If frac{tan(A-B)}{tan A} + frac{sin^{2}C}{sin | vedclass

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(A) Given: $\frac{	an(A-B)}{	an A} + \frac{\sin^{2}C}{\sin^{2}A} = 1$ Using $	an(A-B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$,we have: $\frac{

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$	an A, 	an C, 	an B$ are in $G$.$P$.

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(A) Given: $\frac{	an(A-B)}{	an A} + \frac{\sin^{2}C}{\sin^{2}A} = 1$ Using $	an(A-B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$,we have: $\frac{	an A - 	an B}{	an A(1 + 	an A 	an B)} + \frac{\sin^{2}C}{\sin^{2}A} = 1$ Let $	an A = x, 	an B = y, 	an C = z$. Since $\sin^{2}C = \frac{	an^{2}C}{1 + 	an^{2}C} = \frac{z^{2}}{1 + z^{2}}$ and $\sin^{2}A = \frac{x^{2}}{1 + x^{2}}$,the equation becomes: $\frac{x-y}{x(1+xy)} + \frac{z^{2}(1+x^{2})}{x^{2}(1+z^{2})} = 1$ Multiplying by $x^{2}(1+xy)(1+z^{2})$ and simplifying: $x(x-y)(1+z^{2}) + z^{2}(1+x^{2})(1+xy) = x^{2}(1+xy)(1+z^{2})$ After algebraic expansion and cancellation,we obtain: $z^{2} = xy$ $	herefore 	an^{2}C = 	an A \cdot 	an B$ Thus,$	an A, 	an C, 	an B$ are in $G$.$P$.

If $\frac{\tan(A-B)}{\tan A} + \frac{\sin^{2}C}{\sin^{2}A} = 1,$ where $A, B, C \in (0, \frac{\pi}{2})$,then:

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