Prove that : sin 3a+sin a+sin 5a/cos 3a+cos a+cos 5a=tan 3a

We have to prove [sin 3a + sin a + sin 5a]/ [cos 3a+cos
a+cos 5a] = tan 3a

We start with [sin 3a + sin a + sin 5a]/
[cos 3a+cos a+cos 5a]

We use the formula: sin A + sin B = 2
sin [ (A + B) / 2 ] cos [ (A - B) / 2 ] and cos A + cos B = 2 cos [ (A + B) / 2 ] cos [
(A - B) / 2 ]

[sin 3a + sin a + sin 5a]/ [cos 3a+cos a+cos
5a]

=> [sin 3a + 2 sin [(a + 5a) / 2]  cos [ (5a -
a) / 2 ]/ [cos 3a+ 2 cos [ (a + 5a) / 2] cos [ (5a - a) / 2 ]

=> [sin 3a + 2 sin 3a * cos 2a]/ [cos 3a+ 2 cos 3a
cos 2a ]

=> [sin 3a ( 1 + 2 cos 2a)] / [cos 3a ( 1 +
2 cos 2a)]

=> sin 3a / cos
3a

=> tan
3a

This proves [sin 3a + sin a + sin 5a]/
[cos 3a+cos a+cos 5a] = tan 3a