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Question: If \(\tan \alpha = \dfrac{m}{{m + 1}}\) and \(\tan \beta = \dfrac{1}{{2m + 1}}\) , then \(\alpha + \...

If tan⁡α=mm+1\tan \alpha = \dfrac{m}{{m + 1}} and tan⁡β=12m+1\tan \beta = \dfrac{1}{{2m + 1}} , then α+β\alpha + \beta is equal to
A. π3\dfrac{\pi }{3}
B. π4\dfrac{\pi }{4}
C. 00
D. π2\dfrac{\pi }{2}

Explanation

Solution

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

Formula Used:
tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡α⋅tan⁡β\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⁡α=mm+1\tan \alpha = \dfrac{m}{{m + 1}} and tan⁡β=12m+1\tan \beta = \dfrac{1}{{2m + 1}} .
We have the formula here
tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡α⋅tan⁡β\tan (\alpha + \beta ) = \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \cdot \tan \beta }}
By substituting the values in the formula we have:
mm+1+12m+11−(mm+1)(12m+1)\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:
=m(2m+1)+1(m+1)(m+1)(2m+1)(m+1)(2m+1)−m(m+1)(2m+1)= \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:
=2m2+m+m+1m(2m+1)+1(2m−1)−m= \dfrac{{2{m^2} + m + m + 1}}{{m(2m + 1) + 1(2m - 1) - m}}
=2m2+2m+12m2+3m+1−m= \dfrac{{2{m^2} + 2m + 1}}{{2{m^2} + 3m + 1 - m}}
It gives us value:
⇒2m2+2m+12m2+2m+1=1\Rightarrow \dfrac{{2{m^2} + 2m + 1}}{{2{m^2} + 2m + 1}} = 1
Now we have
tan⁡(α+β)=1\tan (\alpha + \beta ) = 1
We know that the value of tangent is one when we have tan⁡90∘=1\tan 90^\circ = 1
We can also write
⇒tan⁡90=tan⁡π4\Rightarrow \tan 90 = \tan \dfrac{\pi }{4}
Therefore by substituting this in the expression we have;
tan⁡(α+β)=tan⁡π4\tan (\alpha + \beta ) = \tan \dfrac{\pi }{4}
Again the similar term i.e. tan⁡\tan from the left-hand side and right-hand side of the equation will get canceled, so we have
(α+β)=π4(\alpha + \beta ) = \dfrac{\pi }{4} .
Hence the correct option is (B) π4\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⁡(α−β)\tan (\alpha - \beta ) . The formula is:
tan⁡(α−β)=tan⁡α−tan⁡β1+tan⁡αtan⁡β\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+θ)=cot⁡θ\tan (90 + \theta ) = \cot \theta
tan⁡(90−θ)=−cot⁡θ\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.