1 Marks Question

Question 11 Mark

Evaluate the following:
$\big(\sin^230^\circ+4\cot^230-\frac34\sec^260^\circ\big)\big(\text{cosec}^245^\circ\sec^230^\circ\big)$

Answer

On substituting values of various T-ratious, we get:
$\big(\sin^230^\circ+4\cot^230-\frac34\sec^260^\circ\big)\\\big(\text{cosec}^245^\circ\sec^230^\circ\big)$
$=\Big[\Big(\frac12\Big)^2+4\times(1)^2-(2)^2\Big]\bigg[\Big(\sqrt{2}\Big)^2\Big(\frac{2}{\sqrt{3}}\Big)^2\bigg]$
$=\Big(\frac14+4-4\Big)\Big(2\times\frac43\Big)$
$=\frac14\times\frac83=\frac23$

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Question 21 Mark

Evaluate the following:
Show that:
$\frac{1-\sin60^\circ}{\cos60^\circ}=\frac{\tan60^\circ-1}{\tan60^\circ+1}$

Answer

$\text{L.H.S.}=\frac{1-\sin60^\circ}{\cos60^\circ}=\frac{\frac{\sqrt{3}}{2}}{\frac12}$
$=\frac{\Big(\frac{2-\sqrt{3}}{2}\Big)}{\frac{1}{2}}=\Big(\frac{2-\sqrt{3}}{2}\Big)\times2=2-\sqrt{3}$
$\text{R.H.S.}=\frac{\tan60^\circ-1}{\tan60^\circ+1}=\frac{\sqrt{3}-1}{\sqrt{3}+1}$
$=\frac{\sqrt{3}-1}{\sqrt{3}+1}\times\frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{\big(\sqrt{3}-1\big)^2}{\big(\sqrt{3}\big)^2-1^2}$
$=\frac{3+1-2\sqrt{3}}{3-1}=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}$
Hence, L.H.S. = R.H.S.

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Question 31 Mark

Evaluate the following:
$\sin60^\circ\cos30^\circ+\cos60^\circ\sin30^\circ$

Answer

On substituting the values of varios T-ratios, we get:
$\sin60^\circ\cos30^\circ+\cos60^\circ\sin30^\circ$
$=\Big(\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{2}+\frac12\times\frac12\Big)$
$=\Big(\frac34+\frac14\Big)=\frac44=1$

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Question 41 Mark

Evaluate the following:
$\cos60^\circ\cos30^\circ+\sin60^\circ\sin30^\circ$

Answer

On substituting the values of varios T-ratios, we get:
$\cos60^\circ\cos30^\circ+\sin60^\circ\sin30^\circ$
$=\Big(\frac{1}{2}\times\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}\times\frac12\Big)$
$=\Big(\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}\Big)=0$

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Question 51 Mark

Evaluate the following:
$\cos45^\circ\cos30^\circ+\sin45^\circ\sin30^\circ$

Answer

On substituting the values of varios T-ratios, we get:
$\cos45^\circ\cos30^\circ+\sin45^\circ\sin30^\circ$
$=\Big(\frac{1}{\sqrt{2}}\times\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{3}}\times\frac12\Big)$
$=\Big(\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}\Big)=\Big(\frac{\sqrt{3}+1}{2\sqrt{2}}\Big)$

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