Trigonometric Functions

Maths · Class 11

Simulation · Maths · Class 11

tan (A + B) as the slope of a turned line

"tan (A + B) as the slope of a turned line" is a free, interactive Class 11 Maths simulation for JEE, built on the NCERT section "Tangent and cotangent of sums" in Trigonometric Functions. Two angles A and B laid end to end at the origin as lines of slope tan A and tan B; the combined line has slope (tan A + tan B)/(1 − tan A tan B), and it goes vertical when tan A tan B = 1. JEE Main asked 41 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).

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

What it shows

NCERT §3.4

Two angles A and B laid end to end at the origin as lines of slope tan A and tan B; the combined line has slope (tan A + tan B)/(1 − tan A tan B), and it goes vertical when tan A tan B = 1.

Controls

3 to try

  • Slidertan A 0–3
  • Slidertan B 0–3
  • Toggleshow formula working

Try this

Set tan A = 1/2 and tan B = 1/3 and confirm the combined line sits at 45°; then make the combined line vertical (tan A tan B = 1).

Common mistake

Forgetting that tan (A + B) fails when A + B is an odd multiple of π/2.

In JEE Main

42 shifts analysed

41 questions from Trigonometric Functions across the 42 JEE Main shifts Lumi has analysed (2017–2026).

See the past questions

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.

Questions

4 answers

What does the "tan (A + B) as the slope of a turned line" simulation show?
Two angles A and B laid end to end at the origin as lines of slope tan A and tan B; the combined line has slope (tan A + tan B)/(1 − tan A tan B), and it goes vertical when tan A tan B = 1.
What mistake does the "tan (A + B) as the slope of a turned line" simulation fix?
A common mistake it clears up: forgetting that tan (A + B) fails when A + B is an odd multiple of π/2.
Does JEE Main ask questions from Trigonometric Functions?
Yes. JEE Main asked 41 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
Is this simulation free?
Yes. It is free and runs in the browser on a phone or a laptop. It belongs to the lesson "Tangent and cotangent of sums" in Trigonometric Functions (Class 11 Maths).

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