Mathematics - Trigonometric Equation Question with Solution | TestHub

$\begin{array}{rl}& 10{\left({\sin }^2\theta \right)}^2+15{\left(1-{\sin }^2\theta \right)}^2=6\\ & \text{ Let }{\sin }^2\theta =\mathrm{t}\Rightarrow 10{\mathrm{t}}^2+15(1-\mathrm{t}{)}^2=16\\ & 10{\mathrm{t}}^2+15-30\mathrm{t}+15{\mathrm{t}}^2=6\\ & 25{\mathrm{t}}^2-30\mathrm{t}+9=0\\ & (5\mathrm{t}-3{)}^2=0\\ & {\sin }^2\theta =\frac{3}{5}\text{ and }{\cos }^2\theta =\frac{2}{5}\\ & \frac{27\times \frac{125}{27}+8+\frac{125}{8}}{16{\left(\frac{5}{2}\right)}^4}=\frac{250}{125\times 5}=\frac{2}{5}\end{array}$