If $\frac{5 cot θ + \sqrt{3} cosec θ}{2\sqrt{3}cosecθ+3cotθ} $= 1, 0° < θ < 90°, then the value of $\frac{\frac{7}{2}cot^2θ-\frac{3}{4}cosec^2θ}{4sin^2θ+\frac{3}{2}tan^2θ}$ will be :

5

\(\frac{ 5cotθ + √3.cosecθ}{2√3.cosecθ + 3cotθ}\) = 1

5cotθ + √3.cosecθ = 2√3.cosecθ + 3cotθ

2cotθ = 3.cosecθ 

cosθ  = \(\frac{√3}{2}\)

{ we know,  cos30º = \(\frac{√3}{2}\) }

so, θ = 30º

Now,

\(\frac{ 7/2cot²θ - 3/4.cosec²θ}{4sin²θ + 3/2tan²θ}\)

= \(\frac{ 7/2 ×cot²30º - 3/4×cosec²30º}{4×sin²30º + 3/2×tan²30º}\)

= \(\frac{ 7/2 × 3 - 3/4×4}{4× 1/4 + 3/2×1/3}\)

= \(\frac{ 21/2 - 3}{1 + 1/2}\)

= 5