Question
Evaluate the following:
$\big(\text{cosesc}^245^\circ\sec^230^\circ\big)\big(\sin^230^\circ+4\cot^245^\circ-\sec^260^\circ\big)$
$\big(\text{cosesc}^245^\circ\sec^230^\circ\big)\big(\sin^230^\circ+4\cot^245^\circ-\sec^260^\circ\big)$
✓
Answer
$\big(\text{cosesc}^245^\circ\sec^230^\circ\big)\big(\sin^230^\circ+4\cot^245^\circ-\sec^260^\circ\big)\dots(\text{i})$
By trigonometric ration we have
$\text{cosec}45^\circ=\sqrt{2},\ \sec30^\circ=\frac{2}{\sqrt{3}},$ $\sin30^\circ=\frac{1}{2}\ \cot45^\circ=1\ \sec60^\circ=2$
By substituting above values in (i), we get
$\bigg[\Big(\sqrt{2}\Big)^2\cdot\Big(\frac{2}{\sqrt{3}}\Big)^2\bigg]\bigg[\Big(\frac{1}{2}\Big)^2+4(1)^2-(2)^2\bigg]$
$\Rightarrow\Big[2\cdot\frac{4}{3}\Big]\Big[\frac{1}{4}+4-4\Big]\Rightarrow2\cdot\frac{4}{3}\cdot\frac{1}{4}=\frac{2}{3}$
By trigonometric ration we have
$\text{cosec}45^\circ=\sqrt{2},\ \sec30^\circ=\frac{2}{\sqrt{3}},$ $\sin30^\circ=\frac{1}{2}\ \cot45^\circ=1\ \sec60^\circ=2$
By substituting above values in (i), we get
$\bigg[\Big(\sqrt{2}\Big)^2\cdot\Big(\frac{2}{\sqrt{3}}\Big)^2\bigg]\bigg[\Big(\frac{1}{2}\Big)^2+4(1)^2-(2)^2\bigg]$
$\Rightarrow\Big[2\cdot\frac{4}{3}\Big]\Big[\frac{1}{4}+4-4\Big]\Rightarrow2\cdot\frac{4}{3}\cdot\frac{1}{4}=\frac{2}{3}$
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