Question

tan 5 x + tan 3 x = 4 cos 2 x cos 4 x.
(iv)
tan 5 x -tan 3 x​

Answer 1

We have,

${tan5x + tan3x} = 4cos2xcos4x \\ = > \frac{sin5x}{cos5x} + \frac{sin3x}{cos3x} = 4cos2xcos4x \\ = > \frac{sin5xcos3x + sin3xcos5x}{cos5xcos3x} = 4cos2xcos4x \\ = > let \: 5x \: be \: a \: and \: 3x \: be \: b \\ = > \frac{sinacosb + cosasinb}{cosacosb} = 4cos2xcos4x \\ = > \frac{sin(a + b) + sin(a - b) + sin(a + b) - sin(a - b)}{2cosacosb} = 4cos2xcos4x \\ = > \frac{2sin(5x + 3x)}{cos(a + b) + cos(a - b)} = 4cos2xcos4x \\ = > \frac{sin(8x)}{cos(8x) + cos(2x)} = 2cos2xcos4x \\ = > \frac{sin(8x)}{2cos2xcos4x} = cos(8x) + cos(2x)$

Again,

$\frac{sin5x}{cos5x} - \frac{sin3x}{cos3x} \\ = \frac{sin(2x)}{cos(8x) + cos(3x)} \\ = \frac{2cos2xcos4x(sin2x)}{sin8x} \\ = \frac{2cos2xcos4xsin2x}{2cos4xsin4x} \\ = \frac{cos2x \times sin2x}{sin4x} \\ = \frac{sin(2x + 2x) - sin(2x - 2x)}{sin4x} \\ = \frac{sin4x}{sin4x} = 1$

Therefore,

$tan5x - tan3x = 1$