If tan A = tan alpha coth x = cot beta tanh x | vedclass

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(D) Given: $	an A = 	an \alpha \coth x = \cot \beta 	anh x$  ...$(i)$From $(i)$,$	an \alpha = 	an A 	anh x$ and $	an \beta = \frac{	anh x}{	a

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$\sinh 2x \operatorname{cosec} 2A$

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(D) Given: $	an A = 	an \alpha \coth x = \cot \beta 	anh x$  ...$(i)$From $(i)$,$	an \alpha = 	an A 	anh x$ and $	an \beta = \frac{	anh x}{	an A}$.We know $	an (\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta}$.Substituting the values:$	an (\alpha + \beta) = \frac{	an A 	anh x + \frac{	anh x}{	an A}}{1 - (	an A 	anh x)(\frac{	anh x}{	an A})}$$= \frac{	anh x (	an A + \frac{1}{	an A})}{1 - 	anh^2 x}$$= \frac{	anh x (\frac{	an^2 A + 1}{	an A})}{\operatorname{sech}^2 x}$$= \frac{\sinh x}{\cosh x} \cdot \frac{\sec^2 A}{	an A} \cdot \cosh^2 x$$= \sinh x \cosh x \cdot \frac{1}{\cos^2 A} \cdot \frac{\cos A}{\sin A}$$= \frac{2 \sinh x \cosh x}{2 \sin A \cos A} = \frac{\sinh 2x}{\sin 2A} = \sinh 2x \operatorname{cosec} 2A$.

If $\tan A = \tan \alpha \coth x = \cot \beta \tanh x$,then $\tan (\alpha + \beta) =$

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