If tan A = 2tan B + cot B, then 2tan (A - B) | vedclass

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(C) Given $	an A = 2	an B + \cot B.$We know that $	an (A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}.$Substituting the value of $	an A$:$2	a

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$\cot B$

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(C) Given $	an A = 2	an B + \cot B.$We know that $	an (A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}.$Substituting the value of $	an A$:$2	an (A - B) = 2 \left( \frac{2	an B + \cot B - 	an B}{1 + (2	an B + \cot B)	an B} ight)$$= 2 \left( \frac{	an B + \cot B}{1 + 2	an^2 B + \cot B 	an B} ight)$Since $\cot B 	an B = 1$:$= 2 \left( \frac{	an B + \cot B}{1 + 2	an^2 B + 1} ight) = 2 \left( \frac{	an B + \cot B}{2 + 2	an^2 B} ight)$$= 2 \left( \frac{	an B + \cot B}{2(1 + 	an^2 B)} ight) = \frac{	an B + \cot B}{\sec^2 B}$$= \frac{	an B}{\sec^2 B} + \frac{\cot B}{\sec^2 B} = \sin B \cos B + \frac{\cos B}{\sin B} \cos^2 B = \sin B \cos B + \frac{\cos^3 B}{\sin B}$Alternatively,simplifying the expression directly:$= \frac{	an B + \frac{1}{	an B}}{1 + 	an^2 B} = \frac{\frac{	an^2 B + 1}{	an B}}{1 + 	an^2 B} = \frac{1}{	an B} = \cot B.$

List-$A$	List-$B$
$(I)$ $\tan \left(\frac{\alpha + \beta}{2}\right) =$	$(a)$ $\frac{b}{a}$
$(II)$ $\cos (\alpha + \beta) =$	$(b)$ $\frac{2ab}{a^2 + b^2}$
$(III)$ $\sin (\alpha + \beta) =$	$(c)$ $\frac{2ab}{a^2 - b^2}$
$(IV)$ $\tan (\alpha + \beta) =$	$(d)$ $\frac{a^2 - b^2}{a^2 + b^2}$

If $\tan A = 2\tan B + \cot B,$ then $2\tan (A - B) = $

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