Question

If $0^{\circ }\le \theta \le {90}^{\circ }$ and $\sqrt{3}tan\theta -sec\theta =1$, then $\theta$ has the value

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

Answer: (C)

${60}^{\circ }$

$\sqrt{3}tan\theta -1=sec\theta$
Squaring and substituting $se{c}^2\theta =1+ta{n}^2\theta$
$2ta{n}^2\theta -2\sqrt{3}tan\theta =0$ $\therefore 2tan\theta (tan\theta -\sqrt{3})=0$
$\theta ={60}^{\circ }$ is one of the values.