If y = (1 + tan A)(1 - tan B) where A - B = f | vedclass

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(C) Given: $A - B = \frac{\pi}{4}$.Taking tangent on both sides: $	an(A - B) = 	an\left(\frac{\pi}{4}ight) = 1$.Using the formula $	an(A - B) = \

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$27$

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(C) Given: $A - B = \frac{\pi}{4}$.Taking tangent on both sides: $	an(A - B) = 	an\left(\frac{\pi}{4}ight) = 1$.Using the formula $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$,we get:$\frac{	an A - 	an B}{1 + 	an A 	an B} = 1$.$	an A - 	an B = 1 + 	an A 	an B$.$	an A - 	an B - 	an A 	an B = 1$.Adding $1$ to both sides:$1 + 	an A - 	an B - 	an A 	an B = 1 + 1$.$(1 + 	an A) - 	an B(1 + 	an A) = 2$.$(1 + 	an A)(1 - 	an B) = 2$.Since $y = (1 + 	an A)(1 - 	an B)$,we have $y = 2$.Therefore,$(y + 1)^{y + 1} = (2 + 1)^{2 + 1} = 3^3 = 27$.

If $y = (1 + \tan A)(1 - \tan B)$ where $A - B = \frac{\pi}{4}$,then $(y + 1)^{y + 1}$ is equal to

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