Question

2(cos⁴ 60$°$ + sin⁴ 30$°$) $-$ (tan² 60$°$ + cot² 45$°$) + 3sec² 30$°$ = $\frac{1}{4}$

Answer 1

$$\begin{gathered} \mathrm{LHS}:2\left(co{s}^{4}60°+si{n}^{4}30°\right)-\left(ta{n}^{2}60°+co{t}^{2}45°\right)+3se{c}^{2}30° \\ =\left[{\left(\frac{1}{2}\right)}^4+{\left(\frac{1}{2}\right)}^4\right]-\left[{\left(\sqrt{3}\right)}^2+1^2\right]+3{\left(\frac{2}{\sqrt{3}}\right)}^2(\mathrm{From}\mathrm{the}\mathrm{table}) \\ =2\left(\frac{1}{16}+\frac{1}{16}\right)-\left(3+1\right)+3\left({\displaystyle \frac{4}{3}}\right) \\ =2\left(\frac{2}{16}\right)-4+4 \\ =\frac{1}{4}+0 \\ =\frac{1}{4}=\mathrm{RHS} \\ \mathrm{Hence},\mathrm{proved}. \end{gathered}$$