The value of √3,cosec,20^-20^20^40^60^80^ is equal to:

Question

The value of $\dfrac{\sqrt3\,	ext{cosec}\,20^\circ-\sec20^\circ}{\cos20^\circ\cos40^\circ\cos60^\circ\cos80^\circ}$ is equal to:

Accepted Answer

From Question 5's method, $\sqrt3\csc20^\circ - \sec20^\circ = 4$ (same computation with $20^\circ$ in place of $10^\circ$, since $\sqrt3\cos20^\circ-\sin20^\circ = 2\sin40^\circ$ and $\sin20^\circ\cos20^\circ = \tfrac12\sin40^\circ$, giving $\tfrac{2\sin40^\circ}{\tfrac12\sin40^\circ}=4$).

For the denominator, use the identity $\cos A\cos(60^\circ-A)\cos(60^\circ+A) = \tfrac14\cos3A$ with $A=20^\circ$:

$$\cos20^\circ\cos40^\circ\cos80^\circ = \tfrac14\cos60^\circ = \tfrac18$$

So $\cos20^\circ\cos40^\circ\cos60^\circ\cos80^\circ = \tfrac18 \cdot \tfrac12 = \tfrac1{16}$.

$$\text{Value} = \frac{4}{1/16} = 64$$

Solutions · 1