Question 13 Marks
Prove that: $\frac{\tan^22\text{x}-\tan^2\text{x}}{1-\tan^22\text{x}\tan^2\text{x}}=\tan3\text{x}\tan\text{x}$
AnswerWe have, $\text{R.H.S}=\tan3\text{x} \tan\text{x}$ $=\tan(2\text{x}+{\text{x})}\times\tan(2\text{x}-\text{x)}$ $=\Big[\frac{\tan2\text{x}+\tan\text{x}}{1-\tan2\text{x}\tan\text{x}}\Big]\times\Big[\frac{\tan2\text{x}-\tan\text{x}}{1+\tan2\text{x}\tan\text{x}}\Big]$ $=\frac{(\tan2\text{x}+\tan\text{x)}(\tan2\text{x}-\tan\text{x)}}{(1-\tan2\text{x}\tan \text{x)}(1+\tan2\text{x}\tan\text{x)}}$ $=\frac{\tan^22\text{x}-\tan^2\text{x}}{1-\tan^22\text{x}\tan\text{x}}$ $=\text{L.H.S}$ $\therefore\text{L.H.S}=\text{R.H.S}$ hence proved.
View full question & answer→Question 23 Marks
If $\sin\text{A}=\frac{1}{2},$ $\cos\text{B}=\frac{\sqrt{3}}{2},$ where $\frac{\pi}{2}<\text{A}<\pi$ and $0<\text{B}<\frac{\pi}{2},$find the following:$\tan{\text{(A+B)}}$
AnswerGiven: $\sin\text{A}=\frac{1}{2}$ and $\cos\text{B}=\frac{\sqrt{3}}{2}$ Here, $\frac\pi2<\text{A}<\pi$ and $0<\text{B}<\frac\pi2.$ That is, A is in the second quadrant and B is in the first quadrant.We know that in the second quadrant, sine function is positive and cosine and tan functions are negative In the first quadrant, all T−functions are positive. $\cos\text{A}==\sqrt{1-\sin^2\text{A}}=-\sqrt{1-\Big(\frac12\Big)^2}=-\sqrt{1-\frac14}=-\sqrt{\frac34}=\frac{-\sqrt{3}}{2}$ $\tan\text{A}=\frac{\sin\text{A}}{\cos\text{A}}=\frac{\frac{1}{2}}{\frac{-\sqrt{3}}{2}}=\frac{-1}{\sqrt{3}}$ $\sin\text{B}==\sqrt{1-\cos^2\text{A}}=-\sqrt{1-\Big(\frac{\sqrt{3}}{2}\Big)^2}=-\sqrt{1-\frac34}=-\sqrt{\frac14}=\frac{{1}}{2}$ $\tan\text{B}=\frac{\sin\text{B}}{\cos\text{B}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}$ Now, $\tan(\text{A + B})=\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}$ $=\frac{\frac{-1}{\sqrt{3}}+\frac{1}{\sqrt{3}}}{1-\frac{-1}{\sqrt{3}}\times\frac{1}{\sqrt{3}}}$ $=\frac{0}{1+\frac13}=0$
View full question & answer→Question 33 Marks
If $\sin\text{A}=\frac{4}{5}$ and $\cos\text{B}=\frac{5}{13}$ Where $0<\text{A},\text{B}<\frac{\pi}{2},$ find the value of the following:
AnswerWe have, $\sin\text{A}=\frac{4}{5}$ and $\cos\text{B}=\frac{5}{13}$ $\therefore\cos\text{A}=\sqrt{1-\sin^2\text{A}}$ and $\sin\text{B}=\sqrt{1-\cos^2\text{B}}$ $\Rightarrow\cos\text{A}=\sqrt{1-\Big(\frac{4}{5}\Big)^2}$ and $\sin\text{B}=\sqrt{1-\Big(\frac{5}{13}\Big)^2}$ $\Rightarrow\cos\text{A}=\sqrt{1-\frac{16}{25}}$ and $\sin\text{B}=\sqrt{1-\frac{25}{169}}$ $\Rightarrow\cos\text{A}=\sqrt{\frac{25-16}{25}}$ and $\sin\text{B}=\sqrt{\frac{169-25}{169}}$ $\Rightarrow\cos\text{A}=\sqrt\frac{9}{25}$ and $\sin\text{B}=\sqrt\frac{144}{169}$ $\Rightarrow\cos\text{A}=\frac{3}{5}$ and $\sin\text{B}=\frac{12}{13}$ Now, $\cos(\text{A-B})=\cos\text{A}\cos{\text{B}+\sin\text{A}}\sin\text{B}$ $=\frac{3}{5}\times\frac{5}{13}+\frac{4}{5}\times\frac{12}{13}$ $=\frac{15}{65}+\frac{48}{65}$ $=\frac{15+48}{48}$ $=\frac{63}{65}$
View full question & answer→Question 43 Marks
Prove that: $\tan36^\circ+\tan9^\circ+\tan36^\circ+\tan9^\circ=1$
AnswerWe have, $45^\circ=9^\circ+36^\circ$ $\Rightarrow\tan45^\circ=\tan(9^\circ+36^\circ)$ $\Rightarrow1=\frac{\tan9^\circ+\tan36^\circ}{1-\tan9^\circ\tan36^\circ}$ $\Big[\because\tan\text{(A}+\text{B)}=\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}\Big]$ $\Rightarrow 1-\tan9^\circ\tan36^\circ=\tan9^\circ+\tan36^\circ$ $\Rightarrow 1=\tan9^\circ+\tan36^\circ+\tan9^\circ\tan36^\circ$ $\Rightarrow \tan9^\circ+\tan36^\circ+\tan9^\circ\tan36^\circ=1$ Hence proved.
View full question & answer→Question 53 Marks
If $\cos\text{A}=-\frac{24}{25},$ $\cos\text{B}=-\frac{12}{13},$ where A and B both lie in second quadrant,find the value of $\sin\text{(A+B)}$.
- $\sin\text{(A+B)}$
- $\cos\text{(A+B)}$
AnswerWe have, $\cos\text{A}=-\frac{24}{25}$ and $\cos\text{B}=\frac{3}{5}$ $\therefore\sin\text{A}=\sqrt{1-\cos^2\text{A}}$ and $\sin\text{B}=\sqrt{1-\cos^2\text{B}}$ [ in the second quadreant $\cos\theta$ in negetive] $\Rightarrow\sin\text{A}=-\sqrt{1-\Big(-\frac{24}{25}\Big)^2}$ and $\sin\text{B}=\sqrt{1-\Big(\frac{-3}{5}\Big)^2}$ $\Rightarrow\sin\text{A}=-\sqrt{1-\frac{576}{625 }}$ and $\sin\text{B}=\sqrt{1-\frac{9}{25}}$ $\Rightarrow\sin\text{A}=\sqrt{\frac{576}{625}}$ and $\sin\text{B}=\sqrt{\frac{9}{25}}$ $\Rightarrow\sin\text{A}=-\sqrt{\frac{49}{625}}$ and $\sin\text{B}=-\sqrt{\frac{16}{25}}$ $\Rightarrow\sin\text{A}=-\frac{7}{25}$ and $\Rightarrow\sin\text{B}=-\frac{4}{5}$ Now,
- $\sin(\text{A+B})=\sin\text{A}\cos{\text{B}+\cos\text{A}}\sin\text{B}$
$=\frac{7}{25}\times\frac{3}{5}-\frac{24}{25}\times\Big(-\frac{4}{5}\Big)$
$=-\frac{21}{25}+\frac{96}{125}$
$=-\frac{75}{125}$
$=\frac{3}{5}$
- $\cos(\text{A+B})=\cos\text{A}\cos{\text{B}-\sin\text{A}}\sin\text{B}$
$=-\frac{24}{25}\times\frac{3}{5}-\Big(-\frac{7}{25}\Big)\times\Big(-\frac{4}{5}\Big)$
$=-\frac{72}{125}-\frac{28}{125}$
$=-\frac{72-28}{125}$
$=-\frac{100}{125}$
$=-\frac{4}{5}$ View full question & answer→Question 63 Marks
If $\tan\text{A}=\text{x}\tan\text{B},$ prove that $\frac{\sin\text{(A}-\text{B)}}{\sin\text{(A}+\text{B)}}=\frac{\text{x}-\text{1}}{\text{x}+\text{1}}$
AnswerWe have, $\tan\text{A}=\text{x}\tan\text{B}$ $\frac{\sin\text{A}}{\cos\text{B}}=\text{x}\frac{\sin\text{B}}{\cos\text{B}}$ $\Big[\because\tan\theta=\frac{\sin\theta}{\cos\theta}\Big]$ $\Rightarrow\sin\text{A}\cos\text{B}=\text{x}\cos\text{A}\sin\text{B} ...(1)$ Now, $\frac{\sin\text{(A}-\text{B)}}{\sin\text{(A}+\text{B)}}=\frac{\sin\text{A}\cos\text{B}-\sin\text{B}\cos\text{A}}{\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}} $ $=\frac{\text{x}\cos\text{A}\sin\text{B}-\cos\text{A}\sin\text{B}}{\text{x}\cos\text{A}\sin\text{B}+\cos\text{A}\sin\text{B}} $ $[$Using equation$]$ $=\frac{\cos\text{A}\sin\text{B}\text{(x}-{1)}}{\cos\text{A}\sin\text{B}\text{(x}+{1)}}$ $=\frac{\text{x}-1}{\text{x}+1}$ $\therefore\frac{\sin\text{(A}-\text{B)}}{\sin\text{(A}+\text{B)}}=\frac{\text{x}-1}{\text{x}+1}$ Hence proved.
View full question & answer→Question 73 Marks
Prove that: $\frac{\tan\text{A}+\tan\text{B}}{\tan\text{A}-\tan\text{B}}=\frac {\sin\text{A+B}}{\sin\text{A-B}}$
Answer$\text{L.H.S}=$ $\frac{\tan\text{A}+\tan\text{B}}{\tan\text{A}-\tan\text{B} }$ $=\frac{\frac{\sin\text{A}}{\cos\text{A}}+\frac{\sin\text{B}}{\cos\text{B}}}{\frac{\sin\text{A}}{\cos\text{A}}-\frac{\sin\text{B}}{\cos\text{B}}}$ $\Big[\because\tan\theta=\frac{\sin\theta}{\cos\theta}\Big]$ $=\frac{\frac{\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}}{\cos\text{A}\cos\text{B}}}{{\frac{\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}}{\cos\text{A}\cos\text{B}}}{}}$ $=\frac{\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}}{\sin\text{A}\cos\text{B}-\cos\text{A}\sin\text{B}}$ $\begin{bmatrix}\because\sin\text{(A+B)}=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}\\\text{and,}\sin\text{(A-B)}=\sin\text{A}\cos\text{B}-\cos\text{A}\sin\text{B} \end{bmatrix}$ $=\frac{\sin\text{(A+B)}}{\sin\text{(A-B)}}$ $\therefore\frac{\tan\text{A}+\tan\text{B}}{\tan\text{A}-\tan\text{B}}=\frac{\sin\text{(A+B)}}{\sin\text{(A-B)}} $ $=\text{R.H.S}$ $\text{L.H.S}=\text{R.H.S}$ Hence proved.
View full question & answer→Question 83 Marks
If $\cos\text{A}=-\frac{12}{13}$ and $\cot\text{B}=\frac{24}{7}$ where A lies in the second quadrant and B in the third quadrant, find the valus of the following:
AnswerWe have, $\cos\text{A}=\frac{-12}{13}$ and $\cot\text{B}=\frac{24}{7}$ $\therefore\sin\text{A}=\sqrt{1-\cos^2\text{A}}$$=\sqrt{1-\Big(\frac{-12}{13}\Big)^2}=\sqrt{1-\frac{144}{169}}$ $=\sqrt{\frac{25}{169}}=\frac{5}{13}$ and, $\text{cosecB}=-\sqrt{1+\cot^2\text{B}}$ [cosec is negetive in third quadrant] $=-\sqrt{1+\Big(\frac{24}{7}\Big)^2}$ $=-\sqrt{1+\frac{576}{49}}=-\sqrt{\frac{49+576}{49}}$ $\sqrt{\frac{625}{49}}=-\frac{25}{7}$ $\Rightarrow\sin\text{B}=\frac{-7}{25}$ $\Big[\therefore\text{cosecB}=\frac{1}{\sin\text{B}}\Big]$ Now, $\cos\text{B}=-\sqrt{1-\sin^2\text{B}}$ $[\cos\theta $ is negative in third quadrant$]$ $=-\sqrt{1-\Big(\frac{-7}{25}\Big)^2}=-\sqrt{1-\frac{49}{625}}$ $=-\sqrt{\frac{625-49}{625}}=-\sqrt{\frac{576}{625}}$ $=\frac{-24}{25}$ Now, $\sin\text{(A+B)}=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}$ $=\frac{5}{13}\times\Big(\frac{-24}{25}\Big)+\Big(\frac{-12}{13}\Big)\times\Big(\frac{-7}{25}\Big)$ $=\frac{-120}{325}+\frac{84}{325}$ $=\frac{-120+84}{325}$ $=\frac{-36}{325}$
View full question & answer→Question 93 Marks
If $\cos\text{A}=-\frac{12}{13}$ and where A lies in the second quadrant and B in the third quadrant, find the value of the following $\cos\text{(A+B)}$
AnswerWe have, $\cos\text{A}=\frac{-12}{13}$ and $\cot\text{B}=\frac{24}{7}$ $\therefore\sin\text{A}=\sqrt{1-\cos^2\text{A}}$ $=\sqrt{1-\Big(\frac{-12}{13}\Big)^2} $ $=\sqrt{1-\Big(\frac{144}{169}\Big)} $ $=\sqrt{\Big(\frac{25}{169}\Big)} $ $=\frac{5}{13}$ and, $\text{cosecB}=-\sqrt{1+\cot^2\text{B}}$ $[\because$ cosec is negative in third quadrant$]$ $=-\sqrt{1+\Big(\frac{24}{7}\Big)^2} $$=-\sqrt{1+\frac{576}{49}} $$=-\sqrt{\frac{49+576}{49}} $ $=-\sqrt{\frac{625}{49}} $ $=-\frac{25}{7}$ $\Rightarrow\sin\text{B}=\frac{-7}{25}$ $\Big[\therefore\text{cosecB}=\frac{1}{\sin\text{B}}\Big]$ Now, $\cos\text{B}=-\sqrt{1-\sin^2\text{B}}$ $\Big[\because\cos\theta $ is negative in third quadrant$\Big]$ $=-\sqrt{1-\Big(\frac{-7}{25}\Big)^2}$ $=-\sqrt{1-\Big(\frac{49}{625}\Big)}$ $=-\sqrt{\frac{625-49}{625}}$ $=-\sqrt{\frac{576}{625}}$ $=\frac{-24}{25}$ Now, $\cos\text{(A+B)}=\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}$ $=\Big(\frac{-12}{13}\Big)\times\Big(\frac{-24}{25}\Big)-\Big(\frac{5}{13}\Big)\times\Big(\frac{-7}{25}\Big)$ $=\frac{288}{325}+\frac{35}{325}$ $=\frac{323}{325}$
View full question & answer→Question 103 Marks
If $\sin\text{A}=\frac{4}{5}$ and $\cos\text{B}=\frac{5}{13}$ Where $0<\text{A},\text{B}<\frac{\pi}{2},$ find the value of the following:
AnswerWe have, $\sin\text{A}=\frac{4}{5}$ and $\cos\text{B}=\frac{5}{13}$ $\therefore\cos\text{A}=\sqrt{1-\sin^2\text{A}}$ and $\sin\text{B}=\sqrt{1-\cos^2\text{B}}$ $\Rightarrow\cos\text{A}=\sqrt{1-\Big(\frac{4}{5}\Big)^2}$ and $\sin\text{B}=\sqrt{1-\Big(\frac{5}{13}\Big)^2}$ $\Rightarrow\cos\text{A}=\sqrt{1-\frac{16}{25}}$ and $\sin\text{B}=\sqrt{1-\frac{25}{169}}$ $\Rightarrow\cos\text{A}=\sqrt{\frac{25-16}{25}}$ and $\sin\text{B}=\sqrt{\frac{169-25}{169}}$ $\Rightarrow\cos\text{A}=\sqrt\frac{9}{25}$ and $\sin\text{B}=\sqrt\frac{144}{169}$ $\Rightarrow\cos\text{A}=\frac{3}{5}$ and $\sin\text{B}=\frac{12}{13}$ Now, $\sin(\text{A+B})=\sin\text{A}\cos{\text{B}+\cos\text{A}}\sin\text{B}$ $=\frac{4}{5}\times\frac{5}{13}+\frac{3}{5}\times\frac{12}{13}$ $=\frac{20}{65}+\frac{36}{65}$ $=\frac{20+36}{65}$ $=\frac{56}{65}$
View full question & answer→Question 113 Marks
If $\cos\text{A}=-\frac{12}{13}$and where A lies in the second quadrant and B in the third quadrant, find the value of the following: $\tan\text{(A+B)}$
AnswerWe have, $\cos\text{A}=\frac{-12}{13}$ and $\cot\text{B}=\frac{24}{7}$ $\therefore\sin\text{A}=\sqrt{1-\cos^2\text{A}}$$=\sqrt{1-\Big(\frac{-12}{13}\Big)^2} $ $=\sqrt{1-\Big(\frac{144}{169}\Big)} $ $=\sqrt{\Big(\frac{25}{169}\Big)} =\frac{5}{13}$ $$ and, $\text{cosecB}=-\sqrt{1+\cot^2\text{B}}$ $\big[\because\text{cosec}$ is negative in third quadrant$\big]$ $=-\sqrt{1+\Big(\frac{24}{7}\Big)^2} =-\sqrt{1+\frac{576}{49}} =-\sqrt{\frac{49+576}{49}} $ $$ $$ $=-\sqrt{\frac{625}{49}} =-\frac{25}{7}$ $$ $\Rightarrow\sin\text{B}=\frac{-7}{25}$ $\Big[\therefore\text{cosecB}=\frac{1}{\sin\text{B}}\Big]$ Now, $\cos\text{B}=-\sqrt{1-\sin^2\text{B}}$ $\big[\because\cos\theta $ is negative in third quadrant$\big]$ $=-\sqrt{1-\Big(\frac{-7}{25}\Big)^2}=-\sqrt{1-\Big(\frac{49}{625}\Big)}=-\sqrt{\frac{625-49}{625}}$ $$ $$ $=-\sqrt{\frac{576}{625}}=\frac{-24}{25}$ $$ Now, $\tan\text{A}=\frac{\sin\text{A}}{\cos\text{A}}=\frac{\frac{5}{13}}{\frac{-12}{13}}=\frac{-5}{12}$ $\Big[\because\tan\theta=\frac{\sin\theta}{\cos\theta}\Big]$ and, $\tan\text{A}=\frac{\sin\text{B}}{\cos\text{B}}=\frac{\frac{-7}{25}}{\frac{-24}{25}}=\frac{7}{24}$ $\Big[\because\tan\theta=\frac{\sin\theta}{\cos\theta}\Big]$ $\tan\text{(A+B)}=\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}+\tan\text{B}}$ $=\frac{\frac{-5}{12}+\frac{7}{24}}{1-\Big(\frac{-5}{12}\Big)\times\frac{7}{24}}$ $=\frac{\frac{-10+7}{24}}{1+\frac{35}{288}}=\frac{\frac{-3}{24}}{\frac{288+35}{288}}$ $=-\frac{36}{323}$
View full question & answer→Question 123 Marks
If $\tan\text{A}=\frac{\text{m}}{\text{m-1}}$ and $\tan\text{B}=\frac{1}{\text{2m-1}},$ then prove that $\text{A-B}=\frac{\pi}{4}$
AnswerWe have, $\tan\text{A}=\frac{\text{m}}{\text{m-1}}$ and $\tan\text{B}=\frac{\text{1}}{2\text{m}-1}$ Now, $\tan\text{(A-B)}=\frac{\tan\text{A}-\tan\text{B}}{1+\tan\text{A}{\tan\text{B}}}$ $=\frac{\frac{\text{m}}{\text{m-1}}-\frac{1}{{\text{2m}-1}}}{1+\frac{\text{m}}{\text{m-1}}\times\frac{1}{\text{2m-1}}}$ $=\frac{\frac{\text{m(2m-1)}-(\text{m-1})}{(\text{m-1})(\text{2m-1})}}{1+\frac{\text{m}}{\text{(m-1)}\text{(2m-1)}}}$ $=\frac{\frac{\text{m(2m-1)}-(\text{m-1})}{(\text{m-1})(\text{2m-1})}}{\frac{\text{(m-1)}\text{(2m-1)}+\text{(m)}}{\text{(m-1)}\text{(2m-1)}}}$ $=\frac{\text{m}(\text{2m-1)}-\text{(m-1)}}{\text{(m-1)}\text{(2m-1)}+\text{(m)}}$ $=\frac{\text{2m}^2-\text{m}-\text{m}+1}{\text{2m}^2-\text{m}-\text{2m}+1+\text{m}}$ $=\frac{\text{2m}^2-\text{m}-\text{m}+1}{\text{2m}^2-\text{2m}+1}$ $=\frac{\text{2m}^2-\text{2m}+1}{\text{2m}^2-\text{2m}+1}$ $=1$ $\therefore\tan\text{(A-B)}=1=\Big(\tan\frac{\pi}{4}\Big)$ $\Big[\because\tan\frac{\pi}{4}=1\Big]$ $\Rightarrow\tan\text{(A-B)}=1=\Big(\tan\frac{\pi}{4}\Big)$ $\Rightarrow\text{A-B}=\frac{\pi}{4}$
View full question & answer→Question 133 Marks
$\text{If} \tan\text{(A}+\text{B)}=\text{x}$ and $\tan\text{(A}-\text{B)}=\text{y},$find the values of $\tan2\text{A}$ and $\tan2\text{B}.$
AnswerWe have, $\tan\text{(A}+\text{B)}=\tan\text{(A}-\text{B)}=\text{y}$ Now, $\tan2\text{A}=\tan\Big[\text{(A}+\text{B})+\text{(A}-\text{B})\Big]$ $=\frac{\tan\text{(A}+\text{B)+}\tan\text{(A}-\text{B)}}{1-\tan\text{(A}+\text{B)}\text{x}\tan\text{(A}-\text{B})}$ $=\frac{\text{x}+\text{y}}{\text{1}-\text{xy}}$ $\tan2\text{A}=\frac{\text{x}+\text{y}}{1-\text{xy}}$ Now, $\tan2\text{A}=\tan\Big[\text{(A}-\text{B)}-\text{(A}-\text{B)}\Big]$ $=\frac{\tan\text{(A}+\text{B)-}\tan\text{(A}-\text{B)}}{1+\tan\text{(A}+\text{B)}\text{x}\tan\text{(A}-\text{B})}$ $=\frac{\text{x}-\text{y}}{\text{1}+\text{xy}}$ $\therefore \tan2\text{B}=\frac{\text{x}-\text{y}}{\text{1}+\text{xy}}$
View full question & answer→Question 143 Marks
If $\sin\text{A}=\frac{12}{13}$ and $\sin\text{B}=\frac{4}{5}$ Where $\frac{\pi}{2}<\text{A}<\pi$and $0<\text{B}<\frac{\pi}{2},$ find the following:
AnswerWe have, $\sin\text{A}=\frac{12}{13}$ and $\sin\text{B}=\frac{4}{5}$ $\therefore\cos\text{A}=\sqrt{1-\sin^2\text{A}}$ and $\cos\text{B}=\sqrt{1-\cos^2\text{B}}$ $[$ in the second quadreant $\cos\theta$ in negetive$]$ $\Rightarrow\cos\text{A}=-\sqrt{1-\Big(\frac{12}{13}\Big)^2}$ and $\cos\text{B}=\sqrt{1-\Big(\frac{4}{5}\Big)^2}$ $\Rightarrow\cos\text{A}=-\sqrt{1-\frac{144}{169 }}$ and $\cos\text{B}=\sqrt{1-\frac{16}{25}}$ $\Rightarrow\cos\text{A}=\sqrt{\frac{25}{169}}$ and $\cos\text{B}=\sqrt{\frac{169-25}{169}}$ $\Rightarrow\cos\text{A}=\sqrt\frac{25}{169}$ and $\cos\text{B}=\sqrt\frac{9}{25}$ $\Rightarrow\cos\text{A}=\frac{-5}{13}$ and $\cos\text{B}=\frac{9}{25}$ Now, $\cos(\text{A+B})=\cos\text{A}\cos{\text{B}-\sin\text{A}}\sin\text{B}$ $=\frac{-5}{13}\times\frac{3}{5}-\frac{5}{13}\times\frac{4}{5}$ $=\frac{-15}{65}-\frac{48}{65}$ $=\frac{-63}{65}$
View full question & answer→Question 153 Marks
If $\sin\text{A}=\frac{12}{13}$ and $\sin\text{B}=\frac{4}{5}$ Where $\frac{\pi}{2}<\text{A}<\pi$and $0<\text{B}<\frac{\pi}{2},$ find the following:
AnswerWe have, $\sin\text{A}=\frac{12}{13}$ and $\sin\text{B}=\frac{4}{5}$ $\therefore\cos\text{A}=\sqrt{1-\sin^2\text{A}}$ and $\cos\text{B}=\sqrt{1-\cos^2\text{B}}$ $[$ In the second quadreant $\cos\theta$ in negetive$]$ $\Rightarrow\cos\text{A}=-\sqrt{1-\Big(\frac{12}{13}\Big)^2}$ and $\cos\text{B}=\sqrt{1-\Big(\frac{4}{5}\Big)^2}$ $\Rightarrow\cos\text{A}=-\sqrt{1-\frac{144}{169 }}$ and $\cos\text{B}=\sqrt{1-\frac{16}{25}}$ $\Rightarrow\cos\text{A}=\sqrt{\frac{25}{169}}$ and $\cos\text{B}=\sqrt{\frac{169-25}{169}}$ $\Rightarrow\cos\text{A}=\sqrt\frac{25}{169}$ and $\cos\text{B}=\sqrt\frac{9}{25}$ $\Rightarrow\cos\text{A}=\frac{-5}{13}$ and $\cos\text{B}=\frac{9}{25}$ Now, $\sin(\text{A+B})=\sin\text{A}\cos{\text{B}+\cos\text{A}}\sin\text{B}$ $=\frac{12}{13}\times\frac{3}{5}-\frac{5}{13}\times\frac{4}{5}$ $=\frac{36}{65}-\frac{20}{65}$ $=\frac{16}{65}$
View full question & answer→Question 163 Marks
Prove that: $\tan13\text{x}-\tan9\text{x}-\tan4\text{x}=\tan13\text{x}\tan9\text{x}\tan4\text{x}$
AnswerWe have, $13\text{x}=9\text{x}+4\text{x}$ $\Rightarrow\tan13\text{x}=\tan(9\text{x}+4\text{x)}$ $\Rightarrow\tan13\text{x}=\frac{\tan9\text{x}+\tan4\text{x}}{1-\tan9\text{x}\tan4\text{x}}$ $\Big[\because\tan\text{(A}+\text{B)}=\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}\Big]$ $\Rightarrow\tan13\text{x}(1-\tan9\text{x}\tan4\text{x)}=\tan9\text{x}+\tan4\text{x}$ $\Rightarrow\tan13\text{x}-\tan13\text{x} \tan9\text{x} \tan4\text{x}=\tan9\text{x}+\tan4\text{x}$ $\Rightarrow\tan13\text{x}-\tan9\text{x}-\tan4\text{x}=\tan13\text{x}\tan9\text{x}\tan4\text{x}$ Hence proved.
View full question & answer→Question 173 Marks
Prove that: $\frac{\cos9^\circ+\sin9^\circ}{\cos9^\circ-\sin9^\circ}=\tan54^\circ$$ $
Answer$\text{L.H.S}=$ $\frac{\cos11^\circ+\sin11^\circ}{\cos11^\circ-\sin11^\circ}$ dividing num erator and demomintor by $\cos9^\circ$we get,
$\frac{\frac{\cos9^\circ}{\cos9^\circ}+\frac{\sin9^\circ}{\cos9^\circ}}{\frac{\cos9^\circ}{\cos9^\circ}-\frac{\sin9^\circ}{\cos9^\circ}}{}$ $\Big[\because\tan\theta=\frac{\sin\theta}{\cos\theta}\Big]$ $=\frac{1+\tan9^\circ}{1-\tan9^\circ}$ $\big[\tan45^\circ=1\big]$ $=\frac{\tan45^\circ+\tan9^\circ}{1-\tan45^\circ\times\tan9^\circ}$ $\Big[\because\tan\text{(A+B)=}\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}\Big]$ $=\tan(45^\circ+9^\circ)$ $=\tan54^\circ$ $=\text{R.H.S}$ $\therefore\text{L.H.S}=\text{R.H.S}$ Hence proved.
View full question & answer→Question 183 Marks
Prove that: $\frac{\sin\text{(A+B)}+\sin\text{(A-B)}}{\cos\text{(A+B)}+\cos\text{(A-B)}}=\tan\text{A}$
Answer$\text{L.H.S}=\frac{\sin\text{(A+B)}+\sin\text{(A-B)}}{\cos\text{(A+B)}+\cos\text{(A-B)}}$ $=\frac{2\sin\text{A}\cos\text{B}}{2\cos\text{A}\cos\text{B}}$ $=\frac{\sin\text{A}}{\cos\text {A}}$ $=\tan\text{A}$ $\therefore\text{L.H.S}=\text{R.H.S}$ Hence proved.
View full question & answer→Question 193 Marks
If $\sin\text{A}=\frac{1}{2},$ $\cos\text{B}=\frac{\sqrt{3}}{2},$ where $\frac{\pi}{2}<\text{A}<\pi$ and $0<\text{B}<\frac{\pi}{2},$ find the following:$\tan{\text{(A - B)}}$
AnswerGiven: $\sin\text{A}=\frac{1}{2}$ and $\cos\text{B}=\frac{\sqrt{3}}{2}$ Here, $\frac\pi2<\text{A}<\pi$ and $0<\text{B}<\frac\pi2.$ That is, A is in the second quadrant and B is in the first quadrant.We know that in the second quadrant, sine function is positive and cosine and tan functions are negative In the first quadrant, all T−functions are positive. $\cos\text{A}==\sqrt{1-\sin^2\text{A}}=-\sqrt{1-\Big(\frac12\Big)^2}=-\sqrt{1-\frac14}=-\sqrt{\frac34}=\frac{-\sqrt{3}}{2}$ $\tan\text{A}=\frac{\sin\text{A}}{\cos\text{A}}=\frac{\frac{1}{2}}{\frac{-\sqrt{3}}{2}}=\frac{-1}{\sqrt{3}}$ $\sin\text{B}==\sqrt{1-\cos^2\text{A}}=-\sqrt{1-\Big(\frac{\sqrt{3}}{2}\Big)^2}=-\sqrt{1-\frac34}=-\sqrt{\frac14}=\frac{{1}}{2}$ $\tan\text{B}=\frac{\sin\text{B}}{\cos\text{B}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}$ Now, $\tan(\text{A - B})=\frac{\tan\text{A}-\tan\text{B}}{1+\tan\text{A}\tan\text{B}}$ $=\frac{\frac{-1}{\sqrt{3}}+\frac{1}{\sqrt{3}}}{1+\frac{-1}{\sqrt{3}}\times\frac{1}{\sqrt{3}}}$ $=\frac{\frac{-2}{\sqrt{3}}}{1-\frac13}$ $=\frac{\frac{-2}{\sqrt{3}}}{\frac23}$ $=-\sqrt{3}$
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If $\tan\text{A}=\frac{3}{4},$ $\cos\text{B}=-\frac{9}{41},$ where $\pi<\text{A}<\frac{3\pi}{2}$ and $0<\text{B}<\frac{\pi}{2},$ find$\tan\text{(A+B)}$.
AnswerWe have, $\tan\text{A}=\frac{3}{4},$ and $\cos\text{B}=\frac{9}{41}$ $\therefore\sin\text{B}=\sqrt{1-\cos^2\text{B}}$ $=\sqrt{1-\Big(\frac{9}{41}\Big)^2}$ $=\sqrt{1-\frac{81}{1681}}$ $=\sqrt{\frac{1600}{1681}}$ $=\frac{40}{41}$ $\tan\text{B}=\frac{\sin\text{B}}{\cos\text{B}}=\frac{\frac{40}{41}}{\frac{9}{41}}$$=\frac{40}{9}$ Now, $\tan\text{(A+B)}=\frac{\tan\text{A}\tan\text{B}}{1-\tan\text{A}\tan\text{B}}$ $=\frac{\frac{3}{4}+\frac{40}{9}}{1-\frac{3}{4}\times\frac{40}{9}}$ $=\frac{\frac{27+160}{36}}{\frac{36-120}{36}}$ $=\frac{\frac{187}{36}}{\frac{-84}{36}}$ $=-\frac{187}{84}$
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Prove that: $\sin^2\text{(n+1)}\text{A}-\sin^2\text{nA}=\sin\text{(2n+1)}\text{A}\sin\text{A}$
AnswerWe have, $\text{L.H.S} =\sin^2\text{(n+1)}\text{A}-\sin^2\text{nA}$ $=\sin[(\text{n+1)}\text{A}+\text{nA}]\sin[(\text{n+1)}\text{A}-\text{nA}]$ $\Big[\because\sin^2\text{A}-\sin^2\text{B}=\sin\text{(A+ B)}\sin\text{(A-B)}\Big]$ $=\sin[\text{nA}+\text{A}+\text{nA}]\sin[\text{nA}+\text{A}-\text{nA}]$ $=\sin\text{(2nA+A)}\sin\text{(A)}$ $=\sin\text{(2n+1)}\text{A}\sin\text{A}$ $=\text{R.H.S}$ $$$\therefore\text{L.H.S}=\text{R.H.S}$ Hence proved.
View full question & answer→Question 223 Marks
Prove that:$\frac{\cos8^\circ-\sin8^\circ}{\cos8^\circ+\sin8^\circ}=\tan37^\circ$$ $
Answer$\text{L.H.S}=\frac{\cos8^\circ-\sin8^\circ}{\cos8^\circ+\sin8^\circ}$dividing num erator and demomintor by $\cos9^\circ$we get,
$\frac{\frac{\cos8^\circ}{\cos8^\circ}-\frac{\sin8^\circ}{\cos8^\circ}}{\frac{\cos8^\circ}{\cos8^\circ}+\frac{\sin8^\circ}{\cos8^\circ}}{}$$\Big[\because\tan\theta=\frac{\sin\theta}{\cos\theta}\Big]$ $=\frac{1-\tan8^\circ}{1+\tan8^\circ}$ $\big[\tan45^\circ=1\big]$ $=\frac{\tan8^\circ-\tan8^\circ}{1+\tan8^\circ\times\tan8^\circ}$ $\Big[\because\tan\text{(A-B)=}\frac{\tan\text{A}-\tan\text{B}}{1+\tan\text{A}\tan\text{B}}\Big]$ $=\tan(45^\circ+8^\circ)$ $=\tan37^\circ$ $=\text{R.H.S}$ $\therefore\text{L.H.S}=\text{R.H.S}$ Hence proved.
View full question & answer→Question 233 Marks
Prove that: $\cos^2\frac{\pi}{4}-\sin^2\frac{\pi}{12}=\frac{\sqrt{3}}{4}$
Answer$\text{L.H.S}=\cos^245^\circ-\sin^215^\circ$ $=\Big(\frac{1}{\sqrt{2}}\Big)^2-\sin^215^\circ$ $=\frac{1}{2}-\Big(\frac{1-\cos^2\times15^\circ}{2}\Big)$ $=\frac{1}{2}-\Big(\frac{1-\cos\times30^\circ}{2}\Big)$ $=\frac{1-1+\cos30^\circ}{2}$ $=\frac{\cos30^\circ}{2}$ $=\frac{\sqrt{3}}{2}\times\frac{1}{2}$ $=\frac{\sqrt{3}}{2}$ $=\text{R.H.S}$ $\therefore\text{L.H.S}=\text{R.H.S}$ Hence proved.
View full question & answer→Question 243 Marks
Find the maximum and minimum values of each of the following trigonometrical expressions: $12\sin\text{x}-5\cos\text{x}$
AnswerLet $\text{f}(\theta)=12\sin\theta-5\cos\theta$ We know that, $-\sqrt{(12)^2+(-5)^2}\leq\text{f}(\theta)\leq\sqrt{(12)^2+(-5)^2}$ $\Rightarrow-\sqrt{144+25}\leq\text{f}(\theta)\leq\sqrt{144+25}$ $\Rightarrow-\sqrt{169}\leq\text{f}(\theta)\leq\sqrt{169}$ $\Rightarrow-13\leq\text{f}(\theta)\leq13$ Hence, minimum and maximum values of $12\sin\theta-5\cos\theta$ are $-13 \text{ and }13$ respectively.
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If $\sin\text{A}=\frac{4}{5}$ and $\cos\text{B}=\frac{5}{13}$ Where $0<\text{A},\text{B}<\frac{\pi}{2}$,find the value of the following:
AnswerWe have, $\sin\text{A}=\frac{4}{5}$ and $\cos\text{B}=\frac{5}{13}$ $\therefore\cos\text{A}=\sqrt{1-\sin^2\text{A}}$ and $\sin\text{B}=\sqrt{1-\cos^2\text{B}}$ $\Rightarrow\cos\text{A}=\sqrt{1-\Big(\frac{4}{5}\Big)^2}$ and $\sin\text{B}=\sqrt{1-\Big(\frac{5}{13}\Big)^2}$ $\Rightarrow\cos\text{A}=\sqrt{1-\frac{16}{25}}$ and $\sin\text{B}=\sqrt{1-\frac{25}{169}}$ $\Rightarrow\cos\text{A}=\sqrt{\frac{25-16}{25}}$ and $\sin\text{B}=\sqrt{\frac{169-25}{169}}$ $\Rightarrow\cos\text{A}=\sqrt\frac{9}{25}$ and $\sin\text{B}=\sqrt\frac{144}{169}$ $\Rightarrow\cos\text{A}=\frac{3}{5}$ and $\sin\text{B}=\frac{12}{13}$ Now, $\cos(\text{A+B})=\cos\text{A}\cos{\text{B}-\sin\text{A}}\sin\text{B}$ $=\frac{3}{5}\times\frac{5}{13}+\frac{4}{5}\times\frac{12}{13}$ $=\frac{15}{65}+\frac{48}{65}$ $=\frac{15+48}{65}$ $=\frac{-33}{65}$
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Prove that: $\tan8\text{x}-\tan6\text{x}-\tan2\text{x}=\tan8\text{x}\tan6\text{x}\tan2\text{x}$
AnswerWe have, $8\text{x}=6\text{x}+2\text{x}$ $\Rightarrow\tan8\text{x}=\tan(6\text{x}+2\text{x)}$ $\Rightarrow\tan8\text{x}=\frac{\tan6\text{x}+\tan2\text{x}}{1-\tan6\text{x}\tan2\text{x}}$ $\Big[\because\tan\text{(A}+\text{B)}=\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}\Big]$ $\Rightarrow\tan8\text{x}(1-\tan6\text{x}\tan2\text{x)}=\tan6\text{x}+\tan2\text{x}$ $\Rightarrow\tan8\text{x}-\tan8\text{x} \tan6\text{x} \tan2\text{x}=\tan6\text{x}+\tan2\text{x}$ $\Rightarrow\tan8\text{x}-\tan6\text{x}-\tan2\text{x}=\tan8\text{x}\tan6\text{x}\tan2\text{x}$ Hence proved.
View full question & answer→Question 273 Marks
Prove that: $\frac{\cos11^\circ+\sin11^\circ}{\cos11^\circ-\sin11^\circ}=\tan56^\circ$$ $
Answer$\text{L.H.S}=$ $\frac{\cos11^\circ+\sin11^\circ}{\cos11^\circ-\sin11^\circ}$ dividing num erator and demomintor by $\cos11^\circ$we get,
$\frac{\frac{\cos11^\circ}{\cos11^\circ}+\frac{\sin11^\circ}{\cos11^\circ}}{\frac{\cos11^\circ}{\cos11^\circ}-\frac{\sin11^\circ}{\cos11^\circ}}{}$ $\Big[\because\tan\theta=\frac{\sin\theta}{\cos\theta}\Big]$ $=\frac{1+\tan11^\circ}{1-\tan11^\circ}$ $\big[\tan45^\circ=1\big]$ $=\frac{\tan45^\circ+\tan11^\circ}{1-\tan45^\circ\times\tan11^\circ}$ $\Big[\because\tan\text{(A+B)=}\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}\Big]$ $=\tan(45^\circ+11^\circ)$ $=\tan56^\circ$ $=\text{R.H.S}$ $\text{L.H.S}=\text{R.H.S}$ Hence proved.
View full question & answer→Question 283 Marks
If $\sin\text{A}=\frac{1}{2},$ $\cos\text{B}=\frac{12}{13},$ where $\frac{\pi}{2}<\text{A}<\pi$ and $\frac{3\pi}{2}<\text{B}<2\pi$ find$\tan\text{(A-B)}$.
AnswerWe have, $\sin\text{A}=\frac{1}{2},$ and $\cos\text{B}=\frac{12}{13}$ $\therefore\cos\text{A}=\sqrt{1-\sin^2\text{A}}$ and $\sin\text{B}=-\sqrt{1-\cos^2\text{B}}$ [Cosin is negative in second quadrant and sine is negative in fourth quadrant] $\Rightarrow\cos\text{A}=-\sqrt{1-\Big(\frac{1}{2}\Big)^2}$ and $\sin\text{B}=-\sqrt{1-\Big(\frac{12}{13}\Big)^2}$ $\Rightarrow\cos\text{A}=-\sqrt{1-\frac{1}{4}}$ and $\sin\text{B}=-\sqrt{1-\frac{144}{169}}$ $\Rightarrow\cos\text{A}=-\sqrt\frac{3}{4}$ and $\sin\text{B}=-\sqrt\frac{25}{169}$ $\Rightarrow\cos\text{A}=-\sqrt\frac{3}{2}$ and $\sin\text{B}-\frac{5}{13}$ $\therefore\tan\text{A}=\frac{\sin\text{A}}{\cos\text{A}}=\frac{\frac{1}{2}}{\frac{-\sqrt{3}}{2}}=\frac{-1}{\sqrt{3}}$ $\tan\text{B}=\frac{\sin\text{B}}{\cos\text{B}}=\frac{-\frac{5}{13}}{\frac{{12}}{13}}=\frac{-5}{12}$ Now, $\tan\text{(A+B)}=\frac{\tan\text{A}-\tan\text{B}}{1+\tan\text{A}\tan\text{B}}$ $=\frac{\frac{-1}{\sqrt{3}}-\Big(\frac{-5}{12}\Big)}{1+\Big(\frac{-1}{\sqrt{3}}\Big)\times\Big(\frac{-5}{12}\Big)}$ $=\frac{\frac{-1}{\sqrt{3}}+\frac{5}{12}}{1+\frac{5}{12\sqrt{3}}}$ $=\frac{\frac{-12+5\sqrt{3}}{12\sqrt{3}}}{\frac{12\sqrt{3}+5}{12\sqrt{3}}}$ $=\frac{5\sqrt{3}-12}{5+12\sqrt{3}}$ $\Rightarrow\tan\text{(A+B)}=\frac{5\sqrt{3}-12}{5+12\sqrt{3}}$
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If $\sin\text{A}=\frac{4}{5}$ and $\cos\text{B}=\frac{5}{13}$ Where $0<\text{A},\text{B}<\frac{\pi}{2},$ find the value of the following:
AnswerWe have, $\sin\text{A}=\frac{4}{5}$ and $\cos\text{B}=\frac{5}{13}$ $\therefore\cos\text{A}=\sqrt{1-\sin^2\text{A}}$ and $\sin\text{B}=\sqrt{1-\cos^2\text{B}}$ $\Rightarrow\cos\text{A}=\sqrt{1-\Big(\frac{4}{5}\Big)^2}$ and $\sin\text{B}=\sqrt{1-\Big(\frac{5}{13}\Big)^2}$ $\Rightarrow\cos\text{A}=\sqrt{1-\frac{16}{25}}$ and $\sin\text{B}=\sqrt{1-\frac{25}{169}}$ $\Rightarrow\cos\text{A}=\sqrt{\frac{25-16}{25}}$ and $\sin\text{B}=\sqrt{\frac{169-25}{169}}$ $\Rightarrow\cos\text{A}=\sqrt\frac{9}{25}$ and $\sin\text{B}=\sqrt\frac{144}{169}$ $\Rightarrow\cos\text{A}=\frac{3}{5}$ and $\sin\text{B}=\frac{12}{13}$ Now, $\sin(\text{A-B})=\sin\text{A}\cos{\text{B}-\cos\text{A}}\sin\text{B}$ $=\frac{4}{5}\times\frac{5}{13}-\frac{3}{5}\times\frac{12}{13}$ $=\frac{20}{65}-\frac{36}{65}$ $=\frac{20-36}{65}$ $=-\frac{16}{65}$
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Find the maximum and minimum values of each of the following trigonometrical expressions: $5\cos\text{x}+ 3\sin\Big(\frac{\pi}{6}-\text{x}\Big)+4$
AnswerLet $\text{f}(\theta)=5\cos\text{x}+ 3\sin\Big(\frac{\pi}{6}-\text{x}\Big)+4$ Then, $\text{f}(\theta)=5\cos\text{x}+ 3\Big[\sin\frac{\pi}{6}\cos\theta-\cos\frac{\pi}{6}\sin\theta\Big]+4$ $=5\cos\text{x}+ 3\Big[\frac{1}{2}\cos\theta-\frac{\sqrt{3}}{2}\sin\theta\Big]+4$ $=5\cos\text{x}+ \frac{3}{2}\cos\theta-\frac{3\sqrt{3}}{2}\sin\theta+4$ $=\Big(5+ \frac{3}{2}\Big)\cos\theta-\frac{3\sqrt{3}}{2}\sin\theta+4$ $=\frac{13}{2}\cos\theta-\Big(\frac{-3\sqrt{3}}{2}\Big)\sin\theta+4$ We know that, $-\sqrt{\Big(\frac{13}{2}\Big)^2+\Big(\frac{-3\sqrt{3}}{2}\Big)^2}\leq\frac{13}{2}\cos\theta-\Big(\frac{-3\sqrt{3}}{2}\Big)\sin\theta\leq\sqrt{\Big(\frac{13}{2}\Big)^2+\Big(\frac{-3\sqrt{3}}{2}\Big)^2}$ $\Rightarrow-\sqrt{\frac{169}{4}+\frac{27}{4}}\leq\frac{13}{2}\cos\theta-\Big(\frac{-3\sqrt{3}}{2}\Big)\sin\theta\leq\sqrt{\Big(\frac{13}{2}\Big)^2+{}{4}}$ $\Rightarrow-\sqrt{\frac{196}{4}}+\leq\frac{13}{2}\cos\theta-\Big(\frac{-3\sqrt{3}}{2}\Big)\sin\theta\leq\sqrt{\frac{196}{4}}$ $\Rightarrow-\frac{14}{2}\leq\frac{13}{2}\cos\theta-\frac{-3\sqrt{3}}{2}\sin\theta\leq\frac{14}{2}$ $\Rightarrow-7\leq\frac{13}{2}\cos\theta-\frac{-3\sqrt{3}}{2}\sin\theta\leq7$ $\Rightarrow-7+4\leq\frac{13}{2}\cos\theta-\frac{-3\sqrt{3}}{2}\sin\theta+4\leq7+4$ $\Rightarrow-3\leq\frac{13}{2}\cos\theta-\Big(\frac{-3\sqrt{3}}{2}\Big)\sin\theta+4\leq11$ $\Rightarrow-3\leq\text{f}(0)\leq11$ Hence, $-3\text{ and }11$ are respectively the minimum and the maximum valus of $5\cos\theta+3\sin\Big(\frac{\pi}{6}-\theta\Big)+4$
View full question & answer→Question 313 Marks
Find the maximum and minimum values of each of the following trigonometrical expressions: $12\cos\text{x}+ 5\sin\text{x}+4$
AnswerLet $\text{f}(\theta)=12\sin\theta+ 5\sin\theta+4$ We know that, $-\sqrt{(12)^2+(-5)^2}\leq12\cos\theta+5\sin\theta\leq\sqrt{(12)^2+(-5)^2}$ $\Rightarrow-\sqrt{144+25}\leq12\cos\theta+5\sin\theta\leq\sqrt{144+25}$ $\Rightarrow-\sqrt{169}\leq12\cos\theta+5\sin\theta\leq\sqrt{169}$ $\Rightarrow-13\leq12\cos+5\sin\theta\leq13$ $\Rightarrow-13+4\leq12\cos\theta+5\sin\theta+4\leq13+4$ $\Rightarrow-9\leq12\cos\theta+5\sin\theta+4\leq17$ $\Rightarrow-9\leq\text{f}(\theta)\leq17$ Hence, minimum and maximum values of $12\sin\theta-5\cos\theta+4$ are $-9 \text{ and }17$ respectively.
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