Question

Use the half-angle identity to determine tan105

Answer 1

Answer:

A

Step-by-step explanation:

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

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]