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If the value of $A + B = 45^\circ $, Prove that $\left( {1 + \tan A} \right)\left( {1 + \tan B} \right) = 2$.

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Hint: We need to know the basic trigonometric formula and basic trigonometric function values to solve this problem.

Complete step-by-step answer:
Given $A + B = 45^\circ $
Applying tan function on both sides of the above equation
$ \Rightarrow \tan (A + B) = \tan 45^\circ $
$ \Rightarrow \tan (A + B) = 1\;\left[ {\because \tan 45^\circ = 1} \right]$
Using tan (A + B) formula, we can write
$ \Rightarrow \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}} = 1$
$ \Rightarrow \tan A + \tan B = 1 - \tan A\tan B$ ... (1)
We need to prove $\left( {1 + \tan A} \right)\left( {1 + \tan B} \right) = 2$
Taking LHS of the above equation
$ \Rightarrow 1 + \tan A + \tan B + \tan A\tan B$
Substituting the equation (1) in the above equation, we get
$ \Rightarrow 1 + \left( {1 - \tan A\tan B} \right) + \tan A\tan B$
$ \Rightarrow 1 + 1$
=2
Hence proved.
$\therefore \left( {1 + \tan A} \right)\left( {1 + \tan B} \right) = 2$

Note: We used the basic trigonometric value $\tan 45^\circ = 1$ and formula of $\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$.
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