Why does tan have vertical asymptotes?
tan θ = sin θcos θ. When cos θ = 0, the denominator is zero → undefined.
This happens at θ = π2, 3π2, −π2, … i.e. at θ = π2 + kπ for any integer k.
At these angles, the terminal ray is vertical, and its slope is undefined (infinite). Each half-turn around the circle creates another vertical asymptote.
π (half of sincos)θ = π2 + kπθ = kπ2πFinding vertical asymptotes of a·tan(b(θ+c)) + d
Asymptotes occur when the argument of tan = π2 + kπ. The vertical shift d and vertical stretch a do not affect asymptote locations — only b and c do.
For tan(bx): set bx = π2 + kπ → solve for x → x = π/(2b) + kπ/b.
For tan(x3): set x3 = π2 + kπ → x = 3π2 + 3kπ.
Worked example: Find the vertical asymptotes of f(θ) = 3·tan(2(θ + π6)) − 5.
Here a = 3, b = 2, c = π6, d = −5. Set the argument equal to π2 + kπ:
2(θ + π6) = π2 + kπ → divide by 2 → θ + π6 = π4 + kπ2 → subtract π6 → θ = π4 − π6 + kπ2.
Common denominator 12: π4 − π6 = 3π12 − 2π12 = π12. So the asymptotes are at θ = π12 + kπ2 for integer k — i.e., θ = π12, 7π12, 13π12, −5π12, … Notice that a = 3 and d = −5 never entered the calculation.
Method: always divide sin by cos
Look up (or recall) the unit circle point (cos θ, sin θ). Then divide: tan θ = sin θ ÷ cos θ. If cos θ = 0, the result is undefined. If sin θ = 0, tan θ = 0. Use ASTC signs to check.