Question

Prove that tan 5 theta + tan 3 theta / tan 5 theta - tan 3 theta = 4 cos 2 theta cos 4 theta

Answer 1

tan(5θ)+tan(3θ) tan(5θ)−tan(3θ) = sin(5θ) cos(5θ) + sin(3θ) cos(3θ) sin(5θ) cos(5θ) − sin(3θ) cos(3θ) = sin(5θ)cos(3θ)+cos(5θ)sin(3θ) cos(5θ)cos(3θ) sin(5θ)cos(3θ)−cos(5θ)sin(3θ) cos(5θ)cos(3θ) = sin(5θ)cos(3θ)+cos(5θ)sin(3θ) sin(5θ)cos(3θ)−cos(5θ)sin(3θ) = sin(5θ+3θ) sin(5θ−3θ) ⎡ ⎣ ⎢ ⎢ ⎢ Since,sin(A±B) =sin(A)cos(B)±cos(A)sin(B) ⎤ ⎦ ⎥ ⎥ ⎥ = sin(8θ) sin(2θ) = 2sin(4θ)cos(4θ) sin(2θ) ⎡ ⎣ ⎢ ⎢ ⎢ Since,sin(2A) =2sin(A)cos(A) ⎤ ⎦ ⎥ ⎥ ⎥ = 2(2sin(2θ)cos(2θ))cos(4θ) sin(2θ) ⎡ ⎣ ⎢ ⎢ ⎢ Since,sin(2A) =2sin(A)cos(A) ⎤ ⎦ ⎥ ⎥ ⎥ = 4sin(2θ)cos(2θ)cos(4θ) sin(2θ) =4cos(2θ)cos(4θ)

Answer 2

let A = θ.

$\frac{tan5A+tan3A}{tan5A-tan3A}=\frac{sin5A\ cos3A+sin3A\ cos5A}{sin5A\ cos3A-sin3A\ cos5A}\\\\=\frac{Sin(3A+5A)}{sin(5A-3A)}=\frac{sin8A}{sin2A}\\\\=\frac{2sin4A\ cos4A}{sin2A}=\frac{2*2sin2A\ cos2A\ cos4A}{sin2A}\\\\=4 cos2A\ cos4A$