trigonometry problems

1.
tan A = 1/3, tan B = 1/2 and tan C = 1.
tan (A + B) = (1/3 + 1/2)/[1 - (1/3)(1/2)] = (5/6)/( 1 - 1/6) = 1
tan ( A + B + C) = tan [(A + B) + C] = ( 1 + 1)/[ 1 - (1)(1)] = 2/0.
Therefore, A + B + C = 90 degree = pi/2 radian.
2.
LHS = (sin A - tan A )( cos A - cot A) = [tan A( sin A/tan A - 1)][ cot A ( cos A/cot A - 1)]
= tan A cot A( cos A - 1)( sin A - 1) = (sin A - 1)( cos A - 1) = RHS.
3.
case 1. sin x is positive.
sinx = sinx + 2cosx
cos x = 0
x = pi/2, 3pi/2 (reject because sin(3pi/2) is not positive.)
case 2. sinx is negative.
-sinx = sinx + 2cos x
-2sinx = 2cosx
tan x = -1
x = 3pi/4( reject because sin(3pi/4) is not negative.) , 7pi/4.
So the answer is x = pi/2 and 7pi/4.