Question

The general solution of the equation $ta{n}^2\alpha +2\sqrt{3}tan\alpha$ = 1 is given by

• A. $\theta =\frac{\pi }{2}$
• B. $(n+\frac{1}{2})\pi$
• C. $(6n+1)\frac{\pi }{12}$
• D. $\frac{n\pi }{12}$

Answer 1

Answer: (C)

$(6n+1)\frac{\pi }{12}$

$ta{n}^2\alpha +2\sqrt{3}tan\alpha =1$

$(tan\alpha +\sqrt{3}{)}^2$ =4

$\therefore$ $tan\alpha =2-\sqrt{3},-2-\sqrt{3}$

or $tan\alpha =tan{15}^{\circ }$, $-tan{75}^{\circ }$ = $tan\frac{\pi }{12}$, $tan(-\frac{5\pi }{12})$

$\therefore$ $\alpha =n\pi +\frac{\pi }{12}=2n\frac{\pi }{2}+\frac{\pi }{12}$= Even $\frac{\pi }{2}+\frac{\pi }{12}$.................(1)

$\therefore$ $\alpha =n\pi +(-\frac{5\pi }{12})=n\pi -\frac{6\pi }{12}+\frac{\pi }{12}$

$=(2n-1)\frac{\pi }{2}+\frac{\pi }{12}=odd\frac{\pi }{2}+\frac{\pi }{12}$....................(2)

From(1) and (2),

$\alpha =k\frac{\pi }{2}+\frac{\pi }{12}=(6k+1)\frac{\pi }{12}$.