If (1 + tan theta )(1 + tan phi ) = 2,then th | vedclass

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(B) Given: $(1 + 	an 	heta )(1 + 	an \phi ) = 2$ Expanding the left side: $1 + 	an \phi + 	an 	heta + 	an 	heta 	an \phi = 2$ Rearranging the

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$45$

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(B) Given: $(1 + 	an 	heta )(1 + 	an \phi ) = 2$ Expanding the left side: $1 + 	an \phi + 	an 	heta + 	an 	heta 	an \phi = 2$ Rearranging the terms: $	an 	heta + 	an \phi = 1 - 	an 	heta 	an \phi$ Dividing both sides by $(1 - 	an 	heta 	an \phi)$: $\frac{	an 	heta + 	an \phi}{1 - 	an 	heta 	an \phi} = 1$ Using the identity $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we get: $	an(	heta + \phi) = 1$ Since $	an(45^\circ) = 1$,we have: $	heta + \phi = 45^\circ$.

If $(1 + \tan \theta )(1 + \tan \phi ) = 2$,then $\theta + \phi = \dots ^\circ$

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