MCQ 11 Mark
The value of $\frac{\sin5\alpha-\sin\beta}{\cos5\alpha+2\cos4\alpha+\cos3\alpha}$ is:
- A
$\cot\frac{\alpha}{2}$
- B
$\cot\alpha$
- ✓
$\tan\frac{\alpha}{2}$
- D
AnswerCorrect option: C. $\tan\frac{\alpha}{2}$
$\frac{\sin5\alpha-\sin3\alpha}{\cos5\alpha+2\cos4\alpha+\cos3\alpha}=\frac{\sin5\alpha-\sin3\alpha}{\cos5\alpha+\cos3\alpha+2\cos4\alpha}$
$=\frac{2\sin\alpha\cos4\alpha}{2\cos4\alpha\cos\alpha+2\cos4\alpha}$
$=\frac{2\sin\alpha\cos4\alpha}{2\cos4\alpha(\cos\alpha+1)}$
$=\frac{\sin\alpha}{\cos\alpha+1}$
$=\frac{2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{\cos^2\frac{\alpha}{2}-\sin^2\frac{\alpha}{2}+\sin^2\frac{\alpha}{2}+\cos^2\frac{\alpha}{2}}$
$=\frac{2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{2\cos^2\frac{\alpha}{2}}$
$=\frac{\sin\frac{\alpha}{2}}{\cos^2\frac{\alpha}{2}}$
$=\tan\frac{\alpha}{2}$
View full question & answer→MCQ 21 Mark
The value of $\tan\text{x}+\tan\Big(\frac{\pi}{3}+\text{x}\Big)+\tan\Big(\frac{2\pi}{3}+\text{x}\Big)$ is:
- ✓
$3\tan3\text{x}$
- B
$\tan3\text{x}$
- C
$3\cot3\text{x}$
- D
$\cot3\text{x}$
AnswerCorrect option: A. $3\tan3\text{x}$
$\frac{\pi}{3}=60^\circ,\frac{2\pi}{3}=120^\circ$
$\tan\text{x}+\tan(60^\circ+\text{x})+\tan(120^\circ+\text{x})\\=\tan\text{x}+\frac{\tan60^\circ+\tan\text{x}}{1-\tan60^\circ\tan\text{x}}+\frac{\tan120^\circ+\tan\text{x}}{1-\tan120^\circ\tan\text{x}}$
$\tan\text{x}+\frac{\sqrt{3}+\tan\text{x}}{1-\sqrt{3}\tan\text{x}}+\frac{(1\sqrt{3}+\tan\text{x})}{1+\sqrt{3}\tan\text{x}}$
$=\frac{\tan\text{x}(1-3\tan^2\text{x})+(\sqrt{3+\tan\text{x}})(1+\sqrt{3}\tan\text{x})+(-\sqrt{3}+\tan\text{x})(1-\sqrt{3}\tan\text{x})}{1-3\tan^2\text{x}}$
$=\frac{\tan\text{x}-3\tan^3\text{x}+\sqrt{3}+3\tan\text{x}+\sqrt{3}\tan^2+\tan^2\text{x}+\tan\text{x}-\sqrt{3}\tan^2\text{x}-\sqrt{3}+3\tan\text{x}}{1-3\tan^2\text{x}}$
$=\frac{9\tan\text{x}-3\tan^3\text{x}}{1-3\tan^2\text{x}}$
$=3\tan3\text{x}$
View full question & answer→MCQ 31 Mark
The value of $\Big(\cot\frac{\text{x}}{2}-\tan\frac{\text{x}}{2}\Big)^2(1-2\tan\text{x}\cot2\text{x})$ is:
AnswerWe have,
$\big(\cot\frac{\text{x}}{2}-\tan\frac{\text{x}}{2}\big)(1-2\tan\text{x}\cot2\text{x})$
$\big(\cot^2\frac{\text{x}}{2}-2\cot\frac{\text{x}}{2}+\tan^2\frac{\text{x}}{2}\big)\Big\{1-2\tan\text{x}\Big(\frac{\cot^2\text{x}-1}{2\cot\text{x}}\Big)\Big\}$
$\big(\cot^2\frac{\text{x}}{2}-2+\tan^2\frac{\text{x}}{2}\big)\Big\{1-\tan\text{x}\Big(\frac{\cot^2\text{x}-1}{\cot\text{x}}\Big)\Big\}$
$\big(\cot^2\frac{\text{x}}{2}+\tan^2\frac{\text{x}}{2}-2\big)\Big(1-\frac{\cot^2\text{x}-\tan\text{x}}{\cot\text{x}}\Big)$
$\big(\cot^2\frac{\text{x}}{2}+\tan^2\frac{\text{x}}{2}-2\big)\big(\tan^2\text{x}\big)$
$\big(\cot^2\frac{\text{x}}{2}+\tan\frac{\text{x}}{2}-2\big)\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1-\tan^2\frac{\text{x}}{2}}\Bigg)^2$
$=\frac{1}{\big(1-\tan^2\frac{\text{x}}{2}\big)^2}\big(4+4\tan^4\frac{\text{x}}{2}-8\tan^2\frac{\text{x}}{2}\big)$
$=\frac{1}{\big(1-\tan^2\frac{\text{x}}{2}\big)^2}\big(4-8\tan^2\frac{\text{x}}{2}+4\tan^4\frac{\text{x}}{2}\big)$
$=\frac{1}{\big(1-\tan^2\frac{\text{x}}{2}\big)^2}\Big\{\big(\tan^2\frac{\text{x}}{2}\big)^2-2\big(\tan^2\frac{\text{x}}{2}+1\big)\Big\}$
$=\frac{4\big(\tan^2\frac{\text{x}}{2}-1\big)^2}{\big(1-\tan^2\frac{\text{x}}{2}\big)^2}$
$=4$
View full question & answer→MCQ 41 Mark
If $\tan\text{X}=\frac{\text{a}}{\text{b}},$ then $\text{b}\cos2\text{x}+\text{a}\sin2\text{x}$ is equal to:
AnswerCorrect option: B. $\text{b}$
Givan: $\tan\text{x}=\frac{\text{a}}{\text{b}}$
Now,
$=\cos2\text{x}+\alpha\sin2\text{x}$
$=\text{b}\Big(\frac{1-\tan^2\text{x}}{1+\tan^2\text{x}}\Big)+\text{a}\Big(\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big)$
$=\text{b}\Bigg(\frac{1-\frac{\text{a}^2}{\text{b}^2}}{1+\frac{\text{a}^2}{\text{b}^2}}\Bigg)+\text{a}\Bigg(\frac{1\times\frac{\text{a}}{\text{b}}}{1+\frac{\text{a}^2}{\text{b}^2}}\Bigg)$
$=\frac{\text{b}(\text{b}^2-\text{a}^2)}{\text{a}^2+\text{b}^2}+\frac{2\text{a}^2\text{b}^2}{\text{a}^2+\text{b}^2}$
$=\frac{\text{b}^3-\text{a}^2\text{b}+2\text{a}^2\text{b}}{\text{a}^2+\text{b}^2}$
$=\frac{\text{b}^3+\text{a}^2\text{b}}{\text{a}^2+\text{b}^2}$
$=\frac{\text{b}(\text{b}^2+\text{a}^2)}{\text{a}^2+\text{b}^2}$
$=\text{b}$
View full question & answer→MCQ 51 Mark
The value of $\sin^2\Big(\frac{\pi}{18}\Big)+\sin^2\Big(\frac{\pi}{9}\Big)+\sin^2\Big(\frac{7\pi}{18}\Big)+\sin^2\Big(\frac{4\pi}{9}\Big)$ is:
AnswerWe have,
$\sin^2\Big(\frac{\pi}{18}\Big)+\sin^2\Big(\frac{7\pi}{18}\Big)+\sin^2\Big(\frac{4\pi}{9}\Big)$
$=\frac{1}{2}\Big[1-\cos\big(\frac{\pi}{9}\big)+1-\cos\big(\frac{2\pi}{9}\big)+1-\cos\frac{7\pi}{9}+1-\cos\frac{8\pi}{9}\Big]$ $\Big(\therefore\sin^2\theta=\frac{1-\cos2\theta}{2}\Big)$
$=\frac{1}{2}\Big[4-\cos\big(\frac{\pi}{9}\big)-\cos\big(\frac{2\pi}{9}\big)-\Big\{-\cos\Big(\pi-\frac{7\pi}{9}\Big)\Big\}-\big\{-\cos(\pi-\frac{8\pi}{9})\big\}\Big]$
$=\frac{1}{2}\Big[4-\cos\big(\frac{\pi}{9}\big)-\cos\big(\frac{2\pi}{9}\big)+\cos\big(\frac{2\pi}{9}\big)+\cos\big(\frac{\pi}{9}\big)\Big]$
$=\frac{4}{2}$
$=2$
View full question & answer→MCQ 61 Mark
If $\cos2\text{x}+2\cos\text{x}=1$ then, $(2-\cos^2\text{x})\sin^2\text{x}$ is equal to:
- ✓
$1$
- B
$-1$
- C
$-\sqrt{5}$
- D
$\sqrt{5}$
AnswerWe have,
$\Rightarrow2\cos^2\text{x}-1+2\cos\text{x}=1$
$\Rightarrow\cos^2\text{x}+\cos\text{x}-1=0$
$\Rightarrow\cos\text{x}=\frac{-1\pm\sqrt{1^2+4}}{2}$
$\Rightarrow\cos\text{x}=\frac{-1\pm\sqrt{5}}{2}$
$\Rightarrow\cos\text{x}=\frac{-1+\sqrt{5}}{2}$
Now,
$(2-\cos^2\text{x})\sin^2\text{x}=\bigg[2-\Big(\frac{-1+\sqrt{5}}{2}\Big)^2\bigg](1-\cos^2\text{x})$
$=\bigg[2-\frac{1}{4}\Big(1-2\sqrt{5}+5\Big)\bigg]\Big(1-\frac{1}{4}\Big(1-2\sqrt{5}+5\Big)\Big)$
$=\frac{1}{4}\Big(1+\sqrt{5}\Big)\Big(\sqrt{5}-1\Big)=\frac{4}{4}=1$
View full question & answer→MCQ 71 Mark
If $2\tan\alpha=3\tan\beta$ then $\tan(\alpha-\beta)=$
- ✓
$\frac{\sin2\beta}{5-\cos2\beta}$
- B
$\frac{\cos2\beta}{5-\cos2\beta}$
- C
$\frac{\sin2\beta}{5+\cos2\beta}$
- D
AnswerCorrect option: A. $\frac{\sin2\beta}{5-\cos2\beta}$
Given:
$2\tan\alpha=3\tan\beta$
Now,
$\tan(\alpha+\beta)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$
$=\frac{\frac{3}{2}\tan\beta-\tan\beta}{1+\Big(\frac{3}{2}\tan\beta\Big)\tan\beta}$
$=\frac{3\tan\beta-2\tan\beta}{2+3\tan^2\beta}$
$=\frac{\tan\beta}{2+3\tan^2\beta}$
$=\frac{\frac{\sin\beta}{\cos\beta}}{2+3\frac{\sin^2\beta}{\cos^2\beta}}$
$=\frac{\sin\beta\cos\beta}{2\cos\beta+3\sin\beta}$
$=\frac{\sin\beta\cos\beta}{2+\sin^2\beta}$
$=\frac{2\sin\beta\cos\beta}{4+2-\cos2\beta}$
$=\frac{\sin2\beta}{4+1-cos2\beta}$
$\therefore\tan(\alpha+\beta)=\frac{\sin2\beta}{5-\cos2\beta}$
View full question & answer→MCQ 81 Mark
If $\tan\alpha=\frac{1}{7},\tan\beta=-\frac{1}{3},$ then $\cos2\alpha$ is equal to:
- A
$\sin2\beta$
- ✓
$\sin4\beta$
- C
$\sin3\beta$
- D
$\cos^2\beta$
AnswerCorrect option: B. $\sin4\beta$
it is given thet $\tan\alpha=\frac{1}{7}$ and $\tan\beta=\frac{1}{3}.$
Now,
$\tan\beta=\frac{2\tan\beta}{1-\tan^2\beta}$
$=\frac{2\times\frac{1}{3}}{1-\frac{1}{9}}$
$=\frac{\frac{2}{3}}{\frac{8}{9}}$
$=\frac{3}{4}$
$\therefore\tan(\alpha+2\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$
$=\frac{\frac{1}{7}+\frac{3}{4}}{1-\frac{1}{7}\times\frac{3}{4}}$
$=\frac{\frac{25}{28}}{\frac{25}{28}}$
$=1$
$\tan(\alpha+2\beta)=1=\tan\frac{\pi}{4}$
$\Rightarrow\alpha+2\beta=\frac{\pi}{4}$
$\Rightarrow\alpha=\frac{\pi}{4}-2\beta$
$\Rightarrow2\alpha=\frac{\pi}{2}-4\beta$
$\Rightarrow\cos2\alpha=\cos\big(\frac{\pi}{2}-4\beta\big)=\sin4\beta$
$\therefore\cos2\alpha=\sin4\beta$
Hence, the correct answer is option B.
View full question & answer→MCQ 91 Mark
If $\tan\frac{\text{x}}{2}=\sqrt{\frac{1-\text{e}}{1+\text{e}}}\tan\frac{\alpha}{2}$ then $\cos\alpha=$
- A
$1-\text{e}\cos(\cos\text{x}+\text{e})$
- B
$\frac{1+\text{e}\cos\text{x}}{\cos\text{x}-\text{e}}$
- C
$\frac{1-\text{e}\cos\text{x}}{\cos\text{x}-\text{e}}$
- ✓
$\frac{\cos\text{x}-\text{e}}{1-\text{e}\cos\text{x}}$
AnswerCorrect option: D. $\frac{\cos\text{x}-\text{e}}{1-\text{e}\cos\text{x}}$
Given:
$\tan\frac{\text{x}}{2}=\sqrt{\frac{1-\text{e}}{1+\text{e}}}\tan\frac{\alpha}{2}$
$\Rightarrow\frac{\tan\frac{\text{x}}{2}}{\tan\frac{\alpha}{2}}=\sqrt{\frac{1-\text{e}}{1+\text{e}}}$
Squaring both sides, we get,
$\frac{\tan^2\frac{\text{x}}{2}}{\tan^2\frac{\alpha}{2}}=\frac{1-\text{e}}{1+\text{e}}$
$\Rightarrow\tan^2\frac{\alpha}{2}(1-\text{e})=\tan^2\frac{\text{x}}{2}(1+\text{e})$
$\Rightarrow\frac{\sin^2\frac{\alpha}{2}}{\cos^2\frac{\alpha}{2}}(1-\text{e})=\frac{\sin^2\frac{\text{x}}{2}}{\cos^2\frac{\text{x}}{2}}(1+\text{e})$
$\Rightarrow\frac{\frac{1}{2}(1-\cos\alpha)}{\frac{1}{2}(1+\cos\alpha)}(1-\text{e})=\frac{\frac{1}{2}(1-\cos\text{x})}{\frac{1}{2}(1+\cos\text{x})}(1+\text{e})$
$\Rightarrow(1-\cos\alpha)(1+\cos\text{x})(1-\text{e})=(1+\cos\alpha)(1-\cos\text{x})(1+\text{x})$
$\Rightarrow(1+\cos\text{x})(1-\text{e})-\cos\alpha(1+\cos\text{x})(1-\text{e})\\=(1-\cos\text{x})(1+\text{e})+\cos\alpha(1-\cos\text{x})(1+\text{e})$
$\Rightarrow\cos\alpha\big\{(1+\cos\text{x})(1-\text{e})+(1-\cos\text{x})(1+\text{e})\big\}\\=(1+\cos\text{x})(1-\text{e})-(1-\cos\text{x})(1+\text{e})$
$\Rightarrow\cos\alpha=\frac{2\cos\text{x}-2\text{e}}{2-2\text{e}\cos\text{x}}=\frac{\cos\text{x}-\text{e}}{1-\text{e}\cos\text{x}}$
View full question & answer→MCQ 101 Mark
If $\cos\text{x}=\frac{1}{2}\Big(\text{a}+\frac{1}{\text{a}}\Big),$ and $3\text{x}=\lambda\Big(\text{a}^3+\frac{1}{\text{a}^3}\Big),$ then $\lambda=$
- A
$\frac{1}{4}$
- ✓
$\frac{1}{2}$
- C
$1$
- D
AnswerCorrect option: B. $\frac{1}{2}$
Given:
$\cos\text{x}=\frac{1}{2}(\text{a}+\frac{1}{\text{a}})$
$\cos3\text{x}=\lambda\Big(\text{a}^2+\frac{1}{a^3}\Big)$
Now,
$\cos^3\text{x}=\frac{1}{8}\Big[\text{a}^3+\frac{1}{\text{a}^2}+3\text{a}\frac{1}{\text{a}^3}+\text{a}\frac{1}{\text{a}}\Big]$
$\cos^3\text{x}=\frac{1}{8}\Big(a^3+\frac{1}{a^3}+32\cos\text{x}\Big)$ $[\because\cos\text{x}=\frac{1}{2}(\text{a}+\frac{1}{\text{a}})]$
$\Rightarrow\cos^3\text{x}=\frac{1}{8}\Big(\frac{\cos^3\text{x}}{\lambda}+6\cos\text{x}\Big)$
$\Rightarrow\cos^3\text{x}=\frac{1}{8}\Big(\frac{4\cos^3\text{x}-3\cos\text{x}}{\lambda}+6\cos\text{x}\Big)$
$\Rightarrow\cos^3\text{x}=\frac{4\cos^3\text{x}}{8\lambda}\frac{3\cos^3\text{x}}{8\lambda}+\frac{6\cos\text{x}}{8}$
On comparing the powers of cos3x on both sides, we get
$1=\frac{4}{8\lambda}$
$1=\frac{4}{8\lambda}$
$\Rightarrow\lambda=\frac{1}{2}$
View full question & answer→MCQ 111 Mark
The value of $2\sin^2\text{B}+4\cos(\text{A+B})\sin\text{A}\sin\text{B}+\cos^2(\text{A+B})$ is:
- A
$0$
- B
$\cos3\text{A}$
- ✓
$\cos2\text{A}$
- D
AnswerCorrect option: C. $\cos2\text{A}$
We have,
$2\sin^2\text{B}+4\cos(\text{A+B})\sin\text{A}\sin\text{B}+\cos2\text{A+B}$
$=1-\cos2\text{B}+\cos^2(\text{A+B})+4\cos(\text{A+B})\sin\text{A}\sin\text{B}$
$=1+(\cos2(\text{A+B})-\cos2\text{B})+4\cos(\text{A+B})\sin\text{A}\sin\text{B}$
$=1-2\sin\text{A}\sin(\text{A+2B})+4\cos(\text{A+B})\sin\text{A}\sin\text{B}\\ \ \ \ \ \ \ \ \ \Big[\because\cos\text{C}\cos\text{D}=-2\sin\frac{\text{C+D}}{2}\sin\frac{\text{C}-\text{D}}{2}\Big]$
$=1-2\sin\text{A}[\sin(\text{A}+2\text{B})-2\sin\text{B}-2\cos(\text{A+B})]$
$=1-2\sin\text{A}[\sin(\text{A}+2\text{B})-\big\{\sin(\text{B+A+B})+\sin(\text{B}-\text{(A+B)})\big\}\\ \ \ \ \ \ [\because2\sin\text{C}\cos\text{D}=\sin(\text{C+D})+\sin(\text{C}-\text{D})]$
$=1-2\sin\text{A}\big[\sin(\text{A+2B})-\big\{\sin(\text{A+2B})+\sin(-\text{A})\big\}\big]$
$=1-2\sin\text{A}[\sin\text{A}]$
$=1-2\sin^2\text{A}$
$=\cos^2\text{A}$
View full question & answer→MCQ 121 Mark
If $\tan\alpha=\frac{1-\cos\beta}{\sin\beta},$ then:
- A
$\tan3\alpha=\tan2\beta$
- ✓
$\tan2\alpha=\tan\beta$
- C
$\tan2\beta=\tan\alpha$
- D
AnswerCorrect option: B. $\tan2\alpha=\tan\beta$
$\tan\alpha=\frac{1-\cos\beta}{\sin\beta}$
$=\frac{2\sin^2\frac{\beta}{2}}{2\sin\frac{\beta}{2}\cos\frac{\beta}{2}}$
$=\frac{\sin\frac{\beta}{2}}{\cos\frac{\beta}{2}}$
$\Rightarrow\tan\alpha=\tan\frac{\beta}{2}$
$\Rightarrow\alpha=\frac{\beta}{2}$
$\Rightarrow2\alpha=\beta$
$\therefore\tan2\alpha=\tan\beta$
View full question & answer→MCQ 131 Mark
The value of $\cos^4\text{x}+\sin^4\text{x}-6\cos^2\sin^2$ is:
- A
$\cos2\text{x}$
- B
$\sin2\text{x}$
- ✓
$\cos4\text{x}$
- D
AnswerCorrect option: C. $\cos4\text{x}$
$\cos^4\text{x}+\sin^4\text{x}-6\cos^2\text{x}\sin^2\text{x}=\cos^4\text{x}\\+\sin^4\text{x}-2\cos^2\text{x}\sin^2\text{x}-4\cos^2\text{x}\sin^2\text{x}$
$=(\cos^2\text{x}-\sin^2\text{x})^2-(2\sin\text{x}\cos\text{x})^2$
$=\cos^22\text{x}-\sin^22\text{x}$
$=\cos4\text{x}$
View full question & answer→MCQ 141 Mark
If $\tan\Big(\frac{\pi}{4}+\text{x}\Big)+\tan\Big(\frac{\pi}{4}-\text{x}\Big)=\lambda\sec2\text{x},$ then:
AnswerGiven:
$\tan\big(\frac{\pi}{4}+\text{x}+\tan\big(\frac{\pi}{4}-\text{x}\big)=\lambda\sec2\text{x}$
$\Rightarrow\frac{\tan\frac{\pi}{4}+\tan\text{x}}{1-\tan\frac{\pi}{4}\times\tan\text{x}}+\frac{\tan\frac{\pi}{4}-\tan\text{x}}{1+\tan\frac{\pi}{4}\times\tan\text{x}}=\lambda2\text{x}$
$\Rightarrow\frac{1+\tan\text{x}}{1-\tan\text{x}}+\frac{1-\tan\text{x}}{1+\tan\text{x}}=\lambda\sec2\text{x}$
$\Rightarrow\frac{(1+\tan\text{x})^2+(1-\tan\text{x})^2}{(1-\tan\text{x})(1+\tan\text{x})}=\lambda\sec2\text{x}$
$\Rightarrow\frac{2(1+\tan^2\text{x})}{1-\tan^2\text{x}}=\lambda\sec2\text{x}$
$\Rightarrow\frac{2\sec^2\text{x}}{1-\tan^2\text{x}}=\lambda\sec2\text{x}$
$\Rightarrow\frac{2}{\cos^2(1-\tan^2\text{x})}=\lambda\sec2\text{x}$
$\Rightarrow\frac{2}{\cos^2\text{x}\Big(1-\frac{\sin^2\text{x}}{\cos^2\text{x}}\Big)}=\lambda\sec2\text{x}$
$\Rightarrow\frac{2}{\cos^2\text{x}-\sin^2\text{x}}=\lambda\sec2\text{x}$
$\Rightarrow\frac{2}{\cos2\text{x}}=\lambda\sec2\text{x}$
$\Rightarrow2\sec2\text{x}=\lambda\sec2\text{x}$
$\Rightarrow2=\lambda$
$\therefore\lambda=2$
View full question & answer→MCQ 151 Mark
If $5\sin\alpha=3\sin(\alpha+2\beta)\not=0,$ then $\tan(\alpha+\beta)$ is equal to:
- A
$2\tan\beta$
- B
$3\tan\beta$
- ✓
$4\tan\beta$
- D
$6\tan\beta$
AnswerCorrect option: C. $4\tan\beta$
We have,
$5\sin\alpha=3\sin(\alpha+2\beta)$
$\Rightarrow\frac{5}{3}=\frac{\sin(\alpha+2\beta)}{\sin\alpha}$
$\Rightarrow\frac{5-3}{5+3}=\frac{\sin(\alpha+2\beta)-\sin\alpha}{\sin(\alpha+2\beta)+\sin\alpha}$ (using componendo and dividendo)
$\Rightarrow\frac{2}{8}=\frac{\sin(\alpha+2\beta)-\sin\alpha}{\sin(\alpha+2\beta)+\sin\alpha}$
$\Rightarrow\frac{1}{4}=\frac{2\cos\frac{\alpha+2\beta+\alpha}{2}\sin\frac{\alpha+2\beta-\alpha}{2}}{2\sin\frac{\alpha+2\beta+\alpha}{2}\cos\frac{\alpha+2\beta-\alpha}{2}}$
$\Rightarrow\frac{1}{4}\frac{\cos(\alpha+\beta)\sin\beta}{\sin(\alpha+\beta)\cos\beta}$
$\Rightarrow\frac{1}{4}=\cot(\alpha+\beta)\tan\beta$
$\Rightarrow\frac{1}{4}=\frac{1}{\tan(\alpha+\beta)}\tan\beta$
$\therefore\tan(\alpha+\beta)=4\tan\beta$
View full question & answer→MCQ 161 Mark
If in a $\Delta\text{ABC},\tan\text{A}+\tan\text{B}+\tan\text{C=0,}$ then $\cot\text{A}\cot\text{B}\cot\text{C}=$
AnswerABC is a tringle.
$\therefore\text{A}+\text{B}+\text{C}=\pi$
$\Rightarrow\text{A}+\text{B}+\pi-\text{C}$
$\Rightarrow\tan(\text{A}+\text{B})=\tan(\pi-\text{C})$
$\Rightarrow\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}=-\tan\text{C}$
$\Rightarrow\tan\text{A}+\tan\text{B}=-\tan\text{C}+\tan\text{A}\tan\text{B}\tan\text{C}$
$\Rightarrow\tan\text{A}+\tan\text{B}+\tan\text{C}=\tan\text{A}\tan\text{B}\tan\text{C}$
$\Rightarrow0=\tan\text{A}+\tan\text{B}\tan\text{C}$ $\big[$ given: $\tan\text{A}\tan\text{B}\tan\text{C}=0\big]$
$\Rightarrow\tan\text{A}\tan\text{B}\tan\text{C}=0$
$\Rightarrow\frac{1}{\tan\text{A}\tan\text{B}\tan\text{C}}=\frac{1}{0}$
$\Rightarrow\cot\text{A}\cot\text{B}\cot\text{C}\rightarrow\infty$
View full question & answer→MCQ 171 Mark
$\frac{\sin5\text{x}}{\sin\text{x}}$ is equal to:
- ✓
$16\cos^4-12\cos^2\text{x}+1$
- B
$16\cos^4\text{x}+12\cos^2\text{x}+1$
- C
$16\cos^4\text{x}-12\cos^2\text{x}-1$
- D
$16\cos^4\text{x}+12\cos^2\text{x}-1$
AnswerCorrect option: A. $16\cos^4-12\cos^2\text{x}+1$
To find: $\frac{\sin5\text{x}}{\sin\text{x}}$
Now,
$\sin5\text{x}=\sin(3\text{x}+2\text{x})$
$=(3\sin\text{x}-4\sin^3\text{x})(1-2\sin^2\text{x})+(4\cos^3\text{x}-3\cos\text{x})(2\sin\text{x}\cos\text{x})$
$=(3\sin\text{x}-4\sin^3\text{x}-4\sin^3\text{x}+8\sin^5\text{x})+2\sin\text{x}\cos^2\text{x}(4\cos^2\text{x}-3)$
$=(3\sin\text{x}-10\sin^3\text{x}+8\sin^5\text{x}+2\sin\text{x}(1-\sin^2\text{x})[2(1-\sin^2\text{x})-3]$
$=(3\sin\text{x}-10\sin^3\text{x}+8\sin^5\text{x}+(2\sin\text{x}-2\sin^3\text{x})(4-4\sin^2\text{x}-3)]$
$=(3\sin\text{x}-10\sin^3\text{x}+8\sin^5\text{x}+(2\sin\text{x}-8\sin^3\text{x}2\sin^3\text{x}+8\sin^5\text{x})]$
$=5\sin\text{x}-20\sin^3+16\sin^5\text{x}$
$\therefore\frac{\sin5\text{x}}{\sin\text{x}}=\frac{5\sin\text{x}-20\sin^3\text{x}+16^5\text{x}}{\sin\text{x}}$
$=5-20\sin^2\text{x}+16\sin^4\text{x}$
$=5-20(1-\cos^2\text{x})+16(1-\cos^2\text{x})^2$
$=5-20+20\cos^2\text{x}+16(1+\cos^4\text{x}-2\cos^4\text{x})$
$=5-20+20\cos^2\text{x}+16+16\cos^4\text{x}-32\cos^2\text{x}$
$=16\cos^4-12\cos^2\text{x}+1$
View full question & answer→MCQ 181 Mark
If $\alpha$ and $\beta$ are acute angles satisfying $\cos2\alpha=\frac{3\cos2\beta-1}{3-\cos2\beta},$ then $\tan\alpha=$
AnswerCorrect option: A. $\sqrt{2}\tan\beta$
Given:
$\cos2\alpha=\frac{3\cos2\beta-1}{3-\cos2\beta}$
$\Rightarrow\frac{\cos2\alpha-1}{\cos2\alpha+1}=\frac{(3\cos2\beta-1)-(3-\cos2\beta)}{(3\cos2\beta-1)+(3-\cos2\beta)}$ (Using componendo and dividendo)
$\Rightarrow\frac{\cos2\alpha-1}{\cos2\alpha+1}=\frac{4\cos2\beta-4}{2\cos2\beta+2}$
$\Rightarrow-=\frac{1-\cos^2\alpha}{1+\cos2\alpha}=\frac{-4(1-\cos2\beta)}{2(1+\cos2\beta)}$
$\Rightarrow\frac{1-\cos2\alpha}{1+\cos2\alpha}=\frac{2(1-\cos2\beta)}{(1+\cos2\beta)}$
$\Rightarrow\frac{2\sin^2\alpha}{2\cos^2\alpha}=\frac{2(2\sin^2\beta)}{2\cos^2\beta}$
$\Rightarrow\tan^2\alpha=2\tan^2\beta$
$\therefore\tan\alpha=\sqrt{2}\tan\beta$
View full question & answer→MCQ 191 Mark
$2(1-2\sin^27\text{x})\sin3\text{x}$ is equal to:
- ✓
$\sin17\text{x}-\sin11\text{x}$
- B
$\sin11\text{x}-\sin17\text{x}$
- C
$\cos17\text{x}-\cos11\text{x}$
- D
$\cos17\text{x}+\cos11\text{x}$
AnswerCorrect option: A. $\sin17\text{x}-\sin11\text{x}$
We have,
$2(1-2\sin^27\text{x})\sin3\text{x}=2(\cos14\text{x})\sin3\text{x}$ $[\because\cos2\text{x}=1-2\sin^2\text{x}]$
$=2\sin3\text{x}\cos14\text{x}$
$=\sin17\text{x}-\sin11\text{x}$ $[\because2\sin\text{A}\cos\text{xB}=\sin(\text{A+B})-\sin(\text{A}-\text{B})]$
$\therefore2(1-2\sin^27\text{x})\sin3\text{x}=\sin17\text{x}-\sin11\text{x}$
View full question & answer→MCQ 201 Mark
If n = 1,2,3, ..., then $\cos\alpha\cos4\alpha\ ...\cos2^{\text{n}-1}\alpha$ is equal to:
- A
$\frac{\sin2\text{n}\alpha}{2\text{n}\sin\alpha}$
- B
$\frac{\sin2^\text{n}\alpha}{2^\text{n}\sin2^{\text{n}-1}\alpha}$
- C
$\frac{\sin4^{\text{n}-1}\alpha}{4^{\text{n}-1}\sin\alpha}$
- ✓
$\frac{\sin4^{\text{n}-1}\alpha}{4^{\text{n}-1}\sin\alpha}$
AnswerCorrect option: D. $\frac{\sin4^{\text{n}-1}\alpha}{4^{\text{n}-1}\sin\alpha}$
$\therefore\cos\alpha\cos2\alpha\cos4\alpha\ ...\cos2^{\text{n}-1}\alpha=\frac{\sin2^\text{n}\alpha}{2^\text{n}\sin\alpha}$
View full question & answer→MCQ 211 Mark
The value of $\frac{2(\sin2\text{x}+2\cos^2\text{x}-1)}{\cos\text{x}-\sin\text{x}-\cos3\text{x}+\sin3\text{x}}$ is:
- A
$\cos\text{x}$
- B
$\sec\text{x}$
- ✓
$\text{cosec}\ \text{x}$
- D
$\sin\text{x}$
AnswerCorrect option: C. $\text{cosec}\ \text{x}$
We have,
$\frac{2(\sin2\text{x}+2\cos^2\text{x}-1)}{\cos\text{x}-\sin\text{x}-\cos3\text{x}+\sin3\text{x}}$
$=\frac{2(\sin2\text{x}+\cos2\text{x})}{\cos\text{x}-\sin\text{x}-4\cos^3\text{x}+3\cos\text{x}\sin\text{x}-4\sin^3\text{x}}$
$=\frac{2(\sin2\text{x}+\cos2\text{x})}{4\cos\text{x}-4\cos^3\text{x}+2\sin\text{x}-4\sin^3\text{x}}$
$=\frac{2(\sin2\text{x}+\cos2\text{x})}{4\cos\text{x}(1-\cos^2\text{x})+2\sin\text{x}(1-2\sin^2\text{x})}$
$=\frac{2(\sin2\text{x}+\cos2\text{x})}{4\cos\text{x}\sin^2\text{x}+2\sin\text{x}\cos2\text{x}}$
$=\frac{2(\sin2\text{x}+\cos2\text{x})}{2\times2\sin\text{x}\cos\text{x}\sin\text{x}+2\sin\text{x}\cos2\text{x}}$
$=\frac{2(\sin2\text{x}+\cos2\text{x})}{2\sin2\text{x}\sin2\text{x}+2\sin\text{x}\cos2\text{x}}$
$=\frac{2(\sin2\text{x}+\cos2\text{x})}{2\sin\text{x}(\sin2\text{x}+\cos2\text{x})}$
$=\frac{1}{\sin\text{x}}$
$=\text{cosec}\ \text{x}$
View full question & answer→MCQ 221 Mark
$\frac{\sin3\text{x}}{1+2\cos2\text{x}}$ is equal to:
- A
$\cos\text{x}$
- ✓
$\sin\text{x}$
- C
$-\cos\text{x}$
- D
$\sin\text{x}$
AnswerCorrect option: B. $\sin\text{x}$
We have,
$\frac{\sin3\text{x}}{1+2\cos2\text{x}}=\frac{3\sin\text{x}-4\sin^2\text{x}}{1+2(1-2\sin^2\text{x})}$
$=\frac{3\sin\text{x}-4\sin^3\text{x}}{1+2-4\sin\text{x}}$
$=\frac{\sin\text{x}(3-4\sin^2\text{x})}{(3-4\sin^2\text{x})}$
$=\sin\text{x}$
View full question & answer→MCQ 231 Mark
The value of $\cos^248^\circ-\sin^212^\circ$ is:
AnswerCorrect option: A. $\frac{\sqrt{5}+1}{8}$
$\cos^248^\circ-\sin^212^\circ$
$=\cos(48^\circ+12^\circ)\cos(48^\circ-12^\circ)$ $[\cos(\text{A+B})\cos(\text{A}-\text{B})=\cos^2\text{A}-\sin^2\text{B}]$
$=\cos60^\circ\cos36^\circ$
$=\frac{1}{2}\times\Big(\frac{\sqrt{5}+1}{4}\Big)$
$=\frac{\sqrt{5}+1}{8}$
Hence, the correct answer is option A.
View full question & answer→MCQ 241 Mark
The value of $\tan\text{x}\sin\Big(\frac{\pi}{2}+\text{x}\Big)\cos\Big(\frac{\pi}{2}-\text{x}\Big)$ is:
AnswerWe have,
$\tan\theta\sin\big(\frac{\pi}{2}+\text{x}\cos\big)\big(\frac{\pi}{2}-\text{x}\big)$
$=\frac{\sin\text{x}}{\cos\text{x}}\cos\text{x}\sin\text{x}$
$=\sin^2\text{x}$
View full question & answer→MCQ 251 Mark
If $\tan\text{x}=\text{t}$ then $\tan2\text{x}+\sec2\text{x}$ is equal to:
- ✓
$\frac{1+\text{t}}{1-\text{t}}$
- B
$\frac{1-\text{t}}{1+\text{t}}$
- C
$\frac{2\text{t}}{1-\text{t}}$
- D
$\frac{2\text{t}}{1+\text{t}}$
AnswerCorrect option: A. $\frac{1+\text{t}}{1-\text{t}}$
$\tan2\text{x}+\sec2\text{x}=\frac{2\tan\text{x}}{1-\tan^2\text{x}}+\frac{1+\tan^2\text{x}}{1-\tan^2\text{x}}$
$=\frac{2\tan\text{x}+1\tan^2\text{x}}{1-\tan^2\text{x}}$
$=\frac{(1+\tan\text{x})^2}{1-\tan^2\text{x}}$
$=\frac{(1+\tan\text{x})(1+\tan\text{x})}{(1+\tan\text{x})(1-\tan\text{x})}$
$=\frac{1+\tan\text{x}}{1-\tan\text{x}}$
$\frac{1+\text{t}}{1-\text{t}}$ $[\tan\text{x}=\text{t}]$ (given)
View full question & answer→MCQ 261 Mark
For all real values of $\text{x},\cot\text{x}-2\cot\text{}$ is equal to:
- A
$\tan2\text{x}$
- ✓
$\tan\text{x}$
- C
$-\cot3\text{x}$
- D
AnswerCorrect option: B. $\tan\text{x}$
We have,
$\cot\text{x}-2\cot2\text{x}=\cot\text{x}-2\frac{\cot^2\text{x}-1}{2\cot\text{x}}$
$=\frac{1}{\cot\text{x}}$
$=\tan\text{x}$
View full question & answer→MCQ 271 Mark
The value of $\cos\frac{\pi}{65}\cos\frac{2\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}$ is:
- A
$\frac{1}{8}$
- B
$\frac{1}{16}$
- C
$\frac{1}{32}$
- ✓
AnswerWe have,
$\cos\frac{\pi}{65}\cos\frac{2\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}$
$=\frac{2\sin\frac{\pi}{65}}{2\sin\frac{\pi}{65}}\cos\frac{\pi}{65}\cos\frac{2\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}$
$\big($dividing and multiplying by $2\sin\frac{\pi}{65}\big)$
$=\frac{2\sin\frac{2\pi}{65}}{2\times2\sin\frac{\pi}{65}}\cos\frac{2\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}$
$=\frac{2\sin\frac{4\pi}{65}}{2\times4\sin\frac{\pi}{65}}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}$
$=\frac{2\sin\frac{8\pi}{65}}{2\times8\sin\frac{\pi}{65}}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}$
$=\frac{2\sin\frac{16\pi}{65}}{2\times16\sin\frac{\pi}{65}}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}$
$=\frac{2\sin\frac{32\pi}{65}}{2\times32\sin\frac{\pi}{65}}\cos\frac{32\pi}{65}$
$=\frac{\sin\frac{64\pi}{65}}{64\sin\frac{\pi}{65}}$
$=\frac{\sin\big(\pi-\frac{\pi}{65}\big)}{64\sin\frac{\pi}{65}}$
$=\frac{\sin\frac{\pi}{65}}{64\sin\frac{\pi}{65}}$
$=\frac{1}{64}$
View full question & answer→MCQ 281 Mark
$\frac{\sec8\text{A}-1}{\sec4\text{A}-1}$ is equal to:
- A
$\frac{\tan2\text{A}}{\tan8\text{A}}$
- ✓
$\frac{\tan8\text{A}}{\tan2\text{A}}$
- C
$\frac{\cot8\text{A}}{\cot2\text{A}}$
- D
AnswerCorrect option: B. $\frac{\tan8\text{A}}{\tan2\text{A}}$
We have,
$\frac{\sec8\text{A}-1}{\sec4\text{A}-1}=\frac{\frac{1}{\cos2\text{A}}-1}{\frac{1}{\cos}4\text{A}-1}$
$=\frac{\cos4\text{A}}{\cos8\text{A}}\times\frac{1-\cos8\text{A}}{1-\cos4\text{A}}$
$=\frac{\cos4\text{A}}{\cos8\text{A}}\times\frac{2\sin^24\text{A}}{2\sin^22\text{A}}(2\sin^2\theta-1-\cos2\theta)$
$=\frac{(2\cos4\text{A}\sin4\text{A})\sin4\text{A}}{2\times\cos8\text{A}\sin^22\text{A}}$
$=\frac{\sin8\text{A}\sin4\text{A}}{\cos8\text{A}\times\text{2}\sin2\text{A}\times\sin2\text{A}}$
$=\tan8\text{A}\times\frac{2\sin2\text{A}\times\cos2\text{A}}{2\sin2\text{A}\times\sin2\text{A}}$
$=\tan8\text{A}\times\cot2\text{A}$
$=\frac{\tan8\text{A}}{\tan2\text{A}}$
View full question & answer→MCQ 291 Mark
The value of $2\cos\text{x}-\cos3\text{x}-\cos5\text{x}-16\cos^3\text{x}\sin^2\text{x}$ is:
AnswerWe have,
$2\cos\text{x}-\cos3\text{x}-\cos5\text{x}-16\cos^3\text{x}\sin^2\text{x}$
$=2\cos\text{x}-\cos3\text{x}-\cos5\text{x}-16\Big[\frac{\cos3\text{x}+3\cos\text{x}}{4}\times\frac{(1-\cos2\text{x})}{2}\Big]$
$=2\cos\text{x}-\cos3\text{x}-\cos5\text{x}-2\big[(\cos3\text{x}+3\cos\text{x})(1-\cos2\text{x})\big]$
$=2\cos\text{x}-\cos3\text{x}-\cos5\text{x}-2\big[\cos3\text{x}-\cos3\text{x}\cos2\text{x}+3\cos\text{x}\cos2\text{x}\big]$
$=2\cos\text{x}-\cos3\text{x}-\cos5\text{x}-2[\cos3\text{x}+3\cos\text{x}]\\+2\cos3\text{x}\cos2\text{x}+3[2\cos\text{x}\cos2\text{x}]$
$=2\cos\text{x}-\cos3\text{x}-\cos5\text{x}-2[\cos3\text{x}+3\cos\text{x}]+\cos5\text{x}+\cos\text{x}\\\ \ \ +3\cos3\text{x}+3\cos\text{x}[2\cos\text{A}\cos\text{B}=\cos(\text{A+B})+\cos(\text{A}-\text{B})]$
$2\cos\text{x}=\cos3\text{x}-\cos5\text{x}-2\cos3\text{x}-6\cos\text{x}\\+\cos5\text{x}+\cos\text{x}+3\cos3\text{x}+3\cos\text{x}=0$
View full question & answer→MCQ 301 Mark
If $(2^\text{n}+1)\text{x}=\pi,$ then $2^\text{n}\cos\text{x}\cos2\text{x}^2\text{x}\cos^\text{n-1}\text{x}=$
AnswerGiven,
$(2^\text{n}+1)\text{x}=\pi$
$\Rightarrow2^\text{n}\text{x}+\text{x}=\pi$
$\Rightarrow2^\text{n}\text{x}=\pi-\text{x}$
$\Rightarrow\sin2^\text{n}\text{x}=\sin(\pi-\text{x})$
$\Rightarrow\sin2^\text{n}\text{x}=\sin\text{x}\ .....(1)$
$2^\text{n}\cos\text{x}\cos2\text{x}\cos2^2\text{x}\ ...\cos2^{\text{n}-1}\text{x}=2^\text{n}\times\frac{\sin2^\text{n}\text{x}}{2^\text{n}\sin\text{x}}$
$=\frac{\sin2^\text{n}\text{x}}{\sin\text{x}}$
$=\frac{\sin\text{x}}{\sin\text{x}}$ [From (1)]
$=1$
View full question & answer→MCQ 311 Mark
If $\text{A}=2\sin^2\text{x}-\cos2\text{x},$ then A lies in the interval:
- ✓
$[-1,3]$
- B
$[1,2]$
- C
$[-2,4]$
- D
AnswerCorrect option: A. $[-1,3]$
$\text{A}=2\sin^2\text{x}-\cos2\text{x}$
$=2\sin^2\text{x}-(1-2\sin^2\text{x})$
$=4\sin^2\text{x}-1$
$\therefore0\leq\sin^2\text{x}\leq1$
$\Rightarrow4\times0\leq4\times\sin^2\text{x}\leq4\times1$
$\Rightarrow0\leq4\sin^2\text{x}\leq4$
$\Rightarrow0-1\leq4\sin^2\text{x}-1\leq4-1$
$\Rightarrow-1\leq4\sin^2\text{x}-1\leq3$
$\Rightarrow-1\leq\text{A}\leq3$
$\Rightarrow\text{A}\in[-1,3]$
View full question & answer→MCQ 321 Mark
The value of $\tan\text{x}\tan\Big(\frac{\pi}{3}-\text{x}\Big)\tan\Big(\frac{\pi}{3}+\text{x}\Big)$ is:
- A
$\cot3\text{x}$
- B
$2\cot3\text{x}$
- ✓
$\tan3\text{x}$
- D
$3\tan3\text{x}$
AnswerCorrect option: C. $\tan3\text{x}$
$\frac{\pi}{3}=60^\circ$
$\tan\text{x}\tan(60^\circ-\text{x})\tan(60^\circ+\text{x})\\=\tan\text{x}\times\frac{\tan60^\circ-\tan\text{x}}{1+\tan60^\circ\tan\text{x}}\times\frac{\tan60^\circ+\tan\text{x}}{1-\tan60^\circ\tan\text{x}}$
$=\tan\text{x}\times\frac{\sqrt{3}-\tan\text{x}}{1+\sqrt{3\tan\text{x}}}\times\frac{\sqrt{3}+\tan\text{x}}{1-\sqrt{3}\tan\text{x}}$
$=\frac{\tan\text{x}(3-\tan^2\text{x})}{1-3\tan^2\text{x}}$
$=\frac{3\tan\text{x}-\tan^3\text{x}}{1-3\tan^2\text{x}}$
$=\tan3\text{x}$
View full question & answer→MCQ 331 Mark
The value of $\cos(36^\circ-\text{A})\cos(36^\circ+\text{A})+\cos(54^\circ-\text{A})\cos(54^\circ+\text{A})$ is:
- ✓
$\cos2\text{A}$
- B
$\sin2\text{A}$
- C
$\cos\text{A}$
- D
$0$
AnswerCorrect option: A. $\cos2\text{A}$
$\cos(36^\circ-\text{A})\cos(36^\circ+\text{A})+\cos(54^\circ-\text{A})\cos(54^\circ+\text{A})$
$=\cos[90^\circ-(54^\circ+\text{A})]\cos[90^\circ-(54^\circ-\text{A})]+\cos(54^\circ\\-\text{A})\cos(54^\circ+\text{A})$
$=\sin(54^\circ+\text{A})\sin(54^\circ-\text{A})+\cos(54^\circ-\text{A})\cos(54^\circ+\text{A})$ $[\cos(90^\circ-\theta)=\sin\theta]$
$=\cos(54^\circ+\text{A}-54^\circ+\text{A})$ $[\cos(\text{A}-\text{B})=\cos\text{A}\cos\text{B}+\sin\text{A}\sin\text{B}]$
$=\cos2\text{A}$
View full question & answer→MCQ 341 Mark
If $\sin\alpha+\sin\beta=\text{a}$ and $\cos\alpha-\cos\beta=\text{b}$ then $\tan\frac{\alpha-\beta}{2}=$
- A
$-\frac{\text{a}}{\text{b}}$
- ✓
$-\frac{\text{b}}{\text{a}}$
- C
$\sqrt{\text{a}^2+\text{b}^2}$
- D
AnswerCorrect option: B. $-\frac{\text{b}}{\text{a}}$
Given:
$\sin\alpha+\sin\beta=\text{a}$
$\Rightarrow2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha+\beta}{2}=\text{a}\ .....(1)$
Also,
$\cos\alpha+\cos\beta=\text{b}$
$\Rightarrow-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha+\beta}{2}=\text{b}\ .....(2)$
On dividing (1) by (2), we get
$\frac{-\cos\frac{\alpha+\beta}{2}}{\sin\frac{\alpha-\beta}{2}}=\frac{\text{a}}{\text{b}}$
$\Rightarrow\frac{-\sin\frac{\alpha+\beta}{2}}{\cos\frac{\alpha-\beta}{2}}=\frac{\text{b}}{\text{a}}$
$\Rightarrow\tan\frac{\alpha-\beta}{2}=-\frac{\text{b}}{\text{a}}$
View full question & answer→MCQ 351 Mark
The value of $\frac{\cos3\text{x}}{2\cos2\text{x}-1}$ is equal to:
- ✓
$\cos\text{x}$
- B
$\sin\text{x}$
- C
$\tan\text{x}$
- D
AnswerCorrect option: A. $\cos\text{x}$
We have,
$\therefore\frac{\cos3\text{x}}{2\cos2\text{x}-1}=\frac{4\cos^3\text{x}-3\cos\text{x}}{1(2\cos^2\text{x}-1)-1}$ $[\therefore\cos3\text{x}=4\cos^3\text{x}-3\cos\text{x}]$
$=\frac{4\cos^3\text{x}-3\cos\text{x}}{4\cos^2\text{x}-2-1}$
$=\frac{4\cos^3\text{x}-3\cos\text{x}}{4\cos^2\text{x}-3}$
$=\cos\text{x}\Big(\frac{4\cos^2\text{x}-3}{4\cos^2\text{x}-3}\Big)$
$=\cos\text{x}$
View full question & answer→MCQ 361 Mark
The value of $\cos^2\Big(\frac{\pi}{6}+\text{x}\Big)-\sin^2\Big(\frac{\pi}{6}-\text{x}\Big)$ is:
AnswerCorrect option: A. $\frac{1}{2}\cos^2\text{x}$
We have,
$\cos^2\Big(\frac{\pi}{6}+\text{x}\Big)-\sin^2\Big(\frac{\pi}{6}-\text{x}\Big)$
$=\cos^2\Big(\frac{\pi}{6}+\text{x}\Big)-\cos^2\big[\frac{\pi}{2}-\big(\frac{\pi}{6}-\text{x}\big)\big]$
$=\cos^2\Big(\frac{\pi}{6}+\text{x}\Big)-\cos^2\Big(\frac{\pi}{3}-\text{x}\Big)$
$=\big[\cos\big(\frac{\pi}{6}+\text{x}\big)+\cos\big(\frac{\pi}{3}+-\text{x}\big)\big]\big[\cos\big(\frac{\pi}{6}+\text{x}\big)-\cos\big(\frac{\pi}{3}+-\text{x}\big)\big]$
$=\cos\bigg(\frac{\frac{\pi}{6}+\text{x}+\frac{\pi}{3}+\text{x}}{2}\bigg)\cos\bigg(\frac{\frac{\pi}{6}+\text{x}-\frac{\pi}{3}-\text{x}}{2}\bigg)2\sin\bigg(\frac{\frac{\pi}{6}+\text{x}+\frac{\pi}{3}+\text{x}}{2}\bigg)\sin\bigg(\frac{\frac{\pi}{6}+\text{x}+\frac{\pi}{6}+\text{x}}{2}\bigg)$
$=4\cos\big(\frac{\pi}{4}+\text{x}\big)\cos\big(-\frac{\pi}{12}\big)\sin\big(\frac{\pi}{4}+\text{x}\big)\sin\big(\frac{\pi}{12}\big)$
$=4\cos\big(\frac{\pi}{4}+\text{x}\big)\cos\big(\frac{\pi}{12}\big)\sin\big(\frac{\pi}{4}+\text{x}\big)\sin\big(\frac{\pi}{12}\big)$
$=\big[2\sin\big(\frac{\pi}{4}+\text{x}\big)\cos\big(\frac{\pi}{4}+\text{x}\big)\big]\big[2\sin\big(\frac{\pi}{12}\big)\cos\big(\frac{\pi}{12}\big)\big]$
$=\sin\big(\frac{\pi}{2}+2\text{x}\big)\sin\frac{\pi}{6}$
$=\cos2\text{x}\times\frac{1}{2}$
$=\frac{1}{2}\cos2\text{x}$
View full question & answer→MCQ 371 Mark
The value of $2\tan\frac{\pi}{10}+3\sec\frac{\pi}{10}-4\cos\frac{\pi}{10}$ is:
AnswerWe have,
$2\tan\frac{\pi}{10}+3\sec\frac{\pi}{10}-4\cos\frac{\pi}{10}$
$=2\tan18^\circ+3\sec18^\circ-4\cos18^\circ$
$=2\frac{\sin18^\circ}{\cos18^\circ}+3\times\frac{1}{\cos18^\circ}-4\cos18^\circ$
$=2\times\frac{\frac{\sqrt{5}-1}{4}}{\frac{\sqrt{10+2\sqrt{5}}}{4}}+3\times\frac{1}{\frac{\sqrt{10+2\sqrt{5}}}{4}}=4\times\frac{\sqrt{10+2\sqrt{5}}}{4}$
$=2\times\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}+3\times\frac{4}{\sqrt{10+2\sqrt{5}}}-\sqrt{10+2\sqrt{5}}$
$=\frac{2\sqrt{5}-2+12-\Big(\sqrt{10+2\sqrt{5}}\Big)^2}{\Big(\sqrt{10+2\sqrt{5}}\Big)}$
$=\frac{2\sqrt{5}+10-10-2\sqrt{5}}{\Big(\sqrt{10+3\sqrt{5}}\Big)}$
$=0$
View full question & answer→MCQ 381 Mark
$8\sin\frac{\text{x}}{8}\cos\frac{\text{x}}{2}\cos\frac{\text{x}}{4}\cos\frac{\text{x}}{8}$ is equal to:
- A
$8\cos\text{x}$
- B
$\cos\text{x}$
- C
$8\sin\text{x}$
- ✓
$\sin\text{x}$
AnswerCorrect option: D. $\sin\text{x}$
we have,
$=8\sin\text{x}8\cos\text{x}2\cos\text{x}4\cos\text{x}8$
$=4\times2\sin\text{x}8\cos\text{x}2\cos\text{x}2$
$=4\times\sin\text{x}\cos\text{x}2\cos\text{x}4$
$=2\times2\sin\text{x}4\cos\text{x}4\cos\text{x}2$
$=2\times\sin2\cos\text{x}2$
$=\sin\text{x}$
View full question & answer→