Question

The value of $(81)^{\sin^{2}\pi/6} + (81)^{\cos^{2}\pi/6} = 30$ in between $(81)^{1/4} + (81)^{3/4} = 30$and $\tan\theta = \sqrt{3} = \tan\frac{\pi}{3} \Rightarrow \theta = n\pi + \frac{\pi}{3}$and satisfying the equation $- \pi < \theta < 0$is equal to.

• A. $n = - 1$and $\theta = - \pi + \frac{\pi}{3} = \frac{- 2\pi}{3}; ⥂ k\ \frac{- 4\pi}{6}$
• B. $\tan\theta + \frac{1}{\sqrt{3}} = 0$and $\tan\theta = - \frac{1}{\sqrt{3}}$
• C. $\because$and $\theta$
• D. $0{^\circ}$and $360{^\circ}$

Answer 1

Answer: (D)

$0{^\circ}$and $360{^\circ}$

Answer 2

We have, $\left| \begin{matrix} 1 & 0 & - 1 \\ 0 & 1 & - 1 \\ \sin^{2}\theta & \cos^{2}\theta & 1 + 4\sin 4\theta \end{matrix} \right| = 0$ or $\Rightarrow$

$1 + 4\sin 4\theta + \cos^{2}\theta + \sin^{2}\theta = 0$ $R_{1})$ lies in between $4\sin 4\theta = - 2$ and $\sin 4\theta = \frac{- 1}{2}$

$\therefore$ $\frac{11\pi}{6}$ and $0 < 4\theta < 2\pi$.