If 0 leq A, B leq frac{pi}{4} and cot A + cot | vedclass

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(C) Given the equation: $\cot A + \cot B + 	an A + 	an B = \cot A \cot B - 	an A 	an B$. Expressing in terms of $\sin$ and $\cos$: $\frac{\cos A}{

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$\frac{1}{\sqrt{2}}$

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(C) Given the equation: $\cot A + \cot B + 	an A + 	an B = \cot A \cot B - 	an A 	an B$. Expressing in terms of $\sin$ and $\cos$: $\frac{\cos A}{\sin A} + \frac{\cos B}{\sin B} + \frac{\sin A}{\cos A} + \frac{\sin B}{\cos B} = \frac{\cos A \cos B}{\sin A \sin B} - \frac{\sin A \sin B}{\cos A \cos B}$. Combining terms on the left: $\frac{\sin(A+B)}{\sin A \sin B} + \frac{\sin(A+B)}{\cos A \cos B} = \frac{\cos^2 A \cos^2 B - \sin^2 A \sin^2 B}{\sin A \sin B \cos A \cos B}$. Using the identity $x^2 - y^2 = (x-y)(x+y)$: $\sin(A+B) \left( \frac{\cos A \cos B + \sin A \sin B}{\sin A \sin B \cos A \cos B} ight) = \frac{(\cos A \cos B - \sin A \sin B)(\cos A \cos B + \sin A \sin B)}{\sin A \sin B \cos A \cos B}$. Since $\cos A \cos B + \sin A \sin B = \cos(A-B)$,we have: $\sin(A+B) \frac{\cos(A-B)}{\sin A \sin B \cos A \cos B} = \frac{\cos(A+B) \cos(A-B)}{\sin A \sin B \cos A \cos B}$. Assuming $\cos(A-B) 
eq 0$,we get $\sin(A+B) = \cos(A+B)$,which implies $	an(A+B) = 1$. Since $0 \leq A, B \leq \frac{\pi}{4}$,$0 \leq A+B \leq \frac{\pi}{2}$. Thus,$A+B = \frac{\pi}{4}$. Therefore,$\sin(A+B) = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.

If $0 \leq A, B \leq \frac{\pi}{4}$ and $\cot A + \cot B + \tan A + \tan B = \cot A \cot B - \tan A \tan B$,then $\sin(A + B) = $

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