Question

Prove that cos 4x + cos 3x + cos2x÷
sin 4x + sin 3x + sin 2x
= cot3x​

Answer 1

Answer:

LHS = RHS

check pic

Explanation:

sin X = cos (90- X)

cos X = sin (90-X)

Answer 2

Question :-- Prove that (cos 4x + cos 3x + cos2x) / (sin 4x + sin 3x + sin 2x) = cot3x

Formula used :-

• cosC + cosD =2*cos[(C+D)/2)]*cos[(C-D)/2)].
• sinC + sinD=2*sin[(C+D)/2)]*cos[(C-D)/2)].
• CosA/sinA = cotA .

Solution :-

using the above told formula in both Numerator and denominator of LHS in 4x and 2x parts , we get,

→ (cos4x+cos2x+cos3x)/(sin4x+sin2x+sin3x)

→ [2cos3x.cosx+cos3x]/[2sin3x.cosx+sin3x]

→ [cos3x(2cosx+1)]/[sin3x(2cosx+1)]

→ (cos3x) / (sin3x)

→ cot3x

✪✪ Hence Proved ✪✪