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If $tanx =\frac{3}{2}$, then the value of $\frac{3sinx+2cosx}{3sinx-2cosx}$ is : |
$\frac{1}{5}$ $\frac{5}{13}$ $\frac{13}{5}$ 5 |
$\frac{13}{5}$ |
tanx = \(\frac{3 }{2}\) { tanx = \(\frac{P }{B}\) } P² + B² = H² 3² + 2² = H² H = √15 Now, $\frac{3sinx+2cosx}{3sinx-2cosx}$ = \(\frac{3 × 3/ √15 + 2 × 2/ √15}{3 × 3/ √15 - 2 × 2/ √15 }\) = \(\frac{ 13 }{5 }\) |
