Find the value of \(\frac{sin^230°\;.\;cos^245°\;+\;4 sin^260°\;+\;\frac{1}{2} sin^290°}{tan^260°\;+\;cos60°}\).
Answer & explanation
Correct answer: option 3
⇒ \(\frac{(\frac{1}{2})^2\;×\;(\frac{1}{\sqrt {2}})^2\;+\;4\;×\;(\frac{\sqrt {3}}{2})^2\;+\;\frac{1}{2}\;×\;1}{(\sqrt {3})^2\;+\;\frac{1}{2}}\)
⇒ \(\frac{\frac{1}{4}\;×\;\frac{1}{2}\;+\;4\;×\;\frac{3}{4}\;+\;\frac{1}{2}}{3\;+\; \frac{1}{2}}\)
= \(\frac{\frac{1}{8}\;+\;3\;+\;\frac{1}{2}}{\frac{7}{2}}\)
= \(\frac{29}{8}\) × \(\frac{2}{7}\) = \(\frac{29}{28}\)