Find $\alpha+\beta$ if $\tan{\alpha} = \dfrac{x}{x+1}$ and $\tan{\beta} = \dfrac{1}{2x+1}$

$\alpha$ and $\beta$ are two angles. The values of $\tan{\alpha}$ is $\dfrac{x}{x+1}$ and $\tan{\beta}$ is equal to $\dfrac{1}{2x+1}$. In this case, the value of $\alpha+\beta$ should be evaluated in this trigonometric problem.

Use angle sum identity

In this trigonometry problem, the values of $\tan{\alpha}$ and $\tan{\beta}$ are given in algebraic form, and asked us to find the value of sum of angles $\alpha$ and $\beta$. The value of $\alpha + \beta$ can be calculated by using tan of angle sum identity.

$\tan{(\alpha+\beta)}$ $\,=\,$ $\dfrac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha}\tan{\beta}}$

Now, substitute values of $\tan{\alpha}$ and $\tan{\beta}$ in the expansion of tan of angle sum formula.

$\tan{(\alpha+\beta)}$ $\,=\,$ $\dfrac{\dfrac{x}{x+1}+\dfrac{1}{2x+1}}{1-\Bigg(\dfrac{x}{x+1}\Bigg)\Bigg(\dfrac{1}{2x+1}\Bigg)}$

Simplify the algebraic expression

The tan of sum of two angles $\alpha$ and $\beta$ is expressed in terms of an algebraic expression. It should be simplified to get the value of $\tan{(\alpha+\beta)}$.

$=\,$ $\dfrac{\dfrac{x}{x+1}+\dfrac{1}{2x+1}}{1-\Bigg(\dfrac{x}{x+1}\Bigg)\Bigg(\dfrac{1}{2x+1}\Bigg)}$

$=\,$ $\dfrac{\dfrac{x(2x+1)+1(x+1)}{(x+1)(2x+1)}}{1-\dfrac{x \times 1}{(x+1)(2x+1)}}$

$=\,$ $\dfrac{\dfrac{x \times 2x+ x \times 1+1 \times x+1 \times 1}{(x+1)(2x+1)}}{1-\dfrac{x}{(x+1)(2x+1)}}$

$=\,$ $\dfrac{\dfrac{2x^2+x+x+1}{(x+1)(2x+1)}}{\dfrac{1 \times (x+1)(2x+1) – x \times 1}{(x+1)(2x+1)}}$

$=\,$ $\dfrac{\dfrac{2x^2+2x+1}{(x+1)(2x+1)}}{\dfrac{(x+1)(2x+1)-x}{(x+1)(2x+1)}}$

$=\,$ $\dfrac{\dfrac{2x^2+2x+1}{(x+1)(2x+1)}}{\dfrac{x \times 2x + x \times 1 +1 \times 2x + 1 \times 1-x}{(x+1)(2x+1)}}$

$=\,$ $\dfrac{\dfrac{2x^2+2x+1}{(x+1)(2x+1)}}{\dfrac{2x^2+x+2x+1-x}{(x+1)(2x+1)}}$

$=\,$ $\dfrac{\dfrac{2x^2+2x+1}{(x+1)(2x+1)}}{\dfrac{2x^2+x+2x-x+1}{(x+1)(2x+1)}}$

$=\,$ $\dfrac{\dfrac{2x^2+2x+1}{(x+1)(2x+1)}}{\dfrac{2x^2+2x+1}{(x+1)(2x+1)}}$

$=\,$ $\dfrac{2x^2+2x+1}{(x+1)(2x+1)}$ $\times$ $\dfrac{(x+1)(2x+1)}{2x^2+2x+1}$

$=\,$ $\dfrac{2x^2+2x+1}{2x^2+2x+1}$ $\times$ $\dfrac{(x+1)(2x+1)}{(x+1)(2x+1)}$

$=\,$ $\require{cancel} \dfrac{\cancel{2x^2+2x+1}}{\cancel{2x^2+2x+1}}$ $\times$ $\require{cancel} \dfrac{\cancel{(x+1)(2x+1)}}{\cancel{(x+1)(2x+1)}}$

$\,\,\, \therefore \,\,\,\,\,\,$ $\tan{(\alpha+\beta)} \,=\, 1$