If $tanA = \dfrac{x}{y}$, find the value of $\dfrac{xsinA - ycosA}{xsinA + ycosA}$. 


Solution:
Given,

$tan A = \dfrac{x}{y}$
To find: $\dfrac{xsinA - ycosA}{xsinA + ycosA} = ?$

Now,

$= \dfrac{xsinA - ycosA}{xsinA + ycosA}$

[Dividing by cosA]

$= \dfrac{ \frac{xsinA}{cosA} - \frac{ycosA}{cosA}}{\frac{xsinA}{cosA} + \frac{ycosA}{cosA}}$

$= \dfrac{x tanA - y }{x tanA + y}$

[Put value of tanA]

$= \dfrac{x (\frac{x}{y})- y}{x (\frac{x}{y} + y}$

$= \dfrac{\dfrac{x^2 - xy}{y}}{\dfrac{x^2 + xy}{y}}$

$= \dfrac{x^2 - y^2}{x^2 + y^2}$

Hence,
The required value of $\dfrac{xsinA - ycosA}{xsinA + ycosA}$ is $ \dfrac{x^2 - y^2}{x^2 + y^2}$.

#Trigonometry