}$.

$	herefore $ Solutions for $0 \le 	heta \le 2\pi $ are $\frac{\pi }{6}$ and $\frac{{7\pi }}{6}$.

Hence there are two solutions.

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

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