Question

If A is in third quadrant and 3 tan A - 4 = 0 then prove that 5 sin 2A + 3 sin A + 4 cos A = 0

Answer 1

We have 3 tan A - 4 = 0

⇒ tan A = \(\frac{4}{3}\) 

Since, A lies in third quadrant

∴ sin A and cos A will be negative

⇒ cos A = \(-\frac{3}{5}\)

⇒ sin A = \(-\frac{4}{5}\)

∴ L.H.S. = 5 sin 2A + 3 sin A + 4 cos A

= 10 sin A cos A + 3 sin A + 4 cos A [∵ sin 2θ = 2 sin θ cos θ]