If $\dfrac{\sin{A}+\cos{A}}{\sin{A}-\cos{A}} = \dfrac{5}{3}$, find the value of $\dfrac{7\tan{A}+2}{2\tan{A}+7}$

$\dfrac{\sin{A}+\cos{A}}{\sin{A}-\cos{A}} = \dfrac{5}{3}$ is the given trigonometric equation and we have to find the value of trigonometric expression $\dfrac{7\tan{A}+2}{2\tan{A}+7}$. The trigonometric expression is in terms of $\tan{A}$ but we don’t know its value. So, it’s essential to find the value of $\tan{A}$ firstly and it can be done by solving the given trigonometric equation.

Find the value of tan(A)

The trigonometric equation can be solved in two different ways to find the value of $\tan{A}$.

Beginner’s Method

If you are a beginner, you can get the value of $\tan{A}$ by using cross multiplication method.

$\dfrac{\sin{A}+\cos{A}}{\sin{A}-\cos{A}}$ $\,=\,$ $\dfrac{5}{3}$

$\implies$ $3 \times (\sin{A}+\cos{A})$ $\,=\,$ $5 \times (\sin{A}-\cos{A})$

$\implies$ $3\sin{A}+3\cos{A}$ $\,=\,$ $5\sin{A}-5\cos{A}$

$\implies$ $5\cos{A}+3\cos{A}$ $\,=\,$ $5\sin{A}-3\sin{A}$

$\implies$ $8\cos{A}$ $\,=\,$ $2\sin{A}$

$\implies$ $\dfrac{8}{2} \,=\, \dfrac{\sin{A}}{\cos{A}}$

$\implies$ $\dfrac{\sin{A}}{\cos{A}} \,=\, \dfrac{8}{2}$

According to ratio or quotient trigonometric identity of sin and cos functions, the ratio of $\sin{A}$ to $\cos{A}$ is equal to $\tan{A}$.

$\implies$ $\tan{A} \,=\, \require{cancel} \dfrac{\cancel{8}}{\cancel{2}}$

$\,\,\, \therefore \,\,\,\,\,\,$ $\tan{A} \,=\, 4$

The value of $\tan{A}$ is equal to $4$. Now, substitute the value of $\tan{A}$ the trigonometric expression to evaluate it.

Advanced Method

If you are an advanced learner, you can solve the trigonometric equation by the componendo and dividendo rule.

$\implies$ $\dfrac{\sin{A}+\cos{A}+\sin{A}-\cos{A}}{\sin{A}+\cos{A}-(\sin{A}-\cos{A})}$ $\,=\,$ $\dfrac{5+3}{5-3}$

$\implies$ $\dfrac{\sin{A}+\sin{A}+\cos{A}-\cos{A}}{\sin{A}+\cos{A}-\sin{A}+\cos{A}}$ $\,=\,$ $\dfrac{8}{2}$

$\implies$ $\dfrac{\sin{A}+\sin{A}+\cos{A}-\cos{A}}{\sin{A}-\sin{A}+\cos{A}+\cos{A}}$ $\,=\,$ $\dfrac{8}{2}$

$\implies$ $\require{cancel} \dfrac{2\sin{A}+\cancel{\cos{A}}-\cancel{\cos{A}}}{\cancel{\sin{A}}-\cancel{\sin{A}}+2\cos{A}}$ $\,=\,$ $\require{cancel} \dfrac{\cancel{8}}{\cancel{2}}$

$\implies$ $\dfrac{2\sin{A}}{2\cos{A}}$ $\,=\,$ $4$

$\implies$ $\require{cancel} \dfrac{\cancel{2}\sin{A}}{\cancel{2}\cos{A}}$ $\,=\,$ $4$

$\implies$ $\dfrac{\sin{A}}{\cos{A}}$ $\,=\,$ $4$

$\,\,\, \therefore \,\,\,\,\,\,$ $\tan{A} \,=\, 4$

It’s true that the value of $\tan{A}$ is equal to $4$.