3 Marks Question

Question 13 Marks

If $\text{a}\cos\theta+\text{b}\sin\theta=\text{m}$ and $\text{a}\sin\theta-\text{b}\cos\theta=\text{n},$ then show that $\text{a}^2+\text{b}^2=\text{m}^2+\text{n}^2.$

Answer

Given that: $\text{a}\cos\theta+\text{b}\sin\theta=\text{m}$
and $\text{a}\sin\theta-\text{b}\cos\theta=\text{n}$
$\text{R.H.S.}=\text{m}^2+\text{n}^2=(\text{a}\cos\theta+\text{b}\sin\theta)^2+(\text{a}\sin\theta-\text{b}\cos\theta)^2$
$=\text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta+2\text{ab}\sin\theta\cos\theta\\+\text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta-2\text{ab}\sin\theta\cos\theta$
$=\text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta+\text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta$
$=\text{a}^2(\cos^2\theta+\sin^2\theta)+\text{b}^2(\sin^2\theta+\cos^2\theta)$
$=\text{a}^2.1+\text{b}^2.1=\text{a}^2+\text{b}^2\text{ L.H.S.}$
$\text{L.H.S. = R.H.S.}$ Hence proved.

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Question 23 Marks

Prove that $\frac{\tan\text{A}+\sec\text{A}-1}{\tan\text{A}-\sec\text{A}+1}=\frac{1+\sin\text{A}}{\cos\text{A}}$

Answer

$\text{L.H.S.}=\frac{\tan\text{A}+\sec\text{A}-1}{\tan\text{A}-\sec\text{A}+1}$
$=\frac{\tan\text{A}+\sec\text{A}-(\sec^2\text{A}-\tan^2\text{A})}{(\tan\text{A}-\sec\text{A}+1)}$ $[\because\sec^2\text{A}-\tan^2\text{A}=1]$
$=\frac{(\tan\text{A}+\sec\text{A})-(\sec\text{A}+\tan\text{A})(\sec\text{A}-\tan\text{A})}{(1-\sec\text{A}+\tan\text{A})}$
$=\frac{(\sec\text{A}+\tan\text{A})(1-\sec\text{A}+\tan\text{A})}{1-\sec\text{A}+\tan\text{A}}$
$=\sec\text{A}+\tan\text{A}=\frac{1}{\cos\text{A}}+\frac{\sin\text{A}}{\cos\text{A}}$
$=\frac{1+\sin\text{A}}{\cos\text{A}}=\text{R.H.S.}$

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Question 33 Marks

Prove that $\cos\theta\cos\frac{\theta}{2}-\cos3\theta\cos\frac{9\theta}{2}=\sin7\theta\sin8\theta$
$\Big[$Hint: Express $\text{L.H.S.}=\frac{1}{2}\Big[2\cos\theta\cos\frac{\theta}{2}-2\cos3\theta\cos\frac{9\theta}{2}\Big]\Big]$

Answer

$\text{L.H.S.}=\cos\theta\cos\frac{\theta}{2}-\cos3\theta.\cos\frac{9\theta}{2}$
$=\frac{1}2\Big[2\cos\theta\cos\frac{\theta}{2}-2\cos3\theta.\cos\frac{9\theta}{2}\Big]$
$=\frac{1}2\Big(\cos\frac{3\theta}{2}+\cos\frac{\theta}{2}-\cos\frac{15\theta}{2}-\cos\frac{3\theta}{2}\Big)$
$=\frac{1}2\Big[\cos\frac{\theta}{2}-\cos\frac{15\theta}{2}\Big]=\frac{1}2\Big[2\sin\Big(\frac{\theta+15\theta}{4}\Big).\sin\Big(\frac{15\theta-\theta}{4}\Big)\Big]$
$=\sin4\theta.\sin\frac{7\theta}{2}=\text{R.H.S.}$

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Question 43 Marks

If $\frac{\sin(\text{x+y})}{\sin(\text{x}-\text{y})}=\frac{\text{a+b}}{\text{a}-\text{b}},$ then show that $\frac{\tan\text{x}}{\tan\text{y}}=\frac{\text{a}}{\text{b}}.$
[Hint: Use Componendo and Dividendo]

Answer

Given that ,$\frac{\sin(\text{x}+\text{y})}{\sin(\text{x}-\text{y})}=\frac{\text{a}+\text{b}}{\text{a}-\text{b}}$
Using componendo and dividendo , we get
$\Rightarrow\frac{\sin(\text{x}+\text{y})+\sin(\text{x}-\text{y})}{\sin(\text{x}+\text{y})-\sin(\text{x}-\text{y})}=\frac{\text{a}+\text{b}+\text{a}-\text{b}}{\text{a}+\text{b}-\text{a}+\text{b}}$
$\Rightarrow\frac{2\sin\Big(\frac{\text{x}+\text{y}+\text{x}-\text{y}}{2}\Big)\cdot\cos\Big(\frac{\text{x}+\text{y}-\text{x}+\text{y}}{2}\Big)}{2\cos\Big(\frac{\text{x}+\text{y}+\text{x}-\text{y}}{2}\Big)\cdot\sin\Big(\frac{\text{x}+\text{y}-\text{x}+\text{y}}{2}\Big)}=\frac{2\text{a}}{2\text{b}}$
$\Rightarrow\frac{\sin\text{x}\cdot\cos\text{y}}{\cos\text{x}\cdot\sin\text{y}}=\frac{\text{a}}{\text{b}}\Rightarrow\frac{\tan\text{x}}{\tan\text{y}}=\frac{\text{a}}{\text{b}}$

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Question 53 Marks

Find the general solution of the equation $\sin\text{x}-3\sin2\text{x}+\sin3\text{x}=\cos\text{x}-3\cos2\text{x}+\cos3\text{x}$

Answer

We have, $(\sin\text{x}+\sin3\text{x})-3\sin2\text{x}=(\cos\text{x}+\cos3\text{x})-3\cos2\text{x}$
$\Rightarrow2\sin2\text{x}\cos\text{x}-3\sin2\text{x}=2\cos2\text{x}.\cos\text{x}-3\cos2\text{x}$
$\Rightarrow\sin2\text{x}(2\cos\text{x}-3)=\cos2\text{x}(2\cos\text{x}-3)$
$\Rightarrow\sin2\text{x}=\cos2\text{x}\Big(\text{As}\cos\text{x}\neq\frac{3}{2}\Big)$
$\Rightarrow\tan2\text{x}=1\Rightarrow\tan2\text{x}=\tan\frac{\pi}{4}$
$\Rightarrow2\text{x}=\text{n}\pi+\frac{\pi}{4},\text{n}\in\text{Z}$
$\text{x}=\text{n}\frac{\pi}{2}+\frac{\pi}{8},\text{n}\in\text{Z}$

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