If 4 cot θ = 3, show that $\Big(\dfrac{\sin θ - \cos θ}{\sin θ+ \cos θ}\Big) = \dfrac{1}{7}$.

cot θ = $\dfrac{\text{base}}{\text{perpendicular}}=\dfrac{3}{4}$

Let base = 3x and perpendicular = 4x

We will find hypotenuse by using pythagoras theorem

Hypotenuse² = Base² + Perpendicular²

Hypotenuse² = (3x)² + (4x)²

Hypotenuse² = 9x² + 16x²

Hypotenuse² = 25x²

Hypotenuse = 5x

Now

sin θ = $\dfrac{\text{perpendicular}}{\text{hypotenuse}}= \dfrac{4x}{5x} = \dfrac{4}{5}$

cos θ = $\dfrac{\text{base}}{\text{hypotenuse}}= \dfrac{3x}{5x} = \dfrac{3}{5}$

Substituting values we get :

$\Rightarrow \dfrac{\sin θ - \cos θ}{\sin θ+ \cos θ}\\[1em] = \dfrac{\dfrac{4}{5} -\dfrac{3}{5}}{\dfrac{4}{5} + \dfrac{3}{5}}\\[1em] = \dfrac{\dfrac{4-3}{5}}{\dfrac{4+3}{5}}\\[1em] = \dfrac{\dfrac{1}{5}}{\dfrac{7}{5}}\\[1em] = \dfrac{1}{5}\times\dfrac{5}{7}\\[1em] = \dfrac{1}{7}.$

Hence, proved that $\dfrac{\sin θ - \cos θ}{\sin θ+ \cos θ}$ = $\dfrac{1}{7}$