We have cosθ = 0.6 = (6k)/(10k) = AC/AB (For some value of k)

By Pythagoras theorem, (hypotenuse)² = (perpendicular)² + (base)²

∴AB² = BC² + AC²

= (10k)² = BC² + (6k)²

= 100k² = BC² + 36k²

= BC² = 64k²

= (8k)²

→ BC = 8k

∴ sinθ = BC/AB = (8k)/(10k) = 0.8

tanθ = sinθ /cosθ = 0.8/0.6

consider, the LHS,

5sinθ - 3tanθ = 5(0.8) - 3(0.8/0.6)

= 4 - 3(0.4/0.3)

= 4(0.3) - 3(0.4)

= 1.2 - 1.2

= 0

= RHS

HENCE PROVED