# Please prove tan A + tan A tan B + tan B = 1. Please do it step by step. [on hold]

I am expecting an answer in degrees and I have no idea how to do this. I just rearranged the tan symbols and done a few values in my calculator.

Please do it step by step:

tanA+tanB+tanAtanB=1\tan A + \tan B + \tan A \tan B = 1 if A+B=45∘A + B = 45^\circ and AA and BB are not multiples of 90∘90^\circ (including 0∘0^\circ)

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We know
tan(A+B)=tanA+tanB1−tanAtanB \tan (A+B) = \frac{ \tan A + \tan B }{1 – \tan A \tan B }

Since A+B=π4A + B = \frac{ \pi }{4} , then

1=tanA+tanB1−tanAtanB⟹1=tanA+tanB+tanAtanB 1 = \frac{ \tan A + \tan B }{1 – \tan A \tan B } \implies \boxed{ 1 = \tan A + \tan B + \tan A \tan B }

Hint:

tan(A+B)=tan(A)+tan(B)1−tan(A)tan(B)\tan(A + B) = \frac{\tan(A) + \tan(B)}{1 – \tan(A)\tan(B)}

Hence your equation becomes, for example by writing the previous identity as tan(A)+tan(B)=tan(A+B)−tan(A+B)tan(A)tan(B)\tan(A) + \tan(B) = \tan(A+B) – \tan(A+B)\tan(A)\tan(B):

tan(A+B)−tan(A+B)tan(A)tan(B)+tan(A)tan(B)\tan(A+B) – \tan(A+B)\tan(A)\tan(B) + \tan(A)\tan(B)

tan(A+B)−tan(A)tan(B)(tan(A+B)−1)\tan(A+B) – \tan(A)\tan(B)(\tan(A+B) – 1)

Since you know A+B=45A + B = 45 degrees, you know that tan(45)=1\tan(45) = 1 hence

1−tan(A)tan(B)(1−1)1 – \tan(A)\tan(B)(1-1)

Which means obviously

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It is a well known result that tan(A+B)=(tan(A)+tan(B))/(1−tan(A)tan(B))tan(A + B) = (tan(A) + tan(B)) / (1 – tan(A) tan(B))
Therefore tan(A) + tan(B) = 1 – tan(A) tan(B) is equivalent to tan(A + B) = 1, hence the result.