If (1+tan alpha)(1+tan 4 alpha)=2 and alpha i | vedclass

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(A) Given that $(1+	an \alpha)(1+	an 4 \alpha)=2$ where $\alpha \in \left(0, \frac{\pi}{16}ight)$.Expanding the expression: $1 + 	an \alpha + 	a

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$\frac{\pi}{20}$

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(A) Given that $(1+	an \alpha)(1+	an 4 \alpha)=2$ where $\alpha \in \left(0, \frac{\pi}{16}ight)$.Expanding the expression: $1 + 	an \alpha + 	an 4 \alpha + 	an \alpha 	an 4 \alpha = 2$.Rearranging the terms: $	an \alpha + 	an 4 \alpha = 1 - 	an \alpha 	an 4 \alpha$.Dividing both sides by $(1 - 	an \alpha 	an 4 \alpha)$,we get: $\frac{	an \alpha + 	an 4 \alpha}{1 - 	an \alpha 	an 4 \alpha} = 1$.Using the identity $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we have: $	an(\alpha + 4 \alpha) = 1$.This simplifies to: $	an(5 \alpha) = 1$.Since $	an(\frac{\pi}{4}) = 1$,we have $5 \alpha = \frac{\pi}{4} + n\pi$.For $\alpha \in \left(0, \frac{\pi}{16}ight)$,we take $n=0$,so $5 \alpha = \frac{\pi}{4}$,which gives $\alpha = \frac{\pi}{20}$.

If $(1+\tan \alpha)(1+\tan 4 \alpha)=2$ and $\alpha \in \left(0, \frac{\pi}{16}\right)$,then $\alpha$ is equal to

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