Question

If cot A=12/5 and A is an acute angle, then find sin3A and cos 3A​

Answer 1

Ans : cos A = 4/5

Solution :

Given

cot A + cosec A = 3

It can be written as

(cos A/sin A) + (1/sin A) = 3

(cos A + 1) / sin A = 3

cos A + 1 = 3 sin A

   cos A = cos^2 (A/2) - sin^2 (A/2)

   cos^2 (A/2) + sin^2 (A/2) = 1

cos^2 (A/2) - sin^2 (A/2) + cos^2 (A/2) + sin^2 (A/2) = 3*2 sin (A/2)cos (A/2),

2 cos^2 (A/2) = 3*2 sin (A/2)cos (A/2)

On solving

cos (A/2) = 3sin (A/2)

tan (A/2) = 1/3

   tan A = 2tan (A/2)/[1 - tan^2 A]

● tan A = 2*(1/3)/(1 - 1/9)

● tan A = (2/3)/(8/9)