tanx = 4/3, x in quadrant II. Find the value of sinx/2, cosx/2, tanx/2

As tanx = 4/3, π/2 < x < π i.e. x lies in 2nd quadrant
Hence tanx = 4/3 ⇒ sinx = 4/√(4²+3²) = 4/5
And cosx = -3/√(4²+3²) = -3/5
Now using 1 - cosx = 2sin²(x/2) ⇒ sinx/2 = ±√(1 - cosx)/2, we get
sinx/2 = ±√(1 - (-3/5))/2 = ±√8/10
As π/2 < x < π ⇒ π/4 < x/2 < π/2 and sine is positive in 1st quadrant
Then sinx/2 = 2/√5
Using 1 + cosx = 2cos²(x/2) ⇒ cosx/2 = ±√(1 + cosx)/2
We get, cosx/2 = ±√(1 + (-3/5))/2 = ±√2/10
π/2 < x < π ⇒ π/4 < x/2 < π/2 and cos is positive 1st quadrant
∴ cosx/2 = 1/√5
Using cosx = (1 - tan²(x/2))/(1 + tan²(x/2)) ⇒ tanx/2 = ±√(1 - cosx)/(1 + cosx)
We get, tanx/2 = ±√(1 - (-3/5))/(1 + (-3/5)) = ±√4/2
As π/2 < x < π ⇒ π/4 < x/2 < π/2 and tan is positive in 1st quadrant
∴ tanx/2 = 2