A geometry problem by Prasmit Biswas

sinA+cosecA=sinA+1/sinA

But by AM-GM inequality a+1/a is always greater than or equal to 2. So the answer is 0.

This problem needn't even involve geometry, and can be solved algebraically.

Take $x = \sin A$

We know $\sin A = \frac{1}{cosec A}$

So $\frac{1}{x} = cosec A$

So, $x + \frac{1}{x} = 1.5$

Multiplying by x, $x^2 + 1 = 1.5x$

Taking 1.5x to the LHS, $x^2 -1.5x + 1 = 0$

So, we get a quadratic equation as the result.

We find the determinant to be: $D = (-1.5)^2 - 4(1)(1)$

$D = 2.25 - 4$

$D = -1.75$

So, since the determinant is negative, all roots are unreal.

Therefore, there are zero real roots.

Since we need to find the sum of real roots, and there are no real roots, the answer is $\boxed{0}$

Considering all angles A, thereby providing real values to the sine and the cosecant, the indicated sum is never between -2 and 2 and so, the problem is invalidated. Yet I know the actual one, so bingo!