3 Marks Question

Question 13 Marks

If $15\cot\text{A}=8,$ find the value of $\sin\text{A}$ and $\sec\text{A}.$

Answer

$15\cot\text{A}=8\Rightarrow\cot\text{A}=\frac{8}{15}$ Cosider $\triangle\text{ABC},$ where $\angle\text{B}=90^\circ$ Then, $\cot\text{A}=\frac{\text{Base}}{\text{Perpendicular}}=\frac{\text{AB}}{\text{BC}}=\frac{8}{15}$ Let AB = 8 and BC = 15 Then, by Pythagoras theoram, $\text{AC}^2 = (\text{AB})^2 + (\text{BC})^2 $ $\Rightarrow(\text{AB})^2 = (\text{AC})^2 - (\text{BC})^2 $ $= 82 + 152 = 64 + 225 = 289$ $\Rightarrow \text{AC} = 17$ Now, $\sin\text{A}=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{15}{17}$ $\sin\text{A}=\frac{\text{Hypotenuse}}{\text{Base}}=\frac{\text{AC}}{\text{AB}}=\frac{17}{8}$

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Question 23 Marks

If $\sin\theta=\frac{\sqrt{3}}{2},$ find the value of the all T-ratios of $\theta.$

Answer

Given: $\sin\theta=\frac{\sqrt{3}}{2}$
Let us draw a $\triangle\text{ABC}$ in which $\angle\text{B}=90^\circ$ and $\angle\text{BAC}=\theta$

Then, $\sin\theta=\frac{\text{BC}}{\text{AC}}=\frac{\sqrt{3}}{2}$
Let $\text{BC}=\sqrt{3}\text{k}$
And AC = 2k,
Where k is Positive
By pythagoras theorem, we have
$\text{AC}^2 = \text{AB}^2 + \text{BC}^2$
$\Rightarrow\text{AB}^2 = \text{AC}^2 - \text{BC}^2$
$\text{AB}^2=\Big[(2\text{k})^2=\big(\sqrt{3}\text{k}\big)^2\Big]$
$=\big(4\text{k}^2-3\text{k}^2\big)$
$\Rightarrow\text{AB}=\sqrt{\text{k}^2}=\text{k}$
$\therefore\sin\theta=\frac{\text{BC}}{\text{AC}}=\frac{\sqrt{3}\text{k}}{2\text{k}}=\frac{\sqrt{3}}{2},$
$\cos\theta=\frac{\text{AB}}{\text{AC}}=\frac{\text{k}}{2\text{k}}=\frac12$
$\tan\theta=\frac{\sin\theta}{\cos\theta}=\Big(\frac{\sqrt{3}}{2}\times\frac21\Big)=\sqrt{3};$
$\text{cosec}\theta=\frac{1}{\sin\theta}=\frac{2}{\sqrt{3}},$
$\sec\theta=\frac{1}{\cos\theta}=\frac21=2$ and $\cot \theta=\frac{1}{\tan\theta}=\frac{1}{\sqrt{3}}$

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Question 33 Marks

If $\sin\theta=\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2},$ find the value of the all T-ratios of $\theta.$

Answer

Consider $\triangle\text{ABC}$ where $\angle\text{B}=90^\circ,\angle\text{A}=\theta.$
Then, $\sin\theta=\frac{\text{Perpendicural}}{\text{Hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2}$
Let $BC = a^2- b^2$ and $AC = a^2 + b^2$
Then, by Pythagoras theoram,
$\text{AC}^2 = (\text{AB})^2 + (\text{BC})^2$
$\Rightarrow(\text{AB})^2 = (\text{AC})^2 - (\text{BC})^2 $
$=(\text{a}^2 + \text{b}^2)^2 - (\text{a}^2 - \text{b}^2)^2$
$ = \text{a}^4 + \text{b}^4 + 2\text{a}^2\text{b}^2 = 4\text{a}^2\text{b}^2$
$\Rightarrow\text{AB} = 2\text{ab}$
Now,
$\cos\theta=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{2\text{ab}}{\text{a}^2+\text{b}^2}$
$\cot\theta=\frac{1}{\tan\theta}=\frac{2\text{ab}}{\text{a}^2-\text{b}^2}$
$\text{cosec}\theta=\frac{1}{\sin\theta}=\frac{\text{a}^2+\text{b}^2}{\text{a}^2-\text{b}^2}$
$\sec\theta=\frac{1}{\cos\theta}=\frac{\text{a}^2+\text{b}^2}{2\text{ab}}$

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Question 43 Marks

If $\text{cosec}\theta=\sqrt{10},$ find the value of the all T-ratios of $\theta.$

Answer

Given: $\text{cosec}\theta=\frac{\text{AC}}{\text{BC}}=\frac{\sqrt{10}}{1}$
Let $\text{AC}=\sqrt{10}\text{k}$
And BC = 1k,
Where k is positive.
Let us draw a $\triangle\text{ABC}$ in which $\angle\text{B}=90^\circ$ and $\angle\text{BAC}=\theta$

By pythagoras theorem, we have
$\text{AC}^2 = (\text{AB})^2 + (\text{BC})^2 $
$⇒ (\text{AB})^2 = (\text{AC})^2 - (\text{BC})^2$
$=\Big[(10\text{k})^2+(\text{k})^2\Big]=\big(10\text{k}^2-1\text{k}^2\big)$
$\Rightarrow(\text{AB})^2 = 9\text{k}^2$
$\Rightarrow\text{AB}=\sqrt{9}\text{k}^2=3\text{k}$
$\therefore\sin\theta=\frac{\text{BC}}{\text{AC}}=\frac{1}{\sqrt{10}}$
$\cos\theta=\frac{\text{AB}}{\text{AC}}=\frac{3\text{k}}{\sqrt{10}\text{k}}=\frac{3}{\sqrt{10}}$
$\text{cosec}\theta=\sqrt{10}$ (Given)
$\sec\theta=\frac{1}{\cos\theta}=\frac{\sqrt{10}}{3}$
$\tan\theta=\frac{\sin\theta}{\cos\theta}=\Big(\frac{1}{\sqrt{10}}\times\frac{\sqrt{10}}{3}\Big)=\frac13$
$\cot\theta=\frac{1}{\tan\theta}=3$

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Question 53 Marks

If $\text{cosec}\theta=2,$ show that $\Big(\cot\theta+\frac{\sin\theta}{1+\cos\theta}\Big)=2.$

Answer

Given: $\text{cosec}\theta=\frac{\text{AC}}{\text{BC}}=\frac21$
Let BC = 1k and AC = 2k Where k is positive Let us draw a
$\triangle\text{ABC}$ in which
$\angle\text{B}=90^\circ$ and $\angle\text{A}=\theta$

By Pythagoras theorem, we have
$(AC)^2 = (AB)^2 + (BC)^2$
$\Rightarrow (AB)^2 = (AC)^2 - (BC)^2$
$=\Big[(2\text{k})^2-(1\text{k})^2\Big]=\big(4\text{k}^2-1\text{k}^2\big)=3\text{k}^2$
$\Rightarrow(\text{AB})=\sqrt{3}\text{k}$
$\sin\theta=\frac{\text{BC}}{\text{AC}}=\frac{1\text{k}}{2\text{k}}=\frac{1}{2}$
$\cos\theta=\frac{\text{AB}}{\text{AC}}=\frac{\sqrt{3}\text{k}}{2\text{k}}=\frac{\sqrt{3}}{2}$
$\cot\theta=\frac{\cos\theta}{\sin\theta}=\Big(\frac{\sqrt{3}}{2}\times\frac21\Big)=\sqrt{3}$
$\Rightarrow\Big[\cot\theta+\frac{\sin\theta}{1+\cos\theta}\Big]\Bigg[\sqrt{3}+\frac{\frac12}{1+\frac{\sqrt{3}}{2}}\Bigg]$
$=\Big(\sqrt{3}+\frac{1}{2+\sqrt{3}}\Big)=\Big(\frac{2\sqrt{3}+3+1}{2+\sqrt{3}}\Big)$
$=\Big(\frac{2\sqrt{3}+4}{2+\sqrt{3}}\Big)=2\Big(\frac{\sqrt{3}+2}{2+\sqrt{3}}\Big)=2$
Hence, $\Big[\cot\theta+\frac{\sin\theta}{1+\cos\theta}\Big]=2$

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Question 63 Marks

If $\cos\theta=0.6,$ show that $(5\sin\theta-3\tan\theta)=0.$

Answer

Given: $\cos\theta=0.6=\frac{6}{10}=\frac{3}{5}$
Let us draw a $\triangle\text{ABC}$ in
which $\angle\text{B}=90^\circ$ and $\angle\text{A}=\theta$

Then, $\cos\theta=\frac{\text{AB}}{\text{AC}}=\frac35$
Let AB = 3k
and AC = 5k
Where k is positive
By Pythagoras theorem, we have
$(AC)^2 = (AB)^2 + (BC)^2$
$\Rightarrow (BC)^2 = (AC)^2 - (AB)^2$
$= [(5k)^2 - (3k)^2] = 16k^2$
$= 16k^2$
$\Rightarrow BC = 4k$
$\sin\theta=\frac{\text{AB}}{\text{AC}}=\frac{4\text{k}}{5\text{k}}=\frac{4}{5}$
$\cos\theta=\frac{\sin\theta}{\cos\theta}=\Big(\frac45\times\frac53\Big)=\frac{4}{3}$
$\Rightarrow(5\sin\theta-3\tan\theta)=\Big(5\times\frac{4}{5}-3\times\frac43\Big)=0$
Hence, $(5\sin\theta-3\tan\theta)=0$

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Question 73 Marks

If $\sec\theta=\frac{17}{8},$ verify that $\frac{\big(3-4\sin^2\theta\big)}{\big(4\cos^2\theta-3\big)}=\frac{\big(3-\tan^2\theta\big)}{\big(1-3\tan^2\theta\big)}.$

Answer

Given: $\sec\theta=\frac{17}{8}=\frac{17\text{k}}{8\text{k}}$ Let us draw a $\triangle\text{ABC},$ in which $\angle\text{B}=90^\circ$ and $\angle\text{A}=\theta$

By pythagoras theoram we have $\text{AC}^2 = \text{AB}^2 + \text{BC}^2$ Or $\text{BC}^2 = \text{AC}^2 + \text{AB}^2$ $\therefore\text{BC}^2=\big(17\text{k}\big)^2-\big(8\text{k}\big)^2$ $=289\text{k}^2-64\text{k}^2=225\text{k}^2$ $=\text{BC}=15\text{k}$ $\therefore\ \sin\theta=\frac{\text{AC}}{\text{BC}}=\frac{15{\text{k}}}{17\text{k}}=\frac{15}{17}$ $\cos\theta=\frac{\text{AB}}{\text{AC}}=\frac{8\text{k}}{17\text{k}}=\frac{8}{17}$ $\tan\theta=\frac{15\text{k}}{8\text{k}}=\frac{15}{8}$ $\text{L.H.S.}=\frac{3-4\sin^2\theta}{4\cos^2\theta-3}$ $=\frac{3-4\times\big(\frac{15}{17}\big)^2}{4\times\big(\frac{8}{17}\big)^2-3}$ $=\frac{3-\frac{4\times225}{289}}{4\times\frac{64}{289}-3}$ $=\frac{\frac{3\times289-4\times225}{289}}{\frac{4\times64-3\times289}{289}}$ $=\frac{867-900}{256-867}=\frac{-33}{-611}=\frac{33}{611}$ $\text{R.H.S.}=\frac{3-\tan^2\theta}{1-3\tan^2\theta}=\frac{3-\big(\frac{15}{8}\big)^2}{1-3\times\big(\frac{15}{8}\big)^2}$ $=\frac{3-\frac{225}{64}}{1-3\times\frac{225}{64}}=\frac{\frac{3\times64-225}{64}}{\frac{64-3\times225}{64}}$ $=\frac{192-225}{64-675}=\frac{-33}{-611}=\frac{33}{611}$ Hence, L.H.S = R.H.S.

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