Question

Find the general solution of $\sec \theta +1=(2+\sqrt{3})\tan \theta$.

Answer 1

$\because \sec \theta +1=(2+\sqrt{3})\tan \theta$
On squaring both sides, we get
${(\sec \theta +1)}^2={(2+\sqrt{3})}^2{tan}^2\theta$
$\Rightarrow {(\sec \theta +1)}^2=(4+3+4\sqrt{3})({sec}^2\theta -1)$
$\Rightarrow (\sec \theta +1)\{(\sec \theta +1)-(7+4\sqrt{3})(\sec \theta -1)\}=0$
$\Rightarrow (\sec \theta +1)\{(8+4\sqrt{3})-(6+4\sqrt{3})\sec \theta \}=0$
$\Rightarrow (\sec \theta +1)=0$
or $(8+4\sqrt{3})-(6+4\sqrt{3})\sec \theta =0$
$\Rightarrow \cos \theta =-1=\cos \pi$
$\Rightarrow \theta =2n\pi \pm \pi$
Now, If $(8+4\sqrt{3})-(6+4\sqrt{3})\sec \theta =0$
$\Rightarrow \cos \theta =\frac{\sqrt{3}}{2}=\cos \frac{\pi }{6}$
$\Rightarrow \theta =2n\pi \pm \frac{\pi }{6}$
But $\theta =2n\pi -\frac{\pi }{6}$ does not satisfying the given equation.
$\therefore \theta =2n\pi \pm \pi ,2n\pi +\frac{\pi }{6},n\in 1$