Question

If `tan theta = p/q` and `theta = 3 phi( 0 < theta < pi/2)`, then `p/(sin phi) - q/ (cos phi)` =

Answer 1

Here, ` tantheta = p/q=>p = qtantheta->(1)`
`=>sec theta = sqrt(1+p^2/q^2) = sqrt(p^2+q^2)/q`
`=>costheta = q/(sqrt(p^2+q^2))->(2)`
`:. p/sinphi - q/cosphi = (qtantheta)/sinphi - q/cosphi`
`=q(tantheta/sinphi - 1/cosphi)`
`=q(sintheta/(costheta sinphi) - 1/cosphi)`
`=q((sinthetacosphi -costhetasinphi)/(costheta sinphicosphi))`
`=2q(sin(theta-phi)/(2sinphicosphi(costheta)))`
`=2q(sin(3phi-phi)/(sin2phi(costheta)))` .....(As `theta = 3phi`)
`=(2q)/costheta`
`=2q(sqrt(p^2+q^2)/q)` .....From(2)
`=2sqrt(p^2+q^2)`
`:. p/sinphi - q/cosphi = 2sqrt(p^2+q^2)`