Question

prove
cot(a+b)cot(a-b)={cos^2(b)-sin^2(a)}/{sin^2(a)-sin^(b)}

Answer 1

You will need these relations:

(1)

$$\cot(x) = \frac{\cos(x)}{\sin(x)} (2)$$

\cos(a + b) \cdot \cos(a - b) = \frac{1}{2} (\cos(2b) + \cos(2a))

(\sin(x) \cdot \sin(y)) = \frac{1}{2} (\cos(2b) - \cos(2a))

$$which come from$$

\cos(x) \cdot \cos(y) = \frac{1}{2} (\cos(x - y) + \cos(x + y))

\sin(x) \cdot \sin(y) = \frac{1}{2} (\cos(x - y) - \cos(x + y))

$$(3)$$

\cos(2x) = 1 - 2 \cdot \sin^2(x)

$$(4)$$

1 - \sin^2(x) = \cos^2(x)

$$Now we have$$

\cot(a + b) \cdot \cot(a - b) = (1) = \frac{\cos(a + b) \cdot \cos(a - b)}{\sin(a + b) \cdot \sin(a - b)} = (2) = \frac{\cos(2b) + \cos(2a)}{\cos(2b) - \cos(2a)} =

= (3) = \frac{1 - 2 \cdot \sin^2(b) + 1 - 2 \sin^2(a)}{1 - 2 \sin^2(b) - 1 + 2 \sin^2(a)} = \frac{(1 - \sin^2(b)) - \sin^2(a)}{\sin^2(a) - \sin^2(b)} = (4) = \frac{\cos^2(b) - \sin^2(a)}{\sin^2(a) - \sin^2(b)}

$$