Question

(3cos²60° + 2cot²30° – 5sin²45°) = ?
(a) 1

(b) 4

(c) $\frac{17}{4}$

(d) $\frac{13}{6}$

Answer 1

As we know that,
$$\begin{gathered} \sin 45°=\frac{1}{\sqrt{2}} \\ \cos 60°=\frac{1}{2} \\ \cot 30°=\sqrt{3} \\ \mathrm{By}\mathrm{substituting}\mathrm{these}\mathrm{values},\mathrm{we}\mathrm{get} \\ \left(3{\cos }^{2}60°+2{\cot }^{2}30°-5{\sin }^{2}45°\right)=3{\left(\frac{1}{2}\right)}^2+2{\left(\sqrt{3}\right)}^2-5{\left(\frac{1}{\sqrt{2}}\right)}^2 \\ =3\left(\frac{1}{4}\right)+2\left(3\right)-5\left(\frac{1}{2}\right) \\ =\frac{3}{4}+6-\frac{5}{2} \\ =\frac{3+24-10}{4} \\ =\frac{17}{4} \\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\left(\mathrm{c}\right). \end{gathered}$$