Question

If we have sum of two angles as ($\theta + \phi = \dfrac{\pi }{4}$), then the value of $\left( {1 + \tan \theta } \right) \times \left( {1 + \tan \phi } \right)$ is....

Answer 1

Hint – In this particular type of question use the concept that from the q=given condition evaluate the value of any variable in terms of other variable and substitute it into the given equation which we have to solve then use the property of tan i.e. $\tan (A – B) =\dfrac {(\tan A – \tan B)}{(1 + \tan A \tan B)}$ and use the value of tan$\dfrac{\pi }{4}$ which is equal to 1 so use these concepts to reach the solution of the question.

Complete step-by-step solution -
We have to find the value of $\left( {1 + \tan \theta } \right) \times \left( {1 + \tan \phi } \right)$............................ (1)
Given equation:
$\theta + \phi = \dfrac{\pi }{4}$
So from here, ($\phi = \dfrac{\pi }{4} - \theta$).
Now substitute this value in equation (1) we have,
$\Rightarrow \left( {1 + \tan \theta } \right) \times \left( {1 + \tan \phi } \right) = \left( {1 + \tan \theta } \right)\left( {1 + \tan \left( {\dfrac{\pi }{4} - \theta } \right)} \right)$
Now as we know that $\tan (A – B) =\dfrac {(\tan A – \tan B)}{(1 + \tan A \tan B)}$ so use this property in the above equation we have,
$\Rightarrow \left( {1 + \tan \theta } \right) \times \left( {1 + \tan \phi } \right) = \left( {1 + \tan \theta } \right)\left( {1 + \dfrac{{\tan \dfrac{\pi }{4} - \tan \theta }}{{1 + \tan \dfrac{\pi }{4}\tan \theta }}} \right)$
Now as we know that the value of ($\tan \dfrac{\pi }{4}$ = 1) so use this value in the above equation we have,
$\Rightarrow \left( {1 + \tan \theta } \right) \times \left( {1 + \tan \phi } \right) = \left( {1 + \tan \theta } \right)\left( {1 + \dfrac{{1 - \tan \theta }}{{1 + \tan \theta }}} \right)$
Now simplify this we have,
$\Rightarrow \left( {1 + \tan \theta } \right) \times \left( {1 + \tan \phi } \right) = \left( {1 + \tan \theta } \right)\left( {\dfrac{{1 + \tan \theta + 1 - \tan \theta }}{{1 + \tan \theta }}} \right)$
$\Rightarrow \left( {1 + \tan \theta } \right) \times \left( {1 + \tan \phi } \right) = \left( {1 + \tan \theta } \right)\left( {\dfrac{{1 + 1}}{{1 + \tan \theta }}} \right)$
$\Rightarrow \left( {1 + \tan \theta } \right) \times \left( {1 + \tan \phi } \right) = 2$
So this is the required value of the given equation.
So this is the required answer.

Note – Whenever we face such types of questions the key concept we have to remember is that always recall all the basic trigonometric properties which can be proven very helpful to solve these types of problems so first use the basic property of tan as above applied then use the standard value of tan$\dfrac{\pi }{4}$ as above and simplify as above, we will get the required answer.