Question 15 Marks
If $\theta$ lies in the first quadrant and $\cos\theta=\frac{8}{17},$ then find the value of $\cos(30^\circ+\theta)+\cos(45^\circ-\theta)+\cos(120^\circ-\theta).$
AnswerGiven that: $\cos\theta=\frac{8}{17}$
$\therefore\sin\theta=\sqrt{1-\Big(\frac{8}{17}\Big)^2}=\sqrt{1-\frac{64}{289}}=\sqrt{\frac{289-64}{289}}=\sqrt{\frac{225}{289}}=\pm\frac{15}{17}$
But $\theta$ lies in I quadrant ans so $\sin\theta$ is positive
$\therefore\sin\theta=\frac{15}{17}$
Now $\cos(30^\circ+\theta)+\cos(45^\circ-\theta)+\cos(120^\circ-\theta)$
$=\cos30^\circ\cos\theta-\sin30^\circ\sin\theta+\cos45^\circ\cos\theta\\+\sin45^\circ\sin\theta+\cos120^\circ\cos\theta+\sin120^\circ\sin\theta$
$=\frac{\sqrt{3}}{2}\cos\theta-\frac{1}{2}\sin\theta+\frac{1}{\sqrt{2}}\cos\theta+\frac{1}{\sqrt{2}}\sin\theta-\frac{1}{2}\cos\theta+\frac{\sqrt{3}}2{}\sin\theta$
$=\frac{\sqrt{3}}2{}(\cos\theta+\sin\theta)-\frac{1}{2}(\sin\theta+\cos\theta)+\frac{1}{\sqrt{2}}(\cos\theta+\sin\theta)$
$=\Big(\frac{\sqrt{3}}{2}-\frac{1}{2}+\frac{1}{\sqrt{2}}\Big)(\cos\theta+\sin\theta)=\Big(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\Big)\Big(\frac{8}{17}+\frac{15}{17}\Big)$
$=\Big(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\Big)\Big(\frac{23}{17}\Big)$
Hence, the required solution $=\frac{23}{17}\Big(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\Big)=\frac{23}{17}\Big(\frac{\sqrt{3}-1+\sqrt{2}}{2}\Big)=\frac{23}{34}(\sqrt{3}-1+\sqrt{2})$
View full question & answer→Question 25 Marks
Find the value of the expression $\cos^4\frac{\pi}{8}+\cos^4\frac{3\pi}{8}+\cos^4\frac{5\pi}{8}+\cos^4\frac{7\pi}{8}$
[Hint: Simplify the expression to $2\Big(\cos^4\frac{\pi}{8}+\cos^4\frac{3\pi}{8}\Big)=2\Big[\Big(\cos^2\frac{\pi}{8}+\cos^2\frac{3\pi}8{}\Big)^2-2\cos^2\frac{\pi}{8}\cos^2\frac{3\pi}{8}\Big]$
Answer$\cos^4\frac{\pi}{8}+\cos^4\frac{3\pi}{8}+\cos^4\frac{5\pi}{8}+\cos^4\frac{7\pi}{8}$
$=\cos^4\frac{\pi}{8}+\cos^4\frac{3\pi}{8}+\cos^4\Big(\pi-\frac{3\pi}{8}\Big)+\cos^4\Big(\pi-\frac{\pi}{8}\Big)$
$=\cos^4\frac{\pi}{8}+\cos^4\frac{3\pi}{8}+\cos^4\frac{3\pi}{8}+\cos^4\frac{\pi}{8}$
$=2\Big[\cos^4\frac{\pi}{8}+\cos^4\frac{3\pi}{8}\Big]$
$=2\Big[\cos^4\frac{\pi}{8}+\cos^4\Big(\frac{\pi}{2}-\frac{\pi}{8}\Big)\Big]$
$=2\Big[\cos^4\frac{\pi}{8}+\sin^4\frac{\pi}{8}\Big]$
$=2\Big[\Big(\cos^2\frac{\pi}{8}+\sin^2\frac{\pi}{8}\Big)^2-2\cos^2\frac{\pi}{8}.\sin^2\frac{\pi}{8}\Big]$
$=2\Big[1-2\cos^2\frac{\pi}{8}.\sin^2\frac{\pi}{8}\Big]$
$=2-\Big(2\sin\frac{\pi}{8}.\cos\frac{\pi}{8}\Big)^2$
$=2-\Big(\sin\frac{2\pi}{8}\Big)^2=2-\Big(\frac{1}{\sqrt{2}}\Big)^2=2-\frac{1}{2}=\frac{3}{2}$
View full question & answer→Question 35 Marks
In the following match each item given under the column C1 to its correct answer given under the column C2: | | Column C1 | | Column C2 |
| (a) | $\sin(\text{x + y})\sin\text{x}-\text{y}$ | (i) | $\cos^2\text{x}-\sin^2\text{y}$ |
| (b) | $\cos(\text{x + y})\cos(\text{x}-\text{y})$ | (ii) | $\frac{1-\tan\theta}{1+\tan\theta}$ |
| (c) | $\cot\Big(\frac{\pi}{4}+\theta\Big)$ | (iii) | $\frac{1+\tan\theta}{1-\tan\theta}$ |
| (d) | $\tan\Big(\frac{\pi}{4}+\theta\Big)$ | (iv) | $\sin^2\text{x}-\sin^2\text{y}$ |
Answer | | Column C1 | | Column C2 |
| (a) | $\sin(\text{x + y})\sin\text{x}-\text{y}$ | (iv) | $\sin^2\text{x}-\sin^2\text{y}$ |
| (b) | $\cos(\text{x + y})\cos(\text{x}-\text{y})$ | (i) | $\cos^2\text{x}-\sin^2\text{y}$ |
| (c) | $\cot\Big(\frac{\pi}{4}+\theta\Big)$ | (ii) | $\frac{1-\tan\theta}{1+\tan\theta}$ |
| (d) | $\tan\Big(\frac{\pi}{4}+\theta\Big)$ | (ii) | $\frac{1+\tan\theta}{1-\tan\theta}$ |
Explanation: - $\sin(\text{x + y})\sin(\text{x}-\text{y})=\sin^2\text{x}-\sin^2\text{y}$
- $\cos(\text{x + y})\cos(\text{x}-\text{y})\cos^2\text{x}-\cos^2\text{y}$
- $\cot\Big(\frac{\pi}{4}+\theta\Big)=\frac{\frac{\cot\pi}{4}\cot\theta-1}{\cot\theta+\cot\frac{\pi}{4}}$
$=\frac{\cot\theta-1}{\cot\theta+1}=\frac{1-\tan\theta}{1+\tan\theta}$ - $\tan\Big(\frac{\pi}{4}+\theta\Big)=\frac{\tan\frac{\pi}{4}+\tan\theta}{1-\tan\frac{\pi}{4}\tan\theta}=\frac{1+\tan\theta}{1-\tan\theta}$
View full question & answer→Question 45 Marks
If $\tan\theta=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha},$ then show that $\sin\alpha+\cos\alpha=\sqrt{2}\cos\theta.$
[Hint: Express $\tan\theta=\tan(\alpha-\frac{\pi}{4})\theta=\alpha-\frac{\pi}{4}$ ]
AnswerGiven that: $\tan\theta=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}$
$\Rightarrow\tan\theta=\frac{\frac{\sin\alpha-\cos\alpha}{\cos\alpha}}{\frac{\sin\alpha+\cos\alpha}{\cos\alpha}}$
$\Rightarrow\tan\theta=\frac{\tan\alpha-1}{\tan\alpha+1}=\frac{\tan\alpha-\tan\frac{\pi}{4}}{1+\tan\frac{\pi}{4}\tan\alpha}$$[\tan(\text{A}-\text{B})=\frac{\tan\text{A}-\tan\text{B}}{1+\tan\text{A}\tan\text{B}}]$
$\Rightarrow\tan\theta=\tan\Big(\alpha-\frac{\pi}{4}\Big)$
$\therefore\theta=\alpha-\frac{\pi}{4}$
$\Rightarrow\cos\theta=\cos\Big(\alpha-\frac{\pi}{4}\Big)$
$\Rightarrow\cos\theta=\cos\alpha\cos\frac{\pi}{4}+\sin\alpha\sin\frac{\pi}{4}$
$[\cos(\text{A}-\text{B})=\cos\text{A}\cdot\cos\text{B}+\sin\text{A}\cdot\sin\text{B}]$
$\Rightarrow\cos\theta=\cos\alpha\cdot\frac{1}{\sqrt2}+\sin\alpha\cdot\frac{1}{\sqrt2}$
$\Rightarrow\sqrt2\cos\theta=\cos\alpha+\sin\alpha$
$\Rightarrow\sin\alpha+\cos\alpha=\sqrt2\cos\theta.$ Hence proved.
View full question & answer→Question 55 Marks
If $\cos(\theta+\phi)=\text{m}\cos(\theta-\phi),$ then prove that $\tan \theta=\frac{1-\text{m}}{1+\text{m}}\cot\phi.$
[Hint: Express $\frac{\cos(\theta+\phi)}{\cos(\theta-\phi)}=\frac{\text{m}}{1}$ and apply Componendo and Dividendo]
AnswerGiven that: $\cos(\theta+\phi)=\text{m}\cos(\theta-\phi)$
$\Rightarrow\frac{\cos(\theta+\phi)}{\cos(\theta-\phi)}=\frac{\text{m}}{1}$
Using componendo and dividend rule, we get
$\frac{\cos(\theta+\phi)+\cos(\theta-\phi)}{\cos(\theta+\phi)-\cos(\theta-\phi)}=\frac{\text{m}+1}{\text{m}-1}$
$\Big[\cos\text{C}-\cos\text{D}=-2\sin\Big(\frac{\text{C+D}}{2}\Big)\sin\Big(\frac{\text{C}-\text{D}}{2}\Big)\Big]$
$\Big[\cos\text{C}+\cos\text{D}=2\cos\Big(\frac{\text{C+D}}{2}\Big)\cos\Big(\frac{\text{C}-\text{D}}{2}\Big)\Big]$
$\Rightarrow\frac{2\cos\Big(\frac{\theta+\phi+\theta-\phi}{2}\Big).\cos\Big(\frac{\theta+\phi-\theta+\phi}{2}\Big)}{-2\sin\Big(\frac{\theta+\phi+\theta-\phi}{2}\Big).\sin\Big(\frac{\theta+\phi-\theta+\phi}{2}\Big)}=\frac{\text{m}+1}{\text{m}-1}$
$\Rightarrow\frac{\cos\theta\cos\phi}{-\sin\theta\sin\phi}=\frac{\text{m}+1}{\text{m}-1}\Rightarrow-\cot\theta.\cot\phi=\frac{\text{m}+1}{\text{m}-1}$
$\Rightarrow\frac{-\cot\phi}{\tan\theta}=\frac{\text{m}+1}{\text{m}-1}=-\frac{1+\text{m}}{1-\text{m}}$
$\Rightarrow\tan\theta(1+\text{m})=(1-\text{m})\cot\phi$
$\Rightarrow\tan\theta=\frac{1-\text{m}}{1+\text{m}}\cot\phi.$ Hence proved.
View full question & answer→Question 65 Marks
Find the general solution of the equation $(\sqrt{3}-1)\cos\theta+(\sqrt{3}+1)\sin\theta=2$
[Hint: Put $\sqrt{3}-1=\text{r}\sin\alpha,\sqrt{3}+1=\text{r}\cos\alpha$ which gives $\tan\alpha=\tan\Big(\frac{\pi}{4}-\frac{\pi}{6}\Big)\alpha=\frac{\pi}{12}$]
Answer$(\sqrt{3}-1)\cos\theta+(\sqrt{3}+1)\sin\theta=2...(\text{i})$
Put $\sqrt{3}-1=\text{r}\sin\alpha$ and $\sqrt{3}+1=\text{r}\cos\alpha$
$\therefore\text{r}^2=(\sqrt{3}-1)^2+(\sqrt{3}+1)^2\Rightarrow\text{r}^2=8\Rightarrow\text{r}=2\sqrt{2}$
Also, $\tan\alpha=\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3}\Rightarrow\alpha=\frac{\pi}{12}$
From eq. (i), we have
$\text{r}\sin\alpha\cos\theta+\text{r}\cos\alpha\sin\theta=2\Rightarrow\text{r}\sin(\theta+\alpha)=2$
$\Rightarrow\sin(\theta+\alpha)=\frac{1}{\sqrt{2}}\Rightarrow\sin(\theta+\alpha)=\sin\frac{\pi}{4}$
$\Rightarrow\theta+\alpha=\text{n}\pi+(-1)^{\text{n}}\frac{\pi}{4},\text{n}\in\text{Z}$
$\Rightarrow\theta=\text{n}\pi+(-1)^{\text{n}}\frac{\pi}{4}-\frac{\pi}{12},\text{n}\in\text{Z}$
View full question & answer→Question 75 Marks
If $\cos\theta+\tan\theta=2\text{cosec}\theta,$ then find the general value of $\theta.$
AnswerGiven that: $\cot\theta+\tan\theta=2\text{cosec}\theta$
$ \Rightarrow\frac{\cos\theta}{\sin\theta}+\frac{\sin\theta}{\cos\theta}=\frac{2}{\sin\theta}.$
$\Rightarrow\frac{\cos^2\theta+\sin^2\theta}{\sin\theta\cos\theta}=\frac{2}{\sin\theta}$
$\Rightarrow\frac{1}{\sin\theta\cos\theta}=\frac{2}{\sin\theta}$
$\Rightarrow2\sin\theta\cos\theta=\sin\theta $
$\Rightarrow2\sin\theta\cos\theta-\sin\theta=0$
$\Rightarrow\sin\theta(2\cos\theta-1)=0$
$\Rightarrow\sin\theta=0$ Or $2\cos\theta-1=0$ Or $\cos\theta=\frac{1}{2}$
Now $\sin\theta=0\Rightarrow\theta=\text{n}\pi,\text{n}\in\text{z}$
$\cos\theta=\frac{1}{2}\Rightarrow\cos\theta=\cos\frac{\pi}{3}$
$\therefore\theta=2\text{n}\pi\pm\frac{\pi}{3}$
Hence ,the general values of $\theta$ is $2\text{n}\pi\pm\frac{\pi}{3}$ and $\text{n}\pi,\text{n}\in\text{z}$
View full question & answer→Question 85 Marks
If $\tan\text{x}=\frac{\text{b}}{\text{a}},$ then find the value of $\sqrt{\frac{\text{a}+\text{b}}{\text{a}-\text{b}}}+\sqrt{\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}$
AnswerGiven that: $\tan\text{x}=\frac{\text{b}}{\text{a}}$
$\sqrt{\frac{\text{a}+\text{b}}{\text{a}-\text{b}}}+\sqrt{\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}=\sqrt{\frac{\text{a}+\text{b}}{\text{a}-\text{b}}}+\sqrt{\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}$
$\frac{\text{a + b + a}-\text{b}}{\sqrt{(\text{a}-\text{b})(\text{a + b})}}=\frac{2\text{a}}{\sqrt{\text{a}^2-\text{b}^2}}=\frac{2\text{a}}{\text{a}\sqrt{1-\frac{\text{b}^2}{\text{a}^2}}}$ $\Big[\because\tan\text{x}=\frac{\text{b}}{\text{a}}\Big]$
$=\frac{2}{1-\tan^2\text{x}}$
$=\frac{2}{\sqrt{1-\frac{\sin^2\text{x}}{\cos^2\text{x}}}}=\frac{2}{\frac{\sqrt{\cos^2\text{x}-\sin^2\text{x}}}{\cos\text{x}}}$
$=\frac{2\cos\text{x}}{\sqrt{\cos2\text{x}}}$ $[\because\cos^2\text{x}-\sin^2\text{x}=\cos2\text{x}]$
Hence, $\sqrt{\frac{\text{a}+\text{b}}{\text{a}-\text{b}}}+\sqrt{\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}=\frac{2\cos\text{x}}{\sqrt{\cos2\text{x}}}$
View full question & answer→Question 95 Marks
If $\sin(\theta+\alpha)=\text{a}$ and $\sin(\theta+\beta)=\text{b},$ then prove that $\cos2(\alpha-\beta)-4\text{ab}\cos(\alpha-\beta)=1-2\text{a}^2-2\text{b}^2$
[Hint: Express $\cos(\alpha-\beta)=\cos((\theta+\alpha)-(\theta+\beta))$]
AnswerWe have $\sin(\theta+\alpha)=\text{a}...(\text{i})$
$\sin(\theta+\beta)=\text{b}...(\text{ii})$
$\therefore\cos(\theta+\alpha)=\sqrt{1-\text{a}^2}$ and $\cos(\theta+\beta)=\sqrt{1-\text{b}^2}$
$\therefore\cos(\alpha-\beta)=\cos[(\theta+\alpha)-(\theta+\beta)]$
$=\cos(\theta+\beta)\cos(\theta+\alpha)+\sin(\theta+\alpha)\sin(\theta+\beta)$
$=\sqrt{1-\text{a}^2}\sqrt{1-\text{b}^2}+\text{ab}=\text{a b}+\sqrt{1-\text{a}^2-\text{b}^2+\text{a}^2\text{b}^2}$
$\Rightarrow\cos2(\alpha-\beta)-4\text{ab}\cos(\alpha-\beta)$
$=2\cos^2(\alpha-\beta)-1-4\text{ab}\cos(\alpha-\beta)$
$=2\cos(\alpha-\beta)[\cos(\alpha-\beta)-2\text{ab}]-1$
$=2\big(\text{ab}+\sqrt{1-\text{a}^2-\text{b}^2+\text{a}^2\text{b}^2}\big)\\\big(\text{ab}+\sqrt{1-\text{a}^2-\text{b}^2+\text{a}^2\text{b}^2}-2\text{ab}\big)-1$
$=2\big[\big(\sqrt{1-\text{a}^2-\text{b}^2+\text{a}^2\text{b}^2+\text{ab}}\big)\\\big(\sqrt{1-\text{a}^2-\text{b}^2+\text{a}^2\text{b}^2}-\text{ab}\big)\big]-1$
$=2[1-\text{a}^2-\text{b}^2+\text{a}^2\text{b}^2-\text{a}^2\text{b}^2]-1\\=2-2\text{a}^2-2\text{b}^2-1=1-2\text{a}^2-2\text{b}^2$
View full question & answer→Question 105 Marks
If $\cos\alpha+\cos\beta=0=\sin\alpha+\sin\beta,$ then prove that $\cos2\alpha+\cos2\beta=-2\cos(\alpha+\beta).$
$\big[$Hint: $(\cos\alpha+\cos\beta)^2-(\sin\alpha+\sin\beta)^2=0\big]$
AnswerGiven that: $\cos\alpha+\cos\beta=0\ \dots\dots(\text{i})$
and $\sin\alpha+\sin\beta=0\ \dots\dots(\text{ii})$
From (i) and (ii) we have
$(\cos\alpha+\cos\beta)^2-(\sin\alpha+\sin\beta)^2=0$
View full question & answer→Question 115 Marks
Find the general solution of the equation $5\cos^2\theta+7\sin^2\theta-6=0$
Answer$5\cos^2\theta+7\sin^2\theta-6=0$
$\Rightarrow5\cos^2\theta+7(1-\cos^2\theta)-6=0$ $[\because\cos^2\alpha+\sin^2\alpha=1]$
$\Rightarrow5\cos^2\theta+7-7\cos^2\theta-6=0\Rightarrow-2\cos^2\theta+1=0$
$\Rightarrow2\cos^2\theta=1\Rightarrow\cos^2\theta=\frac{1}{2}$
$\Rightarrow\cos^2\theta=\cos^2\frac{\pi}{4}$
$\Rightarrow\frac{1+\cos2\theta}{2}=\frac{1+\cos\frac{\pi}{2}}{2}\Rightarrow\cos2\theta=\cos\frac{\pi}{2}$
$\Rightarrow2\theta=2\text{n}\pi\pm\frac{\pi}{2}$ $[\because\cos\theta=\cos\alpha\Rightarrow\theta=2\text{n}\pi\pm\alpha,\text{n}\in\text{z}]$
$\therefore\theta=\text{n}\pi\pm\frac{\pi}{4}$
Hence, the general solution of $\theta=\text{n}\pi\pm\frac{\pi}{4},\text{n}\in\text{Z}.$
View full question & answer→Question 125 Marks
If $\text{a}\cos2\theta+\text{b}\sin2\theta=\text{c}$ has $\alpha$ and $\beta$ as its roots, then prove that $\tan\alpha+\tan\beta=\frac{2\text{b}}{\text{a+b}}.$
[Hint: Use the identities $\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}$ and $\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}\Big].$
AnswerGiven that: $\text{a}\cos2\theta+\text{b}\sin2\theta=\text{c}...(\text{i})$
$\Rightarrow\text{a}\Big[\frac{1-\tan^2\theta}{1+\tan^2\theta}\Big]+\text{b}\Big[\frac{2\tan\theta}{1+\tan^2\theta}\Big]=\text{c}$ $\Big[\because\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta},\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}\Big]$
$\Rightarrow\text{a}-\text{a}\tan^2\theta+2\text{b}\tan\theta=\text{c}(1+\tan^2\theta)$
$\Rightarrow\text{a}-\text{a}\tan^2\theta+2\text{b}\tan\theta=\text{c + c}\tan^2\theta$
$\Rightarrow\text{a}-\text{a}\tan^2\theta+2\text{b}\tan\theta-\text{c}\tan^2\theta-\text{c}=0$
$\Rightarrow-(\text{a}+\text{c})\tan^2\theta+2\text{b}\tan\theta+(\text{a}-\text{c})=0$
$\Rightarrow(\text{a + c})\tan^2\theta-2\text{b}\tan\theta+(\text{c}-\text{a})=0...(\text{ii})$
Since $\alpha$ and $\beta$ are the roots of equation (i) we have $\tan\alpha$ and $\tan\beta$ are the roots of (ii)
$\Rightarrow\tan\alpha+\tan\beta=\frac{-(-2\text{b})}{\text{a + c}}$ [sum of roots of a quadratic equation $\text{ax}^2+\text{bx}+\text{c}=0$ is $\frac{-\text{b}}{\text{a}}$]
$\Rightarrow\tan\alpha+\tan\beta=\frac{2\text{b}}{\text{a + c}}.$ Hence proved.
View full question & answer→Question 135 Marks
If $\frac{2\sin\alpha}{1+\cos\alpha+\sin\alpha}=\text{y},$ then prove that $\frac{1-\cos\alpha+\sin\alpha}{1+\sin\alpha}$ is also equal to y.
$\Big[\text{Hint: Express }\frac{1-\cos\alpha+\sin\alpha}{1+\sin\alpha}=\frac{1-\cos\alpha+\sin\alpha}{1+\sin\alpha}.\frac{1+\cos\alpha+\sin\alpha}{1+\cos\alpha+\sin\alpha}\Big]$
AnswerWe have, $\frac{2\sin\alpha}{1\cos\alpha+\sin\alpha}=\text{y}$
Now, $\frac{1-\cos\alpha+\sin\alpha}{1+\sin\alpha}=\frac{(1-\cos\alpha+\sin\alpha)}{(1+\sin\alpha)}.\frac{(1+\cos\alpha+\sin\alpha)}{(1+\cos\alpha+\sin\alpha)}$
$=\frac{(1+\sin\alpha)^2-\cos^2\alpha}{(1+\sin\alpha)(1+\sin\alpha+\cos\alpha)}$
$=\frac{1+\sin^2\alpha+2\sin\alpha-1+\sin^2\alpha}{(1+\sin\alpha)(1+\sin\alpha+\cos\alpha)}$
$=\frac{2\sin\alpha(1+\sin\alpha)}{(1+\sin\alpha)(1+\sin\alpha+\cos\alpha)}$
$=\frac{2\sin\alpha}{1+\sin\alpha+\cos\alpha}=\text{y}$
View full question & answer→Question 145 Marks
If $\cos(\alpha+\beta)=\frac{4}{5}$ and $\sin(\alpha-\beta)=\frac{5}{13},$ where $\alpha$ lie between 0 and $\frac{\pi}{4},$ find the value of $\tan2\alpha$
$[$Hint: Express $\tan2\alpha$ as $\tan(\alpha+\beta+\alpha-\beta)]$
AnswerWe have $\cos(\alpha+\beta)=\frac{4}{5}$ and $\sin(\alpha-\beta)=\frac{5}{13}$
$\Rightarrow\tan(\alpha+\beta)=\pm\frac{3}{4}$
and $\tan(\alpha-\beta)=\pm\frac{5}{12}$
Since $\alpha\in\Big(0,\frac{\pi}{2}\Big),2\alpha\in(0,\pi),$ for which $\tan2\alpha>0$
Now, $\tan2\alpha=\tan[(\alpha+\beta)+(\alpha-\beta)]$
$=\frac{\tan(\alpha+\beta)+\tan(\alpha-\beta)}{1-\tan(\alpha+\beta).\tan(\alpha-\beta)}=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}.\frac{5}{12}}=\frac{36+20}{48-15}=\frac{56}{33}$
As other values of $\tan(\alpha+\beta)$ and $\tan(\alpha-\beta)$ gives negative value of $\tan2\alpha$
View full question & answer→Question 155 Marks
If $\text{m}\sin\theta=\text{n}\sin(\theta+2\alpha),$ then prove that $\tan(\theta+\alpha)\cot\alpha=\frac{\text{m}+\text{n}}{\text{m}-\text{n}}$
[Hint: Express $\frac{\sin(\theta+2\alpha)}{\sin\theta}=\frac{\text{m}}{\text{n}}$ and apply componendo and dividendo]
AnswerGiven that: $\text{m}\sin\theta=\text{n}\sin(\theta+2\alpha)$
$\Rightarrow\frac{\sin(\theta+2\alpha)}{\sin\theta}=\frac{\text{m}}{\text{n}}$
Using componendo and dividendo rule we get
$\Rightarrow\frac{\sin(\theta+2\alpha)+\sin\theta}{\sin(\theta+2\alpha)-\sin\theta}=\frac{\text{m}+\text{n}}{\text{m}-\text{n}}$
$\Rightarrow\frac{2\cos\big(\frac{\theta+2\alpha+\theta}{2}\big).\cos\big(\frac{\theta+2\alpha-\theta}{2}\big)}{2\sin\big(\frac{\theta+2\alpha+\theta}{2}\big).\sin\big(\frac{\theta+2\alpha-\theta}{2}\big)}=\frac{\text{m}+\text{n}}{\text{m}-\text{n}}$
$\begin{bmatrix}\because\sin\text{A}+\sin\text{B}=2\sin\frac{\text{A + B}}{2}.\cos\frac{\text{A}-\text{B}}{2}\\\sin\text{A}-\sin\text{B}=2\cos\frac{\text{A + B}}{2}.\sin\frac{\text{A}-\text{B}}{2}\end{bmatrix}$
$\Rightarrow\frac{\sin(\theta+\alpha).\cos\alpha}{\cos(\theta+\alpha).\sin\alpha}=\frac{\text{m}+\text{n}}{\text{m}-\text{n}}$
$\Rightarrow\tan(\theta+\alpha).\cot\alpha=\frac{\text{m}+\text{n}}{\text{m}-\text{n}}$ Hence proved.
View full question & answer→Question 165 Marks
If $\text{x}=\sec\phi-\tan\phi$ and $\text{y}=\text{cosec}\phi+\cot\phi$ then show that $\text{xy}+\text{x}-\text{y}+1=0$
[Hint: Find xy + 1 and then show that x - y = -(xy + 1)]
Answer$\text{x}=\sec\phi-\tan\phi\Rightarrow\text{x}=\frac{1-\sin\phi}{\cos\phi}$
$\text{y}=\text{cosec}\phi+\cot\phi\Rightarrow\text{y}=\frac{1+\cos\phi}{\sin\phi}$
$\Rightarrow\text{xy + x}-\text{y}=\frac{1-\sin\phi1+\cos\phi}{\cos\phi\sin\phi}+\frac{1-\sin\phi}{\cos\phi}-\frac{1+\cos\phi}{\sin\phi}$
$=\frac{(1-\sin\phi)(1+\cos\phi)+(1-\sin\phi)\sin\phi-\cos\phi(1+\cos\phi)}{\sin\phi\cos\phi}$
$=\frac{1-\sin\phi+\cos\phi-\sin\phi\cos\phi+\sin\phi-\sin^2\phi-\cos\phi-\cos^2\phi}{\sin\phi\cos\phi}$
$=\frac{1-\sin\phi\cos\phi-(\sin^2\phi+\cos^2\phi)}{\sin\phi\cos\phi}=-1$
$\therefore\text{xy + x}-\text{y}-1=0$
View full question & answer→Question 175 Marks
Find the value of the expression
$3[\sin^4\Big(\frac{3\pi}{2}-\alpha\Big)+\sin^4(3\pi+\alpha)]-2[\sin^6\Big(\frac{\pi}{2}+\alpha\Big)+\sin^6(5\pi-\alpha)]$
Answer$3[\sin^4\Big(\frac{3\pi}{2}-\alpha\Big)+\sin^4(3\pi+\alpha)]-2[\sin^6\Big(\frac{\pi}{2}+\alpha\Big)+\sin^6(5\pi-\alpha)]$
$=3[\cos^4\alpha+\sin^4\alpha]-2[\cos^6\alpha+\sin^6\alpha]$
$=3[(\cos^2\alpha+\sin^2\alpha)^2-2\cos^2\alpha.\sin^2\alpha]-2[(\cos^2\alpha+\sin^2\alpha)^3]\\-3\cos^2\alpha.\sin^2\alpha(\cos^2\alpha+\sin^2\alpha)$
$=3[1-2\cos^2\alpha.\sin^2\alpha]-2[1-3\cos^2\alpha.\sin^2\alpha]=3-2=1$
View full question & answer→Question 185 Marks
If $\sec\text{x}\cos5\text{x}+1=0,$ where $0<\text{x}\leq\frac{\pi}{2},$ then find the value of x.
AnswerGiven that: $\sec\text{x}\cos5\text{x}+1=0$
$\Rightarrow\frac{1}{\cos\text{x}}\cdot\cos5\text{x}+1$
$\Rightarrow\cos5\text{x}+\cos\text{x}=0$ $\Big[\cos\text{C}+\cos\text{D}=2\cos\Big(\frac{\text{C}+\text{D}}{2}\Big)\cos\Big(\frac{\text{C}-\text{D}}{2}\Big)\Big]$
$\Rightarrow2\cos\Big(\frac{5\text{x}+\text{x}}{2}\Big)\cdot\cos\Big(\frac{5\text{x}-\text{x}}{2}\Big)=0$
$\Rightarrow\cos3\text{x}\cdot\cos2\text{x}=0$
$\cos3\text{x}=0$ Or $\cos2\text{x}=0$
$\Rightarrow3\text{x}=\frac{\pi}{2}$ Or $2\text{x}=\frac{\pi}{2}$
$\text{x}=\frac{\pi}{6}$ Or $\text{x}=\frac{\pi}{4}$
Hence, the value of x are $\frac{\pi}{6},\frac{\pi}{4}.$
View full question & answer→Question 195 Marks
If $\tan\theta+\sin\theta=\text{m}$ and $\tan\theta-\sin\theta=\text{n},$ then prove that $\text{m}^2-\text{n}^2=4\sin\theta\tan\theta$
AnswerGiven that: $\tan\theta+\sin\theta=\text{m}$ and $\tan\theta-\sin\theta=\text{n}$
$\text{L.H.S.}=\text{m}^2-\text{n}^2=(\text{m + n})(\text{m}-\text{n})$
$=[(\tan\theta+\sin\theta)+(\tan\theta-\sin\theta)].[(\tan\theta+\sin\theta)-(\tan\theta-\sin\theta)]$
$=(\tan\theta+\sin\theta+\tan\theta-\sin\theta).(\tan\theta+\sin\theta-\tan\theta+\sin\theta)$
$=2\tan\theta.2\sin\theta=4\sin\theta\tan\theta=\text{R.H.S.}$
$\text{L.H.S. = R.H.S.}$ Hence proved.
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