Question

Evaluate the following
$\left(i\right)\frac{1+{\tan }^2{30}^{\circ }}{1-{\tan }^2{30}^{\circ }}+cose{c}^2{60}^{\circ }-{\cos }^2{45}^{\circ }+{\sin }^2{45}^{\circ }+\frac{1+{\cot }^2{60}^{\circ }}{1-{\cot }^2{60}^{\circ }}$
$\left(ii\right)4({\sin }^4{30}^{\circ }+{\cos }^4{60}^{\circ })-3({\cos }^2{45}^{\circ }-{\sin }^2{90}^{\circ })$

• A. $\left(i\right)\frac{16}{3}\left(ii\right)4$
• B. $\left(i\right)\frac{19}{3}\left(ii\right)7$
• C. $\left(i\right)\frac{16}{7}\left(ii\right)11$
• D. $\left(i\right)\frac{17}{5}\left(ii\right)5$

Answer 1

The correct option is A $\left(i\right)\frac{16}{3}\left(ii\right)4$

$\frac{1+ta{n}^2{30}^0}{1-ta{n}^2{30}^0}+cose{c}^2{60}^0-co{s}^2{45}^0+si{n}^2{45}^0+\frac{1+co{t}^2{60}^0}{1-co{t}^2{60}^0}$
$=\frac{3+1}{3-1}+\frac{4}{3}-\frac{1}{2}+\frac{1}{2}+\frac{3+1}{3-1}$
$=2\left(2\right)+\frac{4}{3}$
$=\frac{16}{3}$
$4(si{n}^4{30}^0+co{s}^4{30}^0)-3(co{s}^2{45}^0-si{n}^2{90}^0)$
$=4\left(\frac{1+9}{16}\right)-3(\frac{1}{2}-1)$
$=\frac{10}{4}+\frac{3}{2}$
$=\frac{8}{2}$
$=4$