If tan A+tan B+cot A+cot B=tan A tan B-cot A | vedclass

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(B) Given: $	an A+	an B+\cot A+\cot B=	an A 	an B-\cot A \cot B$We know that $	an A+	an B = 	an(A+B)(1-	an A 	an B)$ and $\cot A+\cot B = \fr

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(B) Given: $	an A+	an B+\cot A+\cot B=	an A 	an B-\cot A \cot B$We know that $	an A+	an B = 	an(A+B)(1-	an A 	an B)$ and $\cot A+\cot B = \frac{\cot A \cot B}{\cot(A+B)}$ is not standard,but $\cot A+\cot B = \frac{\sin(A+B)}{\sin A \sin B}$.Let us rewrite the equation as: $(	an A+	an B) + (\cot A+\cot B) = 	an A 	an B - \cot A \cot B$$\frac{\sin(A+B)}{\cos A \cos B} + \frac{\sin(A+B)}{\sin A \sin B} = \frac{\sin(A-B)}{\cos A \cos B} \cdot \frac{\sin(A-B)}{\sin A \sin B}$ (This is complex).Alternatively,$	an A+\cot A = \frac{1}{\sin A \cos A} = \frac{2}{\sin 2A}$.The given equation is $	an A+	an B+\cot A+\cot B = 	an A 	an B - \cot A \cot B$.Rearranging: $(	an A+\cot A) + (	an B+\cot B) = 	an A 	an B - \cot A \cot B$.Using $	an 	heta + \cot 	heta = \frac{1}{\sin 	heta \cos 	heta} = \frac{2}{\sin 2	heta}$,this does not simplify easily.Let's use $	an A+	an B = \frac{\sin(A+B)}{\cos A \cos B}$ and $\cot A+\cot B = \frac{\sin(A+B)}{\sin A \sin B}$.$\sin(A+B) [\frac{1}{\cos A \cos B} + \frac{1}{\sin A \sin B}] = 	an A 	an B - \cot A \cot B$$\sin(A+B) [\frac{\sin A \sin B + \cos A \cos B}{\cos A \cos B \sin A \sin B}] = \frac{\sin^2 A \sin^2 B - \cos^2 A \cos^2 B}{\cos A \cos B \sin A \sin B}$$\sin(A+B) \cos(A-B) = (\sin A \sin B - \cos A \cos B)(\sin A \sin B + \cos A \cos B)$$\sin(A+B) \cos(A-B) = -\cos(A+B) \cos(A-B)$Since $\cos(A-B) 
eq 0$,we have $\sin(A+B) = -\cos(A+B) \Rightarrow 	an(A+B) = -1$.Given $0^{\circ} < A+B < 270^{\circ}$,$	an(A+B) = -1$ implies $A+B = 135^{\circ}$.

If $\tan A+\tan B+\cot A+\cot B=\tan A \tan B-\cot A \cot B$ and $0^{\circ} < A+B < 270^{\circ}$,then $A+B=$ (in $^{\circ}$)

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