Question

The value of $\sqrt{3} \cot 20^\circ - 4 \cos 20^\circ$ is equal to}

• A. $1$
• B. $-1$
• C. $0$
• D. $\frac{1}{2}$

Hint:

Whenever you see $\sqrt{3}$ with trigonometric functions, try replacing it with $2 \sin 60^\circ$ or $2 \cos 30^\circ$.

Answer 1

The Correct Option is A

Step 1: Concept
Convert cotangent to sine and cosine to simplify the expression using trigonometric identities.

Step 2: Meaning
Use the identities $\sqrt{3} = \frac{\sin 60^\circ}{\cos 60^\circ}$ or $2\sin 60^\circ$, and common denominator simplification.

Step 3: Analysis
$\sqrt{3} \frac{\cos 20^\circ}{\sin 20^\circ} - 4 \cos 20^\circ = \frac{\sqrt{3} \cos 20^\circ - 4 \sin 20^\circ \cos 20^\circ}{\sin 20^\circ}$. Use $4 \sin 20^\circ \cos 20^\circ = 2 \sin 40^\circ$. Expression $= \frac{\sqrt{3} \cos 20^\circ - 2 \sin 40^\circ}{\sin 20^\circ} = \frac{2(\frac{\sqrt{3}}{2} \cos 20^\circ - \sin 40^\circ)}{\sin 20^\circ}$. $= \frac{2(\sin 60^\circ \cos 20^\circ - \sin 40^\circ)}{\sin 20^\circ}$. Since $\sin 40^\circ = \sin(60^\circ - 20^\circ) = \sin 60^\circ \cos 20^\circ - \cos 60^\circ \sin 20^\circ$. Numerator becomes $2(\sin 60^\circ \cos 20^\circ - (\sin 60^\circ \cos 20^\circ - \cos 60^\circ \sin 20^\circ)) = 2 \cos 60^\circ \sin 20^\circ$.

Step 4: Conclusion
$\frac{2 \cos 60^\circ \sin 20^\circ}{\sin 20^\circ} = 2 \cos 60^\circ = 2(1/2) = 1$. Final Answer: (A)