​If triangle ABC is a right angled isosceles triangle, right angled at B, then find the value of

$$\frac{\sin(A-C) + \sin(A+C) - 2\sin B}{\cot A + \cot B + \cot C}$$

Correct option is B

Given:

Triangle ABC is a right-angled isosceles triangle with right angle at B.

$$\frac{\sin(A-C) + \sin(A+C) - 2\sin B}{\cot A + \cot B + \cot C}$$

Formula Used:

Properties of a right-angled isosceles triangle:

In a right-angled isosceles triangle, the two non-right angles are equal.

In a right-angled triangle:

sin(90°) = 1

cos(90°) = 0

cot(A) = $$\frac {cos(A)} {sin(A)}$$

Solution:

Determine the angles of the triangle:

Since A = C and B = 90°,  A = C = 45°

Simplify the given expression:

$$\frac{\sin(A-C) + \sin(A+C) - 2\sin B}{\cot A + \cot B + \cot C}$$​​

=$$\frac{ [sin(0) + sin(2A) - 2sin(B)] }{ [cot(A) + cot(B) + cot(A)]}$$​​

= $$\frac{[sin(2A) - 2sin(B)] }{[2cot(A) + cot(B)]}$$​​

Substitute the values of sin(2A), sin(B), cot(A), and cot(B):

sin(2A) = sin(90°) = 1

sin(B) = sin(90°) = 1

cot(A) = cot(45°) =  1

cot(B) = cot(90°)  = 0

Substitute the values into the simplified expression:

​$$\frac{[1 - 2(1)] }{ [2(1) + 0]}$$​​

= ​$$-\frac{1}{2}$$​

Therefore, the value of the given expression is ​$$-\frac{1}{2}$$.

Option (b) is right.

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