Question

Prove? cotx = sin5x+sin7x / cos5x-cos7x

Answer 1

See the proof below

Explanation:

Reminder

`sina+sinb=2sin((a+b)/2)cos((a-b)/2)`

`cosa-cosb=-2sin((a+b)/2)sin((a-b)/2)`

Therefore,

`RHS=(sin5x+sin7x)/(cos5x-cos7x)`

`=(2sin((5x+7x)/2)cos((5x-7x)/2))/(-2sin((5x+7x)/2)sin((5x-7x)/2))`

`=(cos((5x-7x)/2))/(-sin((5x-7x)/2))`

`=cosx/(-(-sinx))`

`=cotx`

`=LHS`

`QED`

Answer 2

Please see below.

Explanation:

WE have,
`color(red)((1)sinC+sinD=2sin((C+D)/2)cos((C-D)/2)`

`color(red)((2)cosC-cosD=-2sin((C+D)/2)sin((C-D)/2)`
Here,
`cotx=(sin5x+sin7x)/(cos5x-cos7x)`

`RHS=(sin5x+sin7x)/(cos5x-cos7x)`

Using (1) and (2),

`=(cancel(2)cancelsin((5x+7x)/2)cos((5x-7x)/2))/(-cancel2cancelsin((5x+7x)/2)sin((5x-7x)/2))`

`=-cos((5x-7x)/2)/sin((5x-7x)/2)`

`=-cos((-2x)/2)/sin((-2x)/2)`

`=-cos(-x)/sin(-x)`

`=-cosx/(-sinx`

`=cotx`