If theta+phi=frac{pi}{4}, then (1+tan theta)( | vedclass

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(B) Given that $	heta + \phi = \frac{\pi}{4}$.Taking $	an$ on both sides, we get $	an(	heta + \phi) = 	an(\frac{\pi}{4}) = 1$.Using the formula $

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$2$

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(B) Given that $	heta + \phi = \frac{\pi}{4}$.Taking $	an$ on both sides, we get $	an(	heta + \phi) = 	an(\frac{\pi}{4}) = 1$.Using the formula $	an(	heta + \phi) = \frac{	an 	heta + 	an \phi}{1 - 	an 	heta 	an \phi} = 1$.This implies $	an 	heta + 	an \phi = 1 - 	an 	heta 	an \phi$, or $	an 	heta + 	an \phi + 	an 	heta 	an \phi = 1$.Now, consider the expression $(1 + 	an 	heta)(1 + 	an \phi) = 1 + 	an \phi + 	an 	heta + 	an 	heta 	an \phi$.Substituting the value from the previous step: $1 + (	an 	heta + 	an \phi + 	an 	heta 	an \phi) = 1 + 1 = 2$.

If $\theta+\phi=\frac{\pi}{4}$, then $(1+\tan \theta)(1+\tan \phi)$ is equal to

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