Question

Question #83c24

Answer 1

Given `6tanx-5=0=>tanx=5/6`

Now

`(2sinx-cosx)/(5cosx+9sinx)`

Dividing both numerator and denominator by `cosx` the expression becimemes.

`(2sinx-cosx)/(5cosx+9sinx)`

`=((2sinx)/cosx-cosx/cosx)/((5cosx)/cosx+(9sinx)/cosx)`

`=(2tanx-1)/(5+9tanx)`

`=(2xx5/6-1)/(5+9xx5/6)`

`=(5/3-1)/(5+15/2)`

`=(2/3)/(25/2)=2/3xx2/25=4/75`