Simulation · Maths · Class 11
tan (A + B) as the slope of a turned line
From the lesson Tangent and cotangent of sums in Trigonometric Functions. Change the values and watch what happens.
tan (A + B) as the slope of a turned lineMaths · Class 11
The idea behind it
NCERT §3.4
- tan (A + B) = (tan A + tan B)/(1 − tan A tan B), valid when none of A, B and A + B is an odd multiple of π/2, so that cos A, cos B and cos (A + B) are all non-zero.
- It comes from dividing sin (A + B) = sin A cos B + cos A sin B by cos (A + B) = cos A cos B − sin A sin B, then dividing the top and bottom by cos A cos B.
- tan (A − B) = (tan A − tan B)/(1 + tan A tan B), obtained by putting −B for B.
- cot (A + B) = (cot A cot B − 1)/(cot B + cot A), valid when none of A, B and A + B is a multiple of π; and cot (A − B) = (cot A cot B + 1)/(cot B − cot A), valid when none of A, B and A − B is a multiple of π.
- Worked value: tan 105° = tan (60° + 45°) = (√3 + 1)/(1 − √3 × 1) = (√3 + 1)/(1 − √3); multiplying top and bottom by (1 + √3) gives (4 + 2√3)/(−2) = −(2 + √3).
- With A = π/4, where tan A = 1: tan (π/4 + B) = (1 + tan B)/(1 − tan B) and tan (π/4 − B) = (1 − tan B)/(1 + tan B).
- The formula can also identify an angle: if tan A = 1/2 and tan B = 1/3 with A and B acute, then tan (A + B) = (5/6)/(1 − 1/6) = 1, and since 0 < A + B < π, A + B = π/4.