Question

If $12{\cot }^2\left(\theta \right)-31\mathrm{cosec}\left(\theta \right)+32=0,$ the the value of $\sin \left(\theta \right)$ is

• A. $\frac{3}{5}\mathrm{or}1$
• B. $\frac{2}{3}\mathrm{or}\left(-\frac{2}{3}\right)$
• C. $\frac{4}{5}\mathrm{or}\frac{3}{4}$
• D. $\pm \frac{1}{2}$

Answer 1

The correct option is C

$\frac{4}{5}\mathrm{or}\frac{3}{4}$

Explanation for correct option:

Step-1: Simplify the given data.

Given, $12{\cot }^2\left(\theta \right)-31\mathrm{cosec}\left(\theta \right)+32=0,$

$\Rightarrow$ $12\left(\frac{{\cos }^2\left(\theta \right)}{{\sin }^2\left(\theta \right)}\right)-31\left(\frac{1}{\sin \left(\theta \right)}\right)+32=0$

$\Rightarrow$ $12{\cos }^2\left(\theta \right)-31\sin \left(\theta \right)+32{\sin }^2\left(\theta \right)=0$

$\Rightarrow$$12{\cos }^2\left(\theta \right)+12{\sin }^2\left(\theta \right)-31\sin \left(\theta \right)+20{\sin }^2\left(\theta \right)=0\left(\because {\cos }^2\left(\theta \right)+{\sin }^2\left(\theta \right)=1\right)$

$\Rightarrow$ $20{\sin }^2\left(\theta \right)-31\sin \left(\theta \right)+12=0$

Step-2: Solve the quadratic equation $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

$\Rightarrow$$\sin \left(\theta \right)=\frac{-\left(-31\right)\pm \sqrt{{\left(31\right)}^2-4\times 20\times 12}}{2\times 20}$

$\Rightarrow$$\sin \left(\theta \right)=\frac{31\pm \sqrt{961-960}}{2\times 20}$

$\Rightarrow$$\sin \left(\theta \right)=\frac{32}{40}\mathrm{or}\frac{30}{40}$

$\Rightarrow$$\sin \left(\theta \right)=\frac{4}{5}\mathrm{or}\frac{3}{4}$

Hence, correct answer is option $\left(C\right)$