Question

If A+B= 45° and none of A and B is an odd multiple of pie/2 prove that 1
+ tan A)(1+tan B) = 2 and hence deduce that tan 22 1/2= √2-1​

Answer 1

Given:

• $A+B=45^{\circ}$
• None of $A$ and $B$ is an odd multiple of $\frac{\pi}{2}$.

To prove:

• $(1+tanA)(1+tanB)=2$
• $tan22\frac{1}{2}^{\circ}=\sqrt{2}-1$

Proof:

Here, $A+B=45^{\circ}$

Taking $tangent$ to both sides, we get

$\quad tan(A+B)=tan45^{\circ}$

$\Rightarrow \frac{tanA+tanB}{1-tanA\:tanB}=1$

$\Rightarrow tanA+tanB=1-tanA\:tanB$

$\Rightarrow 1+tanA+tanB+tanA\:tanB=2$

$\Rightarrow 1(1+tanA)+tanB(1+tanA)=2$

$\Rightarrow \boxed{(1+tanA)(1+tanB)=2}$

Thus proved.

Since $22\frac{1}{2}^{\circ}+22\frac{1}{2}^{\circ}=45^{\circ}$, we take $A=B$ where $A=22\frac{1}{2}^{\circ}$.

$\Rightarrow (1+tan22\frac{1}{2}^{\circ})^{2}=2$

$\Rightarrow 1+tan22\frac{1}{2}^{\circ}=\sqrt{2}$

$\Rightarrow \boxed{tan22\frac{1}{2}^{\circ}=\sqrt{2}-1}$

Thus proved.