What the above expert said is right in the front and wrong in the back. Let me correct this:

Solution: (1) Because g(x) is increasing function in the domain;

So the first derivative of g(x) is greater than 0, and the second derivative of g(x) is equal to 0 when x = 1;

That is, the second derivative of g (1) =- 1/(x? )+( 1/(x？ ))( 1/sinθ)= 0；

The solution is θ = π/2;

(2) Let f (x) = f (x)-g (x);

f(x)= MX-(m- 1)/x-lnx- 1/x-lnx from( 1)；

So the first derivative of F(x) is (mx? -2x+m)/x？ ;

Because F(x) is a monotone function on the domain;

So the first derivative of F(x) is always greater than or equal to 0 or less than 0 in the definition domain;

So it's mx -2x+m≥0 or mx? -2x+m < 0；

x？ Must be greater than 0, if the function is monotonically increasing, then mx? -2x+m≥0, Delta must be less than or equal to 0, 4-4 * m2 < = 0; And quadratic coefficient m >；; 0，m & gt 1;

If the function is monotonically decreasing, then mx? -2x+m & lt； =0, Delta must be less than or equal to 0, 4-4 * m2 < = 0; And quadratic coefficient m.

Hope useful to you