# If $A+B = 45^°$, then find $(1+\tan{A})(1+\tan{B})$

It’s mentioned that $A$ and $B$ are two angles, and the sum of them is equal to $45$ degrees. In this trigonometric problem, it is asked us to find the product of two factors $1+\tan{A}$ and $1+\tan{B}$.

### Use Tan of Angle sum identity

It’s given that $A+B = 45^°$ and the product of two factors is in terms of tan functions. So, it is essential to take tangent both sides.

$\implies$ $\tan{(A+B)}$ $\,=\,$ $\tan{(45^°)}$

Now, expand tan of sum of two angles function by using tan of angle sum identity‘s formula. It is proved that the value of tan 45 degrees is equal to one.

$\implies$ $\dfrac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}$ $\,=\,$ $1$

### Simplify the Trigonometric equation

Let us try to simplify this trigonometric equation for simplifying it further.

$\implies$ $\tan{A}+\tan{B}$ $\,=\,$ $1-\tan{A}\tan{B}$

$\implies$ $\tan{A}+\tan{B}$ $+$ $\tan{A}\tan{B}$ $\,=\,$ $1$

$\implies$ $\tan{A}$ $+$ $\tan{A}\tan{B}$ $+$ $\tan{B}$ $\,=\,$ $1$

### Factorize the trigonometric expression

The left hand side trigonometric expression is completely in terms of tan functions. So, factorise it for getting the product of $1+\tan{A}$ and $1+\tan{B}$. You can take either $\tan{A}$ or $\tan{B}$ common from two terms of the trigonometric expression.

$\implies$ $\tan{A}{(1+\tan{B})}$ $+$ $\tan{B}$ $\,=\,$ $1$

$1+\tan{B}$ is a factor in the first term of the expression. If the second term is $1+\tan{B}$, then it’s easy to express the left hand side expression as the product of $1+\tan{A}$ and $1+\tan{B}$. So, add $1$ and subtract $1$ in the left hand side expression.

$\implies$ $\tan{A}{(1+\tan{B})}$ $+$ $1-1$ $+$ $\tan{B}$ $\,=\,$ $1$

$\implies$ $\tan{A}{(1+\tan{B})}$ $+$ $1+\tan{B}$ $\,=\,$ $1+1$

$\implies$ $\tan{A}{(1+\tan{B})}$ $+$ $1(1+\tan{B})$ $\,=\,$ $2$

Now, $1+\tan{B}$ is a common factor in the both terms of the trigonometric expression. So, take it common from both terms for the product of $1+\tan{A}$ and $1+\tan{B}$.

$\implies$ ${(1+\tan{B})}$ ${(\tan{A}+1)}$ $\,=\,$ $2$

$\implies$ ${(1+\tan{B})}$ ${(1+\tan{A})}$ $\,=\,$ $2$

$\,\,\, \therefore \,\,\,\,\,\,$ ${(1+\tan{A})}$ ${(1+\tan{B})}$ $\,=\,$ $2$

Therefore, it is proved that the product of ${(1+\tan{A})}$ and ${(1+\tan{B})}$ equals to $2$ if the sum of angles $A$ and $B$ is $45^°$.

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