# If #A+B=315^0#, then what is the value of #(1-tanA)(1-tanB)#?

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We'll use the formula for the tangent of the sum of two angles:

[ \tan(A + B) = \frac{{\tan A + \tan B}}{{1 - \tan A \cdot \tan B}} ]

Given that ( A + B = 315^\circ ), we can substitute ( 315^\circ ) for ( A + B ) and rearrange the formula to solve for ( 1 - \tan A \cdot \tan B ):

[ 1 - \tan A \cdot \tan B = \frac{{\tan A + \tan B}}{{\tan(315^\circ)}} ]

We know that ( \tan(315^\circ) = -1 ), so substituting this value:

[ 1 - \tan A \cdot \tan B = \frac{{\tan A + \tan B}}{{-1}} = -(\tan A + \tan B) ]

Now, we'll multiply ( (1 - \tan A)(1 - \tan B) ):

[ (1 - \tan A)(1 - \tan B) = 1 - \tan A - \tan B + \tan A \cdot \tan B ]

Given that ( 1 - \tan A \cdot \tan B = -(\tan A + \tan B) ), we can substitute this expression:

[ 1 - \tan A - \tan B - (\tan A + \tan B) = 1 - 2(\tan A + \tan B) ]

We know that ( A + B = 315^\circ ), so ( \tan(315^\circ) = -1 ), therefore:

[ \tan A + \tan B = -1 ]

Substituting this into the expression:

[ 1 - 2(-1) = 1 + 2 = 3 ]

So, the value of ( (1 - \tan A)(1 - \tan B) ) is ( 3 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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