# 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.

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 $4$.

### Find the value of trigonometric expression

Substitute the value of $\tan{A}$ in the trigonometric expression and then simplify it mathematically.

$\dfrac{7\tan{A}+2}{2\tan{A}+7}$ $\,=\,$ $\dfrac{7(4)+2}{2(4)+7}$

$=\, \dfrac{28+2}{8+7}$

$=\, \dfrac{30}{15}$

$=\, \require{cancel} \dfrac{\cancel{30}}{\cancel{15}}$

$=\, 2$

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