It $cot θ=\frac{3}{\sqrt{5}}, 0° < θ > 90°,$ then the value of $\frac{6sec^θ-\frac{5}{3}cosec^2θ}{\frac{3}{5}sec^2θ+\frac{4}{3}cosec^2θ}$, is equal to :
Answer & explanation
Correct answer: option 2
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