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

Question

Elijah Alexander · Accepted Answer

# 2#.

#A+B=315^@=360^@-45^@#. #:. tan(A+B)=tan(360^@-45^@)#. #:. (tanA+tanB)/(1-tanAtanB)=-tan45^@=-1#. #:. tanA+tanB=-1+tanAtanB, or, # #tanA+tanB-tanAtanB=-1..................(ast)#. Now, #"the reqd. value="(1-tanA)(1-tanB),# #=1-tanA-tanB+tanAtanB#, #=1-(tanA+tanB-tanAtanB)#, #=1-(-1).................................................[because, (ast)]#, #rArr" The Reqd. Value"=2#.

Brooklyn Anderson · Answer

undefined 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 $.