If \cos x = \frac{-\sqrt3}{2} and \pi < x < \frac{3\pi}{2}, then the value of 2 \cot^2 x - 3 \sec^2 x is:

Question 62

If $$\cos x = \frac{-\sqrt3}{2} and \pi < x < \frac{3\pi}{2}$$, then the value of $$2 \cot^2 x - 3 \sec^2 x$$ is:

Solution

Given, $$\cos x = \frac{-\sqrt3}{2}$$

$$=$$> $$\sec x=\frac{-2}{\sqrt{3}}$$

$$\therefore\ $$ $$2\cot^2x-3\sec^2x=\frac{2}{\tan^2x}-3\sec^2x$$

$$=\frac{2}{\sec^2x-1}-3\sec^2x$$

$$=\frac{2}{\left(\frac{-2}{\sqrt{3}}\right)^2-1}-3\left(\frac{-2}{\sqrt{3}}\right)^2$$

$$=\frac{2}{\frac{4}{3}-1}-3\left(\frac{4}{3}\right)$$

$$=\frac{2}{\frac{4-3}{3}}-4$$

$$=6-4$$

$$=2$$

Hence, the correct answer is Option B

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