If 4 tan θ = 3, then $ \frac{4sinθ-cosθ+1}{4sinθ+cosθ-1} $= _________.

$\frac{13}{11}$

4 tan θ = 3

tan θ = \(\frac{3 }{4}\)

P² + B² = H²

3² + 4² = H²

H = 5

4 sin θ - cosθ + 1 = 4 × \(\frac{3 }{5}\) - \(\frac{4 }{5}\) + 1

= \(\frac{12 - 4 + 5  }{5}\) = \(\frac{13 }{5}\)

&  4 sin θ + cosθ - 1 = 4 × \(\frac{3 }{5}\) + \(\frac{4 }{5}\) - 1 

= \(\frac{12 + 4 - 5  }{5}\) = \(\frac{11 }{5}\)

Now , \(\frac{ 4 sin θ - cosθ + 1}{4 sin θ + cosθ - 1 }\)

= \(\frac{13 }{11}\)