Question

Use the half-angle identity to determine tan105

Answer 1

A

Explanation:

`tan 2x=(2tan x)/(1-tan^2x) \\put x=105\\2x=105*2=210\\tan ~2x=tan~210=tan(180+30)=tan ~30=(1)/(√(3)) \\(1)/(√(3) ) =(2~tan~x)/(1-tan^2x) \\cross~multiply \\2√(3)~tan~x=1-tan^2~x\\tan^2x+2√(3)~tan~x-1=0\\tan~x=\frac{-2√(3) \pm \sqrt{(2√(3))^2-4*1*(-1)} }{2*1} \\=(-2√(3) \pm √(12+4) )/(2) \\=(-2 √(3) \pm 4)/(2)\\=-√(3)} \pm~2\\`

as tan 105 lies in second quadrant,so it is negative.

tan 105=-√3 -2=-√(-√3-2)²=-√[3+4+2(-√3)(-2)]=-√[7+4√3]