Question

If $\cos x=-\frac{1}{2}$ and $\pi <x<\frac{3\pi }{2}$, find the value of $4{\tan }^{2}x-3cose{c}^{2}x$.

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

Answer: (D)

$8$

It is given that x lies in the third quadrant. Therefore, $sinx$ is negative and $tanx$ is positive. Thus,

$\sin x=\pm \sqrt{1-{\cos }^{2}x}$

$\sin x=-\sqrt{1-\frac{1}{4}}=-\frac{\sqrt{3}}{2}$

$cosecx=-\frac{2}{\sqrt{3}}$

And, $\tan x=\frac{\sin x}{\cos x}=\sqrt{3}$

Hence, $4{\tan }^{2}x-3cose{c}^{2}x=4\times 3-3\times \frac{4}{3}=8$