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if 6sec θ = 10, then find the value of $\frac{5cosecθ-3cotθ}{4cosθ+3sinθ}$ . |
$\frac{2}{3}$ $\frac{3}{2}$ $\frac{5}{6}$ $\frac{6}{5}$ |
$\frac{5}{6}$ |
6sec θ = 10 sec θ = \(\frac{ 5 }{3}\) By using pythagoras theorem , P² + B² = H² P² + 3² = 5² P = 4 Now, \(\frac{ 5cosecθ - 3cotθ }{4cosθ + 3sinθ}\) = \(\frac{ 5 × 5/4 - 3 × 3/4 }{4×3/5 + 3×4/5}\) = \(\frac{16/4 }{24/5}\) = \(\frac{5 }{6}\) |
