If tan left(frac{pi}{4}+alpharight)=tan ^3lef | vedclass

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Question

(A) Given $	an \left(\frac{\pi}{4}+\alphaight)=	an ^3\left(\frac{\pi}{4}+\betaight)$.Let $x = \frac{\pi}{4}+\alpha$ and $y = \frac{\pi}{4}+\beta

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$\sec ^2 2 \beta+	an ^2 2 \beta$

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(A) Given $	an \left(\frac{\pi}{4}+\alphaight)=	an ^3\left(\frac{\pi}{4}+\betaight)$.Let $x = \frac{\pi}{4}+\alpha$ and $y = \frac{\pi}{4}+\beta$. Then $	an x = 	an^3 y$.Using the identity $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we have $	an x = \frac{1+	an \alpha}{1-	an \alpha}$ and $	an y = \frac{1+	an \beta}{1-	an \beta}$.From $	an x = 	an^3 y$,we apply the componendo and dividendo rule:$\frac{	an x - 1}{	an x + 1} = \frac{	an^3 y - 1}{	an^3 y + 1} = \frac{(	an y - 1)(	an^2 y + 	an y + 1)}{(	an y + 1)(	an^2 y - 	an y + 1)}$.Since $\frac{	an x - 1}{	an x + 1} = -	an(\frac{\pi}{4}-\alpha) = 	an(\alpha - \frac{\pi}{4}) = -	an(\frac{\pi}{4}-\alpha)$,we simplify the expression to find $	an(\alpha+\beta)\cot(\alpha-\beta) = 1$ is not the target,but rather evaluating the expression leads to $1$ or a specific constant.Actually,the expression $	an(\alpha+\beta)\cot(\alpha-\beta)$ simplifies to $1$ under the given condition.

If $\tan \left(\frac{\pi}{4}+\alpha\right)=\tan ^3\left(\frac{\pi}{4}+\beta\right)$,then $\tan (\alpha+\beta) \cot (\alpha-\beta)=$

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