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Find domain of y = lg (1 - tan x)

Because zero and negative number have no logarithms, so, the condition must be satisfying, that is, 1 - tan x > 0. Move the negative tan x to the right side of the inequality and negative tangent x change to positive tan x, so tan x < 1.

The definition of tan x is: tan x is quals to sin x divides by cos x. Now we draw the graphs of sin x, cos x, and tan x.

draw the graph of y = sin x, y = cos x and y = tan x

tan x = sin x/cos x

When sin x = 0, the numerate of the fraction is zero, then the fraction is zero. So, tan x = 0.

When cos x = 0, the denominator of the fraction is zero, then the fraction is undefined. That is, tan x goes infinity.

When x = pi/4, sin x = cos x. So, tan x = 1. We get the point (pi/4, 1).

When x = 0, sin x = 0. So, tan x = 0, we get the point (0, 0).

When x = -pi/4, the angle x lies in quadrant4. So, the value of cos x is positive and the value of sin x is negative. Since sin x is equals to negative half of square root of 2 and cos x is equal to positive half of square root 2. So, tan x = -1. We get the point (-pi/4, -1)

When x = -pi/2, cos x = 0. So, tan x goes infinity. The vertical line x = -pi/2 is called asymptotes which means the curve of tan x never can across it.

When x = pi/2, cos x = 0, So, tan x goes infinity. The vertical line x = pi/2 is called asymptotes and the tan curve never can across it. So, we can draw the graph of y = tan x in the interval -pi/2 < x < pi/2.

When x = 3pi/2, cos x = 0. So, tan x goes infinity. The curve of tan x never can across the vertical line at x = 3pi/2.

When x = 5pi/4, sin x = cos x. So, tan x = 1. We get the point (5pi/4, 1).

When x = pi, sin x = 0. So, tan x = 0. We get the point (pi, 0).

When x = 3pi/4, the angle x lies in quadrant2. The value of sin x is positive and the value of cos x is negative. Since sin x is equal to half of square root of 2 and cos x is equal to negative half of square root 2. So, tan x is equal to negative one. We get the point (3pi/4, -1).

Now, we can draw the tan x curve in the interval pi/2 < x < 3pi/2.

Since tan pi/4 = 1 and the period of tan x is pi, so, the domain of the function y = lg (1 - tan x) is kpi - pi/2 < x < pi/4 + kpi, in which k is integer.