Question

If $A + B = \dfrac{\pi }{4}$ then $\left( {1 + \tan A} \right)\left( {1 + \tan B} \right)$ is equal to
A. $1$
B. $2$
C. $3$
D. None of these

Answer 1

Hint: In order to solve this type of question, first we will consider the given equation and solve it by taking tan on both the sides. Then, we will apply a suitable trigonometric identity(tangent sum) to it and solve it further in order to get the desired answer.

Formula used:
$\left[ {\because \tan \dfrac{\pi }{4} = 1} \right]$
$\left[ {\because \tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}} \right]$

Complete step by step solution:
We are given that,
$A + B = \dfrac{\pi }{4}$
Taking tan on both sides
$\tan \left( {A + B} \right) = \tan \dfrac{\pi }{4}$
$\tan \left( {A + B} \right) = 1$ $\left[ {\because \tan \dfrac{\pi }{4} = 1} \right]$
Solving it,
$\dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}} = 1$ $\left[ {\because \tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}} \right]$
\[\tan A + \tan B = 1\left( {1 - \tan A\tan B} \right)\]
Add $\left( {1 + \tan A\tan B} \right)$ to both sides,
\[\tan A + 1 + \tan B + \tan A\tan B = 2\]
Taking common,
\[1\left( {1 + \tan A} \right) + \tan B\left( {1 + \tan A} \right) = 2\]
\[\left( {1 + \tan A} \right)\left( {1 + \tan B} \right) = 2\]
$\therefore $ The correct option is B.

Note: The key concept of solving this type of question is to be careful with the simplification part. Choose the suitable trigonometric identities and be very sure while simplifying them. This type of question requires the use of correct application of trigonometric rules to get the correct answer.