Solve the Following Question.(5 Marks)

Question 15 Marks

If $ \tan\text{A}+\tan\text{B}=\text{a}$ and $\cot\text{A}+\cot\text{B}=\text{b},$ prove that $\cot\text{(A}+\text{B)}=\frac{1}{\text{a}}-\frac{1}{\text{b}}.$

Answer

We have,
$\tan\text{A}+\tan\text{B}=\text{a}$ $\text{and}\ \cot\text{A}+\cot\text{B}=\text{b}$
$\cot\text{A}+\cot\text{B}=\text{b}$
$\Rightarrow\frac{1}{\tan\text{A}}+\frac{1}{\tan\text{B}}=\text{b}$ $\Big[\because\cot\theta=\frac{1}{\tan\theta}\Big]$
$\Rightarrow\frac{\tan\text{B}+\tan\text{A}}{\tan\text{A}\tan\text{B}}=\text{b}$ $\big[\because\tan\text{A}+\tan\text{B}=\text{a}\big]$
$\Rightarrow\frac{\text{a}}{\tan\text{A}\tan\text{B}}=\text{b}$
$\Rightarrow\frac{\text{a}}{\text{b}}=\tan\text{A}\tan\text{B}$
$\because\cot\text{(A}+\text{B)}=\frac{1}{\tan\text{(A}+\text{B)}}$
$=\frac{\frac{1}{\tan\text{A}+\tan\text{B}}}{1-\tan\text{A}\tan\text{B}}$
$=\frac{1-\tan\text{A}\tan\text{B}}{\tan\text{A} +\tan\text{B}}$
$=\frac{1-\frac{\text{a}}{\text{b}}}{\text{a}}$
$=\frac{\text{b}-\text{a}}{\text{ab}}$ $\big[\because\tan\text{A}+\tan\text{B}=\frac{\text{a}}{\text{b}}\big]$
$=\frac{\text{b}}{\text{ab}}-\frac{\text{a}}{\text{ab}}$
$=\frac{\text{1}}{\text{a}}-\frac{\text{1}}{\text{b}}$
$\therefore\cot\text{(A}+\text{B)}=\frac{1}{\text{a}}-\frac{1}{\text{b}}$
Hence proved.

View full question & answer→

Question 25 Marks

If $\sin\alpha\sin\beta-\cos\alpha\cos\beta+1=0$ prove that $1+\cot\alpha\tan\beta$

Answer

We have,
$\sin\alpha\sin\beta-\cos\alpha\cos\beta+1=0$
$\Rightarrow-(\cos\alpha\cos\beta-\sin\alpha\sin\beta)=-1$
$\Rightarrow-\cos(\alpha+\beta)=1\ ...(1)$
$\therefore\sin(\alpha+\beta)=\sqrt{1-\cos^2(\alpha+\beta)}$
$=\sqrt{1-1^2}=0$
$\Rightarrow\sin(\alpha+\beta)=0\ ...(2)$
Now,
$1+\cot\alpha\tan\beta=1+\frac{\cos\alpha}{\sin\alpha}\times\frac{\sin\beta}{\cos\beta}$
$=\frac{\sin\alpha\times\cos\beta+\cos\alpha\times\sin\beta}{\sin\alpha\times\cos\beta}$
$=\frac{\sin(\alpha+\beta)}{\sin\alpha\times\cos\beta}=\frac{0}{\sin\alpha\times\cos\beta}$ [Using equation (2)]
$\therefore1+\cot\alpha\tan\beta=0$
Hence proved.

View full question & answer→

Question 35 Marks

Reduce each of the following expressions to the sine and cosin of a single expression:
$\cos\text{x}-\sin\text{x}$

Answer

Let $\text{f(x)}=\cos\text{x}-\sin\text{x}$
Dividing and multiplying by $\sqrt{1^2+1^2},$ i.e. by $\sqrt{2},$ we get:
$\text{f(x)}=\sqrt{2}\Big(\frac{1}{\sqrt{2}}\cos\text{x}-\frac{1}{\sqrt{2}}\sin\text{x}\Big)$
$\Rightarrow\text{f(x)}=\sqrt{2}(\cos45^\circ\cos\text{x}-\sin45^\circ\sin\text{x})$
$\Rightarrow\text{f(x)}=\sqrt{2}\cos\Big(\frac{\pi}{4}+\text{x}\Big)$
Again,
$\text{f(x)}=\sqrt{2}\Big(\frac{1}{\sqrt{2}}\cos\text{x}-\frac{1}{\sqrt{2}}\sin\text{x}\Big)$
$\Rightarrow\text{f(x)}=\sqrt{2}\Big(\sin45^\circ\cos\text{x}-\cos45^\circ\sin\text{x}\Big)$
$\Rightarrow\text{f(x)}=\sqrt{2}\sin\Big(\frac\pi4-\text{x}\Big)$

View full question & answer→

Question 45 Marks

Show that $\sin100^\circ-\sin10^\circ$ is positive.

Answer

We have, $\sin100^\circ-\sin10^\circ$
$=\sqrt{2}\Big(\frac{1}{\sqrt{2}}\times100^\circ-\frac{1}{\sqrt{2}}\times\cos100^\circ\Big)$ $\big[$Multiplying and dividing by $\sqrt{1^2+1^2}$ i.e., by $\sqrt{2}\big]$
$=\sqrt{2}(\cos45^\circ\times\sin100^\circ-\sin45^\circ\times\cos100^\circ)$
$=\sqrt{2}(\sin100^\circ\times\cos45^\circ-\cos100^\circ\times\sin45^\circ)$
$=\sqrt{2}(\sin(100^\circ-45^\circ))$
$=\sqrt{2}\sin55^\circ,$ which is positive real number. $[\because\sin\theta$ is positive in first quadrant$]$

View full question & answer→

Question 55 Marks

If are two different valus of X lying between 0 and which satisfy the equation $6\cos\text{x}+8\sin\text{x}=9$find the value of $\sin(\alpha+\beta).$

Answer

We have,
$6\cos\text{x}+8\sin\text{x}=9\ ...(1)$
$\Rightarrow8\sin\text{x}=9-6\cos\text{x}$
$\Rightarrow\big(8\sin\text{x}\big)^2=\big(9-6\cos\text{x}\big)^2$
$\Rightarrow64\sin^2\text{x}=81+36\cos^2\text{x}-108\cos\text{x}$
$\Rightarrow64\sin^2\text{x}=81+36\cos^2\text{x}-108\cos\text{x}$
$\Rightarrow64(1-\cos^2\text{x})=81+36\cos^2\text{x}-108\cos\text{x}$
$\Rightarrow64-64\cos^2\text{x}=81+36\cos^2\text{x}-108\cos\text{x}$
$\Rightarrow36\cos^2\text{x}+64\cos^2\text{x}-108\cos\text{x}-108\cos\text{x}$
$\Rightarrow100\cos^2\text{x}-108\cos\text{x}+17=0\ ...(2)$
Since $\alpha,\beta$ are roots of equation ...(i)
$$$\therefore\cos\alpha$ and $\cos\beta$ roots of equation ....(ii)
$\because\cos\alpha+\cos\beta=\frac{17}{100}\ ...(3)$
$\text{Again},6\cos\text{x}+8\sin\text{x}=9$
$\Rightarrow6\cos\text{x}=9-8\sin\text{x}$
$\Rightarrow\big(8\cos\text{x}\big)^2=\big(9-6\sin\text{x}\big)^2$
$\Rightarrow36\cos^2\text{x}=81+64\cos^2\text{x}-144\cos\text{x}$
$\Rightarrow36(1-\sin^2\text{x})=81+64\sin^2\text{x}-144\sin\text{x}$
$\Rightarrow36-36\sin^2\text{x}=81+36\cos^2\text{x}-108\cos\text{x}$ $[\because$ squaring both sides$]$
$\Rightarrow64\sin^2\text{x}+36\sin^2\text{x}-144\sin\text{x}+81-36=0$
$\Rightarrow100\sin^2\text{x}-144\sin\text{x}+45=0\ ...(4)$
$\because\sin\alpha\times\sin\beta=\frac{45}{100}\ ...(5)$
Now, $\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$
$=\frac{17}{100}-\frac{45}{100}$$=-\frac{28}{100}$$=-\frac{7}{25}$ [Using equation (3) and (5)]
Now, $\sin(\alpha+\beta)=\sqrt{1-(\cos\text{x})^2}$
$=\sqrt{1-\Big(-\frac{7}{25}\Big)^2}$$=\sqrt{1-\frac{49}{625}}$ $=\sqrt{\frac{625-49}{625}}$
$=\sqrt{\frac{576}{625}}$$=\frac{24}{25}$ $\therefore\sin(\alpha+\beta)=\frac{24}{25}$

View full question & answer→

Question 65 Marks

Reduce each of the following expressions to the sine and cosin of a single expression:
$\sqrt{3}\sin\text{x}-\cos\text{x}$

Answer

Let $\text{f(x)}=\sqrt{3}\sin\text{x}-\cos\text{x}$
Dividing and multiplying by $\sqrt{3+1},$ i.e. by 2, We get:
$\text{f(x)}=2\Big(\frac{\sqrt{3}}{2}\sin\text{x}-\frac{1}{2}\cos\text{x}\Big)$
$\Rightarrow\text{f(x)}=2\Big(\cos\frac\pi6\sin\text{x}-\sin\frac\pi6\cos\text{x}\Big)$
$\Rightarrow\text{f(x)}=2\sin\Big(\text{x}-\frac{\pi}{6}\Big)$
Again,
$\text{f(x)}=2\Big(\frac{\sqrt{3}}{2}\sin\text{x}-\frac{1}{2}\cos\text{x}\Big)$
$\Rightarrow\text{f(x)}=2\Big(\sin\frac{\pi}{3}\sin\text{x}-\cos\frac\pi3\cos\text{x}\Big)$
$\Rightarrow\text{f(x)}=-2\cos\Big(\frac\pi3+\text{x}\Big)$

View full question & answer→

Question 75 Marks

Prove that $(2\sqrt{3}+3)\sin\text{x}+2\sqrt{3}\cos\text{x}$ lies between $-(2\sqrt{3}+\sqrt{15})$ and $(2\sqrt{3}+\sqrt{15}).$

Answer

Let $\text{f(x)}=(2\sqrt{3}+3)\sin\text{x}+2\sqrt{3}\cos\text{x}$
We know that,
$-\sqrt{(2\sqrt{3}+3)^2+(2\sqrt{3})^2}\le\text{f(x)}\le\sqrt{(2\sqrt{3}+3)^2+(2\sqrt{3})^2}$
$\Rightarrow-\sqrt{12+9+12\sqrt{3}+12}\le\text{f(x)}\le\sqrt{12+9+12\sqrt{3}+12}$
$\Rightarrow-\sqrt{33+12\sqrt{3}}\le\text{f(x)}\le\sqrt{33+12\sqrt{3}}$
Disclaimer: Instead of $-(2\sqrt{3}+\sqrt{15})$ and $(2\sqrt{3}+\sqrt{15}),$ it should be $-\sqrt{33+12\sqrt{3}}$ and $\sqrt{33+12\sqrt{3}}.$

View full question & answer→

Question 85 Marks

If X lies in the first quadrant and $\cos\text{x}=\frac{8}{17},$ then prove that
$\cos\Big(\frac{\pi}{6}+\text{x}\Big) +\cos\Big(\frac{\pi}{4}-\text{x}\Big)+\cos\Big(\frac{2\pi}{3}-\text{x}\Big)=\Big(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\Big)\frac{23}{17}$

Answer

We have,
$\cos\text{x}=\frac{8}{17}$
$\therefore\sin\text{x}=\sqrt{1-\cos^2\text{x}}=\sqrt{1-\frac{64}{289}}$
$=\sqrt{\frac{225}{289}}$
$=\frac{15}{17}$
Now, $\cos\Big(\frac{\pi}{6}+\text{x}\Big) +\cos\Big(\frac{\pi}{4}-\text{x}\Big)+\cos\Big(\frac{2\pi}{3}-\text{x}\Big)$
$=\Big[\cos\frac{\pi}{6}\cos\text{x}-\sin\frac{\pi}{6}\sin\text{x}\Big]+\Big[\cos\frac{\pi}{4}\cos\text{x}+\sin\frac{\pi}{4}\sin\text{x}\Big]\\\ \ +\Big[\cos\frac{\pi}{4}\cos\text{x}+\sin\frac{\pi}{4}\sin\text{x}\Big]$
$$$=\Big[\cos\frac{\pi}{6}+\cos\frac{\pi}{4}+\cos\frac{2\pi}{3}\Big]\cos\text{x}\\\ \ +\sin\text{x}\Big[-\sin\frac{\pi}{6}+\sin\frac{\pi}{4}+\sin\frac{2\pi}{3}\Big]$
$$$=\Big[\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}+\cos\Big(\frac{\pi}{2}+\frac{\pi}{6}\Big)\Big]\times\frac{8}{17}\\\ +\frac{15}{17}\times\Big[\frac{1}{2}+\frac{1}{\sqrt{2}}+\sin\Big(\frac{\pi}{2}+\frac{\pi}{6}\Big)\Big]$
$$$=\Big[\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}-\sin\frac{\pi}{6}\Big]\times\frac{8}{17}+\frac{15}{17}\times\Big[-\frac{1}{2}+\frac{1}{\sqrt{2}} +\cos\frac{\pi}{6}\Big]$
$=\Big[\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}-\frac{1}{2}\Big]\times\frac{8}{17}+\frac{15}{17}\times\Big[-\frac{1}{2}+\frac{1}{\sqrt{2}} +\frac{\sqrt{3}}{2}\Big]$ $\big[\because\cos\text{A}$ is negative in second quadrant$\big]$
$=\Big[\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\Big]\times\frac{8}{17}\times\frac{15}{17}\times\Big[\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\Big]$
$=\Big(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\Big)\Big(\frac{8}{17}+\frac{15}{17}\Big)$
$=\Big(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\Big)\times\frac{23}{17}$
$\therefore\cos\Big(\frac{\pi}{6}+\text{x}\Big) +\cos\Big(\frac{\pi}{4}-\text{x}\Big)+\cos\Big(\frac{2\pi}{3}-\text{x}\Big)=\Big(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\Big)\frac{23}{17}$
Hence proved.

View full question & answer→

Question 95 Marks

If than $\alpha=\text{x}+1,\tan\beta=\text{x}-1,$ prove that $2\cot(\alpha-\beta)=\text{x}^2$

Answer

We have,
$\tan\alpha=\text{x}+1\text{ and }\tan\beta=\text{x}-1$
Now, $2\cot(\alpha-\beta)$
$=\frac{2}{\tan(\alpha-\beta)}$
$=\frac{\frac{2}{\tan\alpha-\tan\beta}}{1+\tan\alpha\tan\beta}$
$=\frac{2(1+\tan\alpha\tan\beta)}{\tan\alpha-\tan\beta}$
$=\frac{2[1+(\text{x}+1)(\text{x}-1)]}{\text{x}+1-(\text{x}-1)}$
$=\frac{2[1+\text{x}^2-1]}{\text{x}+1-\text{x}+1}$
$=\frac{2\text{x}\text{x}^2}{2}=\text{x}^2$
$\therefore2+\cot(\alpha+\beta )=\text{x}^2$
Hence proved.

View full question & answer→

Question 105 Marks

If $\sin\alpha+\sin\beta=\text{a}$ and $\cos\alpha+\cos\beta=\text{b},$ show that
$\sin(\alpha+\beta)=\frac{2\text{ab}}{\text{a}^2+\text{b}^2}$

Answer

$\text{a}^2+\text{b}^2=(\sin\alpha+\sin\beta)^2+(\cos\alpha+\cos\beta)^2$
$\Rightarrow\text{a}^2+\text{b}^2=\sin^2\alpha+\sin^2\beta+2\sin^2\alpha\sin^2\beta+\cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta$
$\Rightarrow\text{a}^2+\text{b}^2=\sin^2\alpha+\cos^2\alpha+\sin^2\beta+\cos^2\beta+2(\sin\alpha+\sin\beta+\cos\alpha\cos\beta)$
$\Rightarrow\text{a}^2+\text{b}^2=2+2\cos(\alpha-\beta)\cdots(1)$
Now,
$\text{b}^2-\text{a}^2=(\cos\alpha+\cos\beta)^2-(\sin\alpha-\sin\beta)^2$
$\Rightarrow\text{b}^2-\text{a}^2=\cos^2+\cos^2\beta+\sin^2\alpha-\sin^2\beta+2\cos\alpha+\cos\beta-2\sin\alpha\sin\beta$
$\Rightarrow\text{b}^2-\text{a}^2=(\cos^2\alpha-\sin^2\beta)+(\cos^2\beta-\sin^2\alpha)-2\cos(\alpha+\beta)$
$\Rightarrow\text{b}^2-\text{a}^2=2\cos(\alpha+\beta)\cos(\alpha-\beta)+2\cos(\alpha-\beta)$
$\Rightarrow\text{b}^2-\text{a}^2=\cos(\alpha+\beta)\Big(2+2\cos(\alpha-\beta)\Big)\cdots(2)$
From (1) and (2), we have
$\text{b}^2-\text{a}^2=\cos(\alpha+\beta)(\text{a}^2+\text{b}^2)$
$\Rightarrow\frac{\text{b}^2-\text{a}^2}{\text{a}^2-\text{b}^2}=\cos(\alpha+\beta)$
$\Rightarrow\sin(\alpha+\beta)=\sqrt{1-\cos^2(\alpha+\beta)}$
$\Rightarrow\sin(\alpha+\beta)=\sqrt{1-\Big(\frac{\text{b}^2-\text{a}^2}{\text{b}^2+\text{a}^2}\Big)^2}=\sqrt{\frac{\text{b}^4+\text{a}^4-\text{b}^4-\text{a}^4+4\text{a}^2\text{b}^2}{(\text{b}^2+\text{a}^2)}}$
$\Rightarrow\sin(\alpha+\beta)=\frac{2\text{ab}}{\text{a}^2+\text{b}^2}$

View full question & answer→

Question 115 Marks

If then $\text{x}+\tan\Big(\frac{\pi}{3}\Big)+\tan\Big(\text{x}=\frac{2\pi}{3}\Big)=3,$prove that $\frac{3\tan\text{x}-\tan^3\text{x}}{1-3\tan^2\text{x}}=1$

Answer

We have,
$\tan\text{x}+\tan\Big(\text{x}+\frac{\pi}{3}\Big)+\tan\Big(\text{x}+\frac{2\pi}{3}\Big)=3$
$\Rightarrow \tan\text{x}+\Bigg[\frac{\tan\text{x}+\tan\frac{\pi}{3}}{1-\tan\text{x}\tan\frac{\pi}{3}}\Bigg]+\Bigg[\frac{\tan\text{x}+\tan\Big(\frac{2\pi}{3}\Big)}{1-\tan\text{x}\tan\frac{2\pi}{3}}\Bigg]=3$
$\Rightarrow \tan\text{x}+\Bigg[\frac{\tan\text{x}+\tan\frac{\pi}{3}}{1-\sqrt{3}\tan\text{x}}\Bigg]+\Bigg[\frac{\tan\text{x}+\tan\Big(\frac{\pi}{2}+\frac{\pi}{3}\Big)}{1-\tan\text{x}\tan\Big(\frac{\pi}{2}+\frac{\pi}{3}\Big)}\Bigg]=3$
$\Rightarrow \tan\text{x}+\frac{\tan\text{x}+ \sqrt{3}}{1-\sqrt{3}\tan\text{x}}+\frac{\tan\text{x}-\cot\frac{\pi}{3}}{1+\tan\text{x}\cot\frac{\pi}{3}}=3$
$\Rightarrow \tan\text{x}+\frac{\tan\text{x}+ \sqrt{3}}{1-\sqrt{3}\tan\text{x}}+\frac{\tan\text{x}-\sqrt{3}}{1+\sqrt{3}\tan\text{x}}=3$
$\Rightarrow\tan\text{x}+\frac{\Big(\tan\text{x}+\sqrt{3}\Big)\Big(1+\sqrt{3}\tan\text{x}\Big ) +\Big(\tan\text{x}-\sqrt{3}\Big)\Big(1-\sqrt{3}\tan\text{x}\Big)}{\Big(1-\sqrt{3}\tan\text{x}\Big)\Big(1+ \sqrt{3}\tan\text{x}\Big)}=3$
$\Rightarrow\tan\text{x}+\frac{\tan\text{x}+\sqrt{3}\tan^2\text{x}+\sqrt{3}\tan\text{x} +\tan\text{x}-\sqrt{3}\tan^2\text{x}-\sqrt{3}+3\tan\text{x}}{1-\Big(\sqrt{3}\tan\text{x}\Big)^2}$
$\Rightarrow\tan\text{x}+\frac{8\tan\text{x}}{1-3\tan^2\text{x}}=3$
$\Rightarrow\tan\text{x}+\frac{\Big(1-3\tan^2\text{x}\Big)+8\tan\text{x}}{1-3\tan^2\text{x}}=3$
$\Rightarrow\tan\text{x}-\frac{3\tan^3\text{x}+8\tan\text{x}}{1-3\tan^2\text{x}}=3$
$\Rightarrow\frac{3\tan\text{x}-\tan^3\text{x}}{1-3\tan^2\text{x}}=1$
Hence proved.

View full question & answer→

Maths Solve the Following Question.(5 Marks)

Test yourself on this topic