If 0° < θ < 90° find \(\frac{{sin}^{2}52+2+sin^238}{4sin^247+5+4cos^243}\)

\(\frac{2}{3}\) \(\frac{1}{3}\) -3 -\(\frac{1}{3}\)

\(\frac{1}{3}\)

If sum of θ = 90° Put then equal Put θ = 45° ⇒ \(\frac{\frac{1}{2}+2+\frac{1}{2}}{\frac{4}{2}+5+\frac{4}{2}}\)=\(\frac{3}{9}\) = \(\frac{1}{3}\)