Math Doubts

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 $\tan{(\alpha+\beta)}$.

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

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

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

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

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

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

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

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

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

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

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

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

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

Find the value of sum of angles

According to trigonometry, the value of tan 45 degrees is one.

$\implies$ $\tan{(\alpha+\beta)} \,=\, \tan{(45^°)}$

Therefore, the value of $\alpha+\beta$ is equal to $45^°$.

Math Doubts

A best free mathematics education website for students, teachers and researchers.

Maths Topics

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Maths Problems

Learn how to solve the maths problems in different methods with understandable steps.

Learn solutions

Subscribe us

You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites.

Copyright © 2012 - 2021 Math Doubts, All Rights Reserved