Question

${\sin }^{2}30°{\cos }^{2}45°+4{\tan }^{2}30°+\frac{1}{2}{\sin }^{2}90°+\frac{1}{8}{\cot }^{2}60°$

Answer 1

As we know that,
​$$\begin{gathered} \sin 30°=\frac{1}{2} \\ \cos 45°=\frac{1}{\sqrt{2}} \\ \tan 30°=\frac{1}{\sqrt{3}} \\ \sin 90°=1 \\ \cot 60°=\frac{1}{\sqrt{3}} \\ \mathrm{By}\mathrm{substituting}\mathrm{these}\mathrm{values},\mathrm{we}\mathrm{get} \\ {\sin }^{2}30°{\cos }^{2}45°+4{\tan }^{2}30°+\frac{1}{2}{\sin }^{2}90°+\frac{1}{8}{\cot }^{2}60°={\left(\frac{1}{2}\right)}^2\times {\left(\frac{1}{\sqrt{2}}\right)}^2+4{\left(\frac{1}{\sqrt{3}}\right)}^2+\frac{1}{2}{\left(1\right)}^2+\frac{1}{8}{\left(\frac{1}{\sqrt{3}}\right)}^2 \\ =\left(\frac{1}{4}\right)\times \left(\frac{1}{2}\right)+4\left(\frac{1}{3}\right)+\frac{1}{2}\left(1\right)+\frac{1}{8}\left(\frac{1}{3}\right) \\ =\frac{1}{8}+\frac{4}{3}+\frac{1}{2}+\frac{1}{24} \\ =\frac{3+32+12+1}{24} \\ =\frac{48}{24} \\ =2 \\ \mathrm{Hence},{\sin }^{2}30°{\cos }^{2}45°+4{\tan }^{2}30°+\frac{1}{2}{\sin }^{2}90°+\frac{1}{8}{\cot }^{2}60°=2. \end{gathered}$$