Question

Evaluate:

$\left(i\right)si{n}^2{30}^{\circ }.co{s}^2{45}^{\circ }+4ta{n}^2{30}^{\circ }+\frac{1}{2}si{n}^2{90}^{\circ }+\frac{1}{8}co{t}^2{60}^{\circ }$

$\left(ii\right)\frac{ta{n}^2{60}^{\circ }+4si{n}^2{45}^{\circ }+3se{c}^2{30}^{\circ }+5co{s}^2{90}^{\circ }}{cosec{30}^{\circ }+sec{60}^{\circ }-co{t}^2{30}^{\circ }}$ [4 MARKS]

Answer 1

Solution: 2 Marks each

$\left(i\right)si{n}^2{30}^{\circ }.co{s}^2{45}^{\circ }+4ta{n}^2{30}^{\circ }+\frac{1}{2}si{n}^2{90}^{\circ }+\frac{1}{8}co{t}^2{60}^{\circ }$

$={\left(\frac{1}{2}\right)}^2.{\left(\frac{1}{\sqrt{2}}\right)}^2+4\times {\left(\frac{1}{\sqrt{3}}\right)}^2+\left(\frac{1}{2}\right)\times (1{)}^2+\frac{1}{8}\times {\left(\frac{1}{\sqrt{3}}\right)}^2$

$=(\frac{1}{4}\times \frac{1}{2}+4\times \frac{1}{3}+\frac{1}{2}\times 1+\frac{1}{8}\times \frac{1}{3})$

$=(\frac{1}{8}+\frac{4}{3}+\frac{1}{2}+\frac{1}{24})=\frac{48}{24}=2$


$\left(ii\right)\frac{ta{n}^2{60}^{\circ }+4si{n}^2{45}^{\circ }+3se{c}^2{30}^{\circ }+5co{s}^2{90}^{\circ }}{cosec{30}^{\circ }+sec{60}^{\circ }-co{t}^2{30}^{\circ }}$

$=\frac{(\sqrt{3}{)}^2+4\times {\left(\frac{1}{\sqrt{2}}\right)}^2+3\times {\left(\frac{2}{\sqrt{3}}\right)}^2+5\times 0^2}{2+2-(\sqrt{3}{)}^2}$

$=\left(\frac{3+4\times \frac{1}{2}+3\times \frac{4}{3}+5\times 0}{4-3}\right)=\left(\frac{3+2+4+0}{1}\right)=9$