If A and B are angles in the second quadrant, then prove that 4cosA +

1.	If A and B are angles in the second quadrant, then prove that 4cosA + 3 cos B = -5
Answer» Given, \(\frac{sin A}{3} = \frac{sin B}{4} = \frac{1}{5}\) ∴ sin A = \(\frac{3}{5}\) and sin B = \(\frac{4}{5}\) We know that, cos² A = 1 - sin² = 1 - \((\frac{3}{5})^2 = 1 - \frac{9}{25} = \frac{16}{25}\) ∴ Cos A = ± 4/5 Since A lies in the second quadrant, cos A < 0 ∴ Cos A = – 4/5 Sin B = 4/5 We know that, cos² B = 1 - sin² B = 1 - \((\frac{4}{5})^2 = 1 - \frac{16}{25} = \frac {9}{25}\) ∴ Cos B = ± 4/5 Since B lies in the second quadrant, cos B < 0 ∴ cos B = - 3/5 ∴ 4cos A + 3 cos B = \(4(-\frac{4}{5}) + 3 (-\frac{3}{5})\) = \(-\frac{16}{5} - \frac {9}{5} = -\frac{25}{5}= -5\)

If A and B are angles in the second quadrant, then prove that 4cosA + 3 cos B = -5

Answer»

Given, \(\frac{sin A}{3} = \frac{sin B}{4} = \frac{1}{5}\)

∴ sin A = \(\frac{3}{5}\) and sin B = \(\frac{4}{5}\)

We know that,

cos² A = 1 - sin² = 1 - \((\frac{3}{5})^2 = 1 - \frac{9}{25} = \frac{16}{25}\)

∴ Cos A = ± 4/5

Since A lies in the second quadrant,

cos A < 0

∴ Cos A = – 4/5

Sin B = 4/5

We know that,

cos² B = 1 - sin² B = 1 - \((\frac{4}{5})^2 = 1 - \frac{16}{25} = \frac {9}{25}\)

∴ Cos B = ± 4/5

Since B lies in the second quadrant, cos B < 0

∴ cos B = - 3/5

∴ 4cos A + 3 cos B = \(4(-\frac{4}{5}) + 3 (-\frac{3}{5})\)

= \(-\frac{16}{5} - \frac {9}{5} = -\frac{25}{5}= -5\)