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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.

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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