Question

If `8sintheta=4+costheta`, what is the value of `tantheta` ?

Answer 1

`rarrtanx= 3/4`

Explanation:

Let `theta=x` then

`rarr8sinx=4+cosx`

`rarr(8sinx)/cosx=(4+cosx)/cosx=4/cosx+1`

`rarr8tanx=4secx+1`

`rarr8tanx-1=4secx`

Squaring both sides, we get,

`rarr64tan^2x-16tanx+1=16sec^2x=16(1+tan^2x)`

`rarr64tan^2x-16tanx+1=16+16tan^2x`

`rarr48tan^2x-16tanx-15=0`

`rarr48tan^2x+20tanx-36tanx-15=0`

`rarr4tanx(12tanx+5)-3(12tanx+5)=0`

`rarr(12tanx+5)(4tanx-3)=0`

`rarrtanx=-5/12 or 3/4`

But `x=tan^(-1)(-5/12)` does not satisfy the equation so `tanx=3/4`