Question
Evaluate the following:
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^230^\circ}-2\cos^245^\circ-\sin^20^\circ$
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^230^\circ}-2\cos^245^\circ-\sin^20^\circ$
✓
Answer
On substituting values of various T-ratious, we get:
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^230^\circ}-2\cos^245^\circ-\sin^20^\circ$
$=\frac{4}{\big(\sqrt{3}\big)^2}+\frac{1}{\big(\frac12\big)^2}-2\times\Big(\frac{1}{\sqrt{2}}\Big)^2-(0)^2$
$=\frac43+\frac11-2\times\frac12-0$
$=\frac43+4-1$
$=\frac43+3=\frac{4+9}{3}=\frac{13}{3}$
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^230^\circ}-2\cos^245^\circ-\sin^20^\circ$
$=\frac{4}{\big(\sqrt{3}\big)^2}+\frac{1}{\big(\frac12\big)^2}-2\times\Big(\frac{1}{\sqrt{2}}\Big)^2-(0)^2$
$=\frac43+\frac11-2\times\frac12-0$
$=\frac43+4-1$
$=\frac43+3=\frac{4+9}{3}=\frac{13}{3}$
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