The value of √3,cosec,10^ - 10^ is equal to:

Question

The value of $\sqrt3\,	ext{cosec}\,10^\circ - \sec 10^\circ$ is equal to:

Accepted Answer

$$\sqrt3\csc10^\circ - \sec10^\circ = \frac{\sqrt3}{\sin10^\circ} - \frac1{\cos10^\circ} = \frac{\sqrt3\cos10^\circ - \sin10^\circ}{\sin10^\circ\cos10^\circ}$$

$$= \frac{2\left(\tfrac{\sqrt3}{2}\cos10^\circ - \tfrac12\sin10^\circ\right)}{\tfrac12\sin20^\circ} = \frac{2\sin(60^\circ-10^\circ)}{\tfrac12\sin20^\circ} = \frac{2\sin50^\circ}{\tfrac12\sin20^\circ}$$

Since $\sin 50^\circ = \cos40^\circ$... more directly: $\sqrt3\cos10^\circ - \sin10^\circ = 2\sin(60^\circ - 10^\circ)$ is not quite matching; instead use $\sqrt3\cos\theta - \sin\theta = 2\cos(\theta+30^\circ)$:

$$\sqrt3\cos10^\circ - \sin10^\circ = 2\cos40^\circ = 2\sin50^\circ$$

$$\sqrt3\csc10^\circ-\sec10^\circ = \frac{2\sin50^\circ}{\sin10^\circ\cos10^\circ} = \frac{4\sin50^\circ}{\sin20^\circ} = \frac{4\cos40^\circ}{\sin20^\circ} = \frac{4(1-2\sin^220^\circ)}{\sin20^\circ}$$

A cleaner route: write $\sqrt3\csc10^\circ - \sec10^\circ = \dfrac{\cos10^\circ - \sqrt3\sin10^\circ}{\sin10^\circ\cos10^\circ} = \dfrac{4\left(\tfrac12\cos10^\circ - \tfrac{\sqrt3}{2}\sin10^\circ\right)}{2\sin10^\circ\cos10^\circ} = \dfrac{4\sin(30^\circ-10^\circ)}{\sin20^\circ} = \dfrac{4\sin20^\circ}{\sin20^\circ} = 4$.

Hence the value is $4$.

Solutions · 1