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If \[A + B = 225\], prove that \[\tan A + \tan B = 1 - \tan A\tan B\].

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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.
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