ExamHabit
Practise free

1979

JEE Main 1979 Mathematics question on Trigonometry

If α+β+γ=2π\alpha + \beta + \gamma = 2\pi, then
  1. A.
    tan⁡α2+tan⁡β2+tan⁡γ2=tan⁡α2tan⁡β2tan⁡γ2\tan\frac{\alpha}{2} + \tan\frac{\beta}{2} + \tan\frac{\gamma}{2} = \tan\frac{\alpha}{2}\tan\frac{\beta}{2}\tan\frac{\gamma}{2}
  2. B.
    tan⁡α2tan⁡β2+tan⁡β2tan⁡γ2+tan⁡γ2tan⁡α2=1\tan\frac{\alpha}{2}\tan\frac{\beta}{2} + \tan\frac{\beta}{2}\tan\frac{\gamma}{2} + \tan\frac{\gamma}{2}\tan\frac{\alpha}{2} = 1
  3. C.
    tan⁡α2+tan⁡β2+tan⁡γ2=−tan⁡α2tan⁡β2tan⁡γ2\tan\frac{\alpha}{2} + \tan\frac{\beta}{2} + \tan\frac{\gamma}{2} = -\tan\frac{\alpha}{2}\tan\frac{\beta}{2}\tan\frac{\gamma}{2}
  4. D.
    None of the above
Show answer and solution

Correct option: A

Solution

Since α + β + γ = 2π, we have (α/2) + (β/2) + (γ/2) = π.

Let A = α/2, B = β/2, C = γ/2, so A + B + C = π (same as angles of a triangle).

Since A + B = π − C, we get tan(A + B) = tan(π − C) = −tan C.

Expanding: (tan A + tan B)/(1 − tan A·tan B) = −tan C

So tan A + tan B = −tan C(1 − tan A·tan B) = −tan C + tan A·tan B·tan C

Rearranging: tan A + tan B + tan C = tan A·tan B·tan C

This is the well-known identity for angles summing to π, giving tan(α/2) + tan(β/2) + tan(γ/2) = tan(α/2)·tan(β/2)·tan(γ/2), which is option 1.

[1979]

Practise Trigonometry the way it is asked

A timed set of real JEE Main Trigonometry questions, graded as the exam grades. No sign-up.

Start a Trigonometry test

More: all JEE Main Trigonometry questions · JEE Main 1979 · Mathematics chapter list