Question

If $\tan \alpha = \dfrac{m}{{m + 1}}$ and $\tan \beta = \dfrac{1}{{2m + 1}}$ , then $\alpha + \beta $ is equal to
A. $\dfrac{\pi }{3}$
B. $\dfrac{\pi }{4}$
C. $0$
D. $\dfrac{\pi }{2}$

Answer 1

Hint: Here we have given two trigonometric values of tangent or $\tan $ function. We will apply trigonometric identities and formulas to solve this question. So we will use the $\tan (\alpha + \beta )$ formula here to simplify this value. We should keep in mind that we have to only find the value of $\alpha + \beta $ , so we have to eliminate $\tan $ from the expression.

Formula Used:
$\tan (\alpha + \beta ) = \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \cdot \tan \beta }}$

Complete step-by-step answer:
Here we have been given in the question $\tan \alpha = \dfrac{m}{{m + 1}}$ and $\tan \beta = \dfrac{1}{{2m + 1}}$ .
We have the formula here
 $\tan (\alpha + \beta ) = \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \cdot \tan \beta }}$
By substituting the values in the formula we have:
 $\dfrac{{\dfrac{m}{{m + 1}} + \dfrac{1}{{2m + 1}}}}{{1 - \left( {\dfrac{m}{{m + 1}}} \right)\left( {\dfrac{1}{{2m + 1}}} \right)}}$ .
We will now simplify this expression by taking the LCM and we have:
$ = \dfrac{{\dfrac{{m(2m + 1) + 1(m + 1)}}{{(m + 1)(2m + 1)}}}}{{\dfrac{{(m + 1)(2m + 1) - m}}{{(m + 1)(2m + 1)}}}}$
We can see that in the denominator of both the fraction, we have the same value, so we will cancel it.
On further simplifying we have:
$ = \dfrac{{2{m^2} + m + m + 1}}{{m(2m + 1) + 1(2m - 1) - m}}$
$ = \dfrac{{2{m^2} + 2m + 1}}{{2{m^2} + 3m + 1 - m}}$
It gives us value:
$ \Rightarrow \dfrac{{2{m^2} + 2m + 1}}{{2{m^2} + 2m + 1}} = 1$
Now we have
 $\tan (\alpha + \beta ) = 1$
We know that the value of tangent is one when we have $\tan 90^\circ = 1$
We can also write
$ \Rightarrow \tan 90 = \tan \dfrac{\pi }{4}$
Therefore by substituting this in the expression we have;
 $\tan (\alpha + \beta ) = \tan \dfrac{\pi }{4}$
Again the similar term i.e. $\tan $ from the left-hand side and right-hand side of the equation will get canceled, so we have
$(\alpha + \beta ) = \dfrac{\pi }{4}$ .
Hence the correct option is (B) $\dfrac{\pi }{4}$

So, the correct answer is “Option (B)”.

Note: We should also know another formula of the tangent which includes the difference between i.e. $\tan (\alpha - \beta )$ . The formula is:
$\tan (\alpha - \beta ) = \dfrac{{\tan \alpha - \tan \beta }}{{1 + \tan \alpha \tan \beta }}$ .
Some of the other basic trigonometric formulas of tangent functions are as follow;
$\tan (90 + \theta ) = \cot \theta $
$\tan (90 - \theta ) = - \cot \theta $ .
We should know that the tangent function is positive in the first and third quadrant and negative in the second and fourth quadrant.