Question

# If $A + B = 225$, prove that $\tan A + \tan B = 1 - \tan A\tan B$.

Hint:- $\tan (x + y) = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}$
We are given with,
$\Rightarrow A + B = 225$ (1)
So, for proving the given result.
Taking tan both sides of equation 1. We get,
$\Rightarrow \tan (A + B) = \tan (225)$ (2)
Now, angle 225 in the RHS of equation 2, can also be written as $180 + 45$.
So, $\tan (A + B) = \tan (180 + 45)$ (3)
And, as we know that, according to trigonometric identities.
$\Rightarrow \tan (180 + \theta )$ can be written as $\tan \theta$.
Now, equation 3 becomes,
$\Rightarrow \tan (A + B) = \tan (45)$
And according to trigonometric identities $\tan (45) = 1$.
So, above equation becomes,
$\Rightarrow \tan (A + B) = 1$ (4)
Now, we have to use $\tan (x + y)$identity. To solve equation 4.
As we know that,
$\Rightarrow \tan (x + y) = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}$
So, $\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
So, equation 4 becomes,
$\Rightarrow \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}} = 1$
Now, cross-multiplying both sides of the above equation. We will get $\tan A + \tan B = 1 - \tan A\tan B$.
Hence, $\tan A + \tan B = 1 - \tan A\tan B$
Note:- Whenever we came up with this type of problem where we are given sum of two
numbers and had to prove a result in which tangent of angle is present. Then we apply tan
to both sides of a given equation and then use $\tan (x + y)$identity to get the required
result.