If $tanx =\frac{3}{2}$, then the value of $\frac{3sinx+2cosx}{3sinx-2cosx}$ is :
Answer & explanation
Correct answer: option 3
tanx = \(\frac{3 }{2}\)
{ tanx = \(\frac{P }{B}\) }
P² + B² = H²
3² + 2² = H²
H = √15
Now,
$\frac{3sinx+2cosx}{3sinx-2cosx}$
= \(\frac{3 × 3/ √15 + 2 × 2/ √15}{3 × 3/ √15 - 2 × 2/ √15 }\)
= \(\frac{ 13 }{5 }\)