Question

tangents of two acute angles are 2 and 3 find sine of twice their difference

Answer 1

Let A and B are two acute angles ..
A/C to question,
TanA = 2 ⇒A = tan⁻¹(2)
and tanB = 3 ⇒ B = tan⁻¹(3)

Now, B - A = tan⁻¹(3) - tan⁻¹(2)
Take tan both sides,
tan(B - A) = tan[(an⁻¹(3) - tan⁻¹(2)]

We know,
Tan[tan⁻¹x - tan⁻¹y] = (x - y)/(1 + xy)
Use this here,

Tan(B - A) = (3 - 2)/(1 + 3 × 2) = 1/7
(B - A) = tan⁻¹(1/7)
2(B - A) = 2tan⁻¹(1/7)
sin[2(B - A) ] = sin(2tan⁻¹(1/7))

2Tan⁻¹(1/7) = P [ let ]
tanP/2 = 1/7
tanP = 2tanP/2/(1 - tan²P/2) [ from formula ]
TanP = 2×1/7/(1 - 1/49) = 2/7/(48/49) = 7/24
TanP = 7/24
∴sinP = 7/25
sin(2tan⁻¹(1/7)) = 7/25

Hence, answer is 7/25