Trigonometric function based sum

Question

If $A+B=45$, then show that $(1+\tan A)(1+\tan B)=2$.

Please show all the steps, and explain how to solve this. I tried to solve this by multiplying both sides by tan but couldn't proceed further

Answer 1

I don't want to give a full solution yet, since I want to give you a chance to give it a go.

Hint:

Use $$\tan(A+B)=\tan(45^\circ)=1=\frac{\tan A+\tan B}{1-\tan A\tan B}$$ Note that $$(1+\tan A)(1+\tan B)=1+\tan A+\tan B+\tan A \tan B$$ Then try to rearrange the first equation so that it gives $(1+\tan A)(1+\tan B)$.

Answer 2

$A+B =45^\circ$

$\tan(A+B)=\tan(45^\circ)$

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

$\tan(A)+\tan(B)=1-\tan(A)\cdot\tan(B)$

$\tan(A)+\tan(B)+1+\tan(A)\cdot\tan(B)=2$

$\tan(A)(1+\tan(B))+1(1+\tan(B))=2$

$(1+\tan(A))(1+\tan(B)) =2$

Answer 3

With $$B=\frac{\pi}{4}-A$$ we get $$(1+\tan(A))(1+\cot(\frac{\pi}{4}+A))$$ and $$\cot(\frac{\pi}{4}+A)=\frac{\cot(A)-1}{1+\cot(A)}$$

Answer 4

$$A=45-B=\frac {\pi}{4}-B $$

$$\implies \tan (A)=\frac {1-\tan (B)}{1+\tan (B)} $$

since $$\tan (\frac {\pi}{4})=1$$ and $$\tan (a-b)=\frac {\tan (a)-\tan (b)}{1+\tan (a)\tan (b)} $$ thus

$$\tan (A)=\frac {2-(1+\tan (B))}{1+\tan (B)}=\frac {2}{1+\tan (B)}-1$$

Answer 5

Let $g = (1+tan (A))(1+tan (B))$ then we can expand to get $\dots$ $$(1+tan (A))(1+tan (B))= g = 1+tan(A)+tan(B)+tan(A)tan(B)$$ Now we turn to the trigonometric identity that states that $\dots$ $$tan(A+B) = \frac{tan(A)+tan(B)}{1-tan(A)tan(B)}$$ Now we take our expansion of $g$ and multiply by $-1$ and add $2$ resulting in $\dots$ $$-g+2= 1-tan(A)-tan(B)-tan(A)tan(B) $$$$ 2-g = -[tan(A) + tan(B)] + 1-tan(A)tan(B)$$ Dividing both sides by $1-tan(A)tan(B)$ we get $\dots $ $$\frac{2-g}{1-tan(A)tan(B)} = - \frac{tan(A)+tan(B)}{1-tan(A)tan(B)} +1$$ $$\frac{2-g}{1-tan(A)tan(B)} = 1-tan(A+B) = 1-tan(45) = 0 $$ $$2-g = 0$$ $$g=(1+tan (A))(1+tan (B)) = 2 $$