Given that cos 30$^o$ = sin 60$^o$ = $\frac{3}{2}$ and sin 30$^o$ = cos 60$^o$ = $\frac{1}{2}$, evaluate $\frac{tan 60^o - 1}{1 - tan 30^o}$

The question has a typo: cos 30° = sin 60° = √3/2, not 3/2. I'll use the correct value.

Step 1: Find the tangent values

tan 60° = sin 60° / cos 60° = (√3/2)/(1/2) = √3

tan 30° = sin 30° / cos 30° = (1/2)/(√3/2) = 1/√3 = √3/3

Step 2: Substitute

(tan 60° − 1)/(1 − tan 30°) = (√3 − 1)/(1 − 1/√3)

Step 3: Simplify the denominator

1 − 1/√3 = (√3 − 1)/√3

Step 4: Divide

(√3 − 1) ÷ [(√3 − 1)/√3] = (√3 − 1) × √3/(√3 − 1) = √3

Check with the typo as written: if cos 30° = 3/2 were used literally, tan 60° would be (3/2)/(1/2) = 3 and tan 30° = (1/2)/(3/2) = 1/3. The expression would be (3 − 1)/(1 − 1/3) = 2/(2/3) = 3, which is not among the options.

Why not the others?

• A (√3 − 2) and B (2 − √3) are different numbers: they are negative or small, but our result is about 1.73.
• D (−2) is negative, but the numerator and denominator are both positive.

Tip: Whenever you see an expression like (a − 1)/(1 − 1/a), rewrite the denominator as (a − 1)/a. The factor (a − 1) cancels and leaves a. Here a = tan 60° = √3. Also remember that tan 30° = 1/tan 60°.