MCQ 1511 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 1521 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 1531 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 1541 Mark
If $\cos(\text{A}-\text{B})=\frac35$ and $\tan\text{A}\tan\text{B}=2,$ then
- ✓
$\cos\text{A}\cos\text{B}=\frac15$
- B
$\cos\text{A}\cos\text{B}=-\frac15$
- C
$\sin\text{A}\sin\text{B}=-\frac15$
- D
$\sin\text{A}\sin\text{B}=-\frac15$
AnswerCorrect option: A. $\cos\text{A}\cos\text{B}=\frac15$
$\tan\text{A}\tan\text{B}=\frac{\sin\text{A}\sin\text{B}}{\cos\text{A}\cos\text{B}}=2\ (\text{Given})\cdots(1)$
Also,
$\cos(\text{A - B})=\frac35$
$\Rightarrow\cos\text{A}\cos\text{B}=\frac35+\sin\text{A}\sin\text{A}=\frac35$
$\therefore\sin\text{A}\sin\text{B}=\frac35-\cos\text{A}\cos\text{B}\cdots(2)$
Substituting eq $(2)$ in eq $(1),$ we get:
$\Rightarrow\frac{\frac35-\cos\text{A}\cos\text{B}}{\cos\text{A}\cos\text{B}}=2$
$\Rightarrow3\cos\text{A}\cos\text{B}=\frac35$
$\Rightarrow\cos\text{A}\cos\text{B}=\frac15$
View full question & answer→MCQ 1551 Mark
Choose the correct answer. If $\sin\theta+\cos\theta=1,$ then the value of $\sin2\theta$ is equal to:
- A
$1$
- B
$\frac{1}{2}$
- ✓
$0$
- D
$-1$
AnswerGiven that, $\sin\theta+\cos\theta=1$
Squaring both sides, we get,
$\Rightarrow(\sin\theta+\cos\theta)^2=(1)^2$
$\Rightarrow\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=1$
$\Rightarrow1+\sin2\theta=1$
$\Rightarrow\sin2\theta=1-1=0$
Hence, the correct option is $(c).$
View full question & answer→MCQ 1561 Mark
If $\text{A}=\sin2\text{x}+\cos4\text{x},$ then for all real $x:$
- A
$\frac{13}{16}\leq\text{A}\leq1$
- B
$1\leq\text{A}\leq2$
- C
$\frac{3}{4}\leq\text{A}\leq\frac{13}{16}$
- ✓
$\frac{3}{4}\leq\text{A}\leq1$
AnswerCorrect option: D. $\frac{3}{4}\leq\text{A}\leq1$
View full question & answer→MCQ 1571 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 1581 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. $\frac{1}{2}\cos2\text{x}$
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 1591 Mark
The value of $\tan1^\circ\tan2^\circ\tan3^\circ\dots\ \tan89^\circ$ is:
AnswerWe know that, $\tan(90^\circ-\theta) = \cot\theta$
so,
$\tan89^\circ = \tan(90^\circ-1^\circ) = \cot 1^\circ$
$\tan88^\circ = \tan(90^\circ-2^\circ) = \cot2^\circ$
$\tan87^\circ = \tan(90^\circ-2^\circ) = \cot3^\circ\\\ \ \ .\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .\ \ \ \ \ \ \ \ \ \ \ \ \ \ .\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .\\\ \ \ .\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .\ \ \ \ \ \ \ \ \ \ \ \ \ \ .\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .$
$\tan46^\circ = \tan(90^\circ-44^\circ)=\cot44^\circ$
$\therefore\tan1^\circ\tan2^\circ\tan3^\circ...\tan89^\circ$
$=\tan1^\circ\tan2^\circ\tan3^\circ...\ \tan44^\circ\tan45^\circ\tan46^\circ\\...\ \tan87^\circ\tan88^\circ\tan89^\circ$
$= \tan1^\circ\tan2^\circ\tan3^\circ\ ...\ \tan44^\circ\tan45^\circ\tan46^\circ\\...\ \cot3^\circ\cot2^\circ\cot1^\circ$
$ =(\tan1^\circ \cot1^\circ)(\tan2^\circ\cot2^\circ)(\tan3^\circ \cot3^\circ)\\\ ... (\tan44^\circ \cot44^\circ) \tan45^\circ$
$=1$ $(\tan45^\circ =1 \text{ and }\tan\theta\cot\theta = 1)$
Hence, the correct answer is option $B.$
View full question & answer→MCQ 1601 Mark
If $\pi<\text{x}<2\pi,$ then $\sqrt{\frac{1+\cos\text{x}}{1-\cos\text{x}}}$ is equal to:
- A
$\text{cosec x}+\cot \text{x}$
- B
$\text{cosec x}-\cot \text{x}$
- C
$-\text{cosec x}+\cot \text{x}$
- ✓
$-\text{cosec x}-\cot \text{x}$
AnswerCorrect option: D. $-\text{cosec x}-\cot \text{x}$
$-\text{cosec}\text{x} -\cot\text{x}$
$=\sqrt{\frac{1+\cos\text{x}}{1-\cos\text{x}}}$
$=\sqrt{\frac{(1+\cos\text{x})(1+\cos\text{x})}{(1+\cos\text{x})(1+\cos\text{x})}}$
$=\sqrt{\frac{(1+\cos\text{x})^2}{1-\cos^2\text{x}}}$
$=\sqrt{\frac{(1+\cos\text{x})^2}{\sin^2\text{x}}}$
$=\frac{(1+\cos\text{x})}{-\sin\text{x}}$ $ [\text{as},\pi<\text{x}<2\pi,\text{ so }\sin\text{x} \text{ will}\text{ be}\text{ negative}]$
$=-(\text{cosec}\text{ x}+\cot\text{x})$
$= -\text{cosec}\text{ x } -\cot\text{x}$
View full question & answer→MCQ 1611 Mark
$\frac{\cos10^\circ+\sin10^\circ}{\cos10^\circ-\sin10^\circ}$ is equal to:
- ✓
$\tan55^\circ$
- B
$\cot55^\circ$
- C
$-\tan35^\circ$
- D
$-\cot35^\circ$
AnswerCorrect option: A. $\tan55^\circ$
$\frac{\cos10^\circ+\sin10^\circ}{\cos10^\circ-\sin10^\circ}$
$=\frac{1+\tan10^\circ}{1-\tan10^\circ}$
$[$Dividing the numerator and denominator by $\cos10^\circ]$
$=\frac{\tan45^\circ+\tan10^\circ}{1-\tan45^\circ\times\tan10^\circ}$
$=\tan(45^\circ+10^\circ)$ $\Big[$Using $\tan(\text{A+B})=\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}\Big]$
$=\tan55^\circ$
View full question & answer→MCQ 1621 Mark
Choose the correct answer. If for real values of $x, \cos\theta=\text{x}+\frac{1}{\text{x}},$ then:
- A
$\theta$ is an acute angle
- B
$\theta$ is right angle
- C
$\theta$ is an obtuse angle
- ✓
No value of $\theta$ is possible
AnswerCorrect option: D. No value of $\theta$ is possible
given that, $\cos\theta=\text{x}+\frac{1}{\text{x}}\Rightarrow\cos\theta=\frac{\text{x}^2+1}{\text{x}}$
$\Rightarrow\text{x}^2+1=\text{x}\cos\theta\Rightarrow\text{x}^2-\text{x}\cos\theta+1=0$
$$For real value of $\text{x},\text{b}^2-4\text{a}\text{c}\geq0$
$\Rightarrow(-\cos\theta)^2-4\times1\times1\geq0$
$\Rightarrow\cos^2\theta-4\geq0\Rightarrow\cos^2\theta\geq4$
$\Rightarrow\cos\theta\geq\pm2[-1\leq\cos\theta\leq1]$
so, the value of $\theta$ is not possible.
Hence, the correct options $(d).$
View full question & answer→MCQ 1631 Mark
The solution of the equation $\cos^2\text{q}+\sin\text{q}+1=0, $ lies in the interval:
- A
$\Big(\frac{\pi}{4},\frac{\pi}{4}\Big)$
- B
$\Big(\frac{-\pi}{4},\frac{3\pi}{4}\Big)$
- C
$\Big(\frac{-3\pi}{4}, \frac{5\pi}{4}\Big)$
- ✓
$\Big(\frac{5{\pi}}{4},\frac{7\pi}{4}\Big)$
AnswerCorrect option: D. $\Big(\frac{5{\pi}}{4},\frac{7\pi}{4}\Big)$
View full question & answer→MCQ 1641 Mark
A circular wire of radius $7\ cm$ is cut and bent again into an arc of a circle of radius $12\ cm.$ The angle subtended by the arc at the centre is:
- A
$50^\circ$
- ✓
$210^\circ$
- C
$100^\circ$
- D
$60^\circ$
AnswerCorrect option: B. $210^\circ$
Length of the arc of radius $=$ Circumference of the circle of radium
Now,
Angle subtended by the arc $=\frac{14\pi}{12}$
View full question & answer→MCQ 1651 Mark
The value of $\cos 1^\circ \cos 2^\circ \cos 3^\circ ...\cos 179^\circ$ is:
- A
$1\sqrt{2}$
- ✓
$0$
- C
$1$
- D
$-1$
View full question & answer→MCQ 1661 Mark
Choose the correct answer. 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
$\cos2\beta$
AnswerCorrect option: B. $\sin4\beta$
Given that, $\tan\alpha=\frac{1}{7}$ and $\tan\beta=\frac{1}{3}$
$\cos2\alpha=\frac{1-\tan^2\alpha}{1+\tan^2\alpha}=\frac{1-\Big(\frac{1}{7}\Big)^2}{1+\Big(\frac{1}{7}\Big)^2}=\frac{1-\frac{1}{49}}{1+\frac{1}{49}}$
$=\frac{48}{50}=\frac{24}{25}$
Now $\tan2\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{2}{3}\times\frac{9}{8}=\frac{3}{4}$
$\therefore\tan^2\beta=\frac{3}{4}$
$\sin4\beta=\frac{2\tan2\beta}{1+\tan^22\beta}$
$=\frac{2\times\frac{3}{4}}{1+\Big(\frac{3}{4}\Big)^2}=\frac{\frac{3}{2}}{1+\frac{9}{16}}=\frac{3}{2}\times\frac{16}{25}=\frac{24}{25}$
$\cos2\alpha=\sin4\beta=\frac{24}{25}$
Hence, the correct option is $(b).$
View full question & answer→MCQ 1671 Mark
In any $\triangle\text{ABC},$ the value of $2\text{ac}\sin\Big(\frac{\text{A}-\text{B + C}}{2}\Big)$ is:
- A
$\text{a}^2+\text{b}^2-\text{c}^2$
- ✓
$\text{c}^2+\text{a}^2-\text{b}^2$
- C
$\text{b}^2-\text{c}^2-\text{a}^2$
- D
$\text{c}^2-\text{a}^2-\text{b}^2$
AnswerCorrect option: B. $\text{c}^2+\text{a}^2-\text{b}^2$
In $\triangle\text{ABC},$
$\text{A + B + C}=\pi ($Angle sum property$)$
$\Rightarrow\text{A + C}=\pi-\text{B}$
$\therefore2\text{ac}\sin\Big(\frac{\text{A}-\text{B + C}}{2}\Big)$
$=2\text{ac}\sin\Big(\frac{\pi-2\text{B}}{2}\Big)$
$=2\text{ac}\sin\Big(\frac{\pi}{2}-\text{B}\Big)$
$=2\text{ac}\cos\text{B}$
$=2\text{ac}\Big(\frac{\text{c}^2+\text{a}^2-\text{b}^2}{2\text{ca}}\Big) ($Using cosine rule$)$
$=\text{c}^2+\text{a}^2-\text{b}^2$
Hence, the correct answer is option $(b).$
View full question & answer→MCQ 1681 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 1691 Mark
If $3\sin\text{x}+4\cos\text{x}=5,$ then $4\sin\text{x}-3\cos\text{x}=$
Answer$3\sin\text{x}+4\cos\text{x}=5$
$\frac{3}{5}\sin\text{x}+\frac{4}{5}\cos\text{x}=1$
Let $\cos\alpha=\frac35$ and $\sin\alpha=\frac45.$
$\therefore\cos\alpha\sin\text{x}+\sin\alpha\cos\text{x}=1$
$\Rightarrow\sin(\alpha+\text{x})=\sin\frac\pi2$
$\Rightarrow\alpha+\text{x}=\frac\pi2$
$\Rightarrow\text{x}=\frac{\pi}{2}-\alpha\cdots(1)$
We have to find the value of $4\sin\text{x}-3\cos\text{x}.$
$4\sin\Big(\frac\pi2-\alpha\Big)-3\cos\Big(\frac\pi3-\alpha\Big) ...\{$From eq $(1)\}$
$=4\cos\alpha-3\sin\alpha$
$=4\times\frac35-3\times\frac45$ $\Big(\because\cos\alpha=\frac35$ and $\sin\alpha=\frac{4}{5}\Big)$
$=0$
View full question & answer→MCQ 1701 Mark
If $\text{cosec x}+\cot \text{x}=\frac{1}{2},0<\text{x}<\frac{\pi}{2},$ then $\cos\text{x}$ is equal to
- A
$\frac{5}{3}$
- ✓
$\frac{3}{5}$
- C
$-\frac{3}{5}$
- D
$-\frac{5}{3}$
AnswerCorrect option: B. $\frac{3}{5}$
$2\text{cosec}=\frac{1}{2}+2$
$\Rightarrow 2\text{cosec}\text{ x} = \frac{5}{2}$
$\Rightarrow \text{cosec}\text{ x} =\frac{5}{4}$
$\Rightarrow\frac{1}{\sin\text{x}}=\frac{5}{4}$
$\Rightarrow\sin\text{x}=\frac{4}{5}$
Now, $0<\theta<\frac{\pi}{2}$
$\therefore\cos\theta=\sqrt{1-\sin^2\theta}$
$=\sqrt{1-\Big(\frac{4}{5}\Big)^2}$
$=\frac{3}{5}$
View full question & answer→MCQ 1711 Mark
What is the correct sequence of the following values?
$1. \sin\Big(\frac{\pi}{12}\Big)$
$2. \cos\Big(\frac{\pi}{12}\Big)$
$3. \cot\Big(\frac{\pi}{12}\Big)$
Select the correct answer using the code given below:
- ✓
$3 > 2 > 1$
- B
$1 > 2 > 3$
- C
$1 > 3 > 2$
- D
$3 > 1 > 2$
AnswerCorrect option: A. $3 > 2 > 1$
View full question & answer→MCQ 1721 Mark
The number of solution in $\big[0,\ \frac{\pi}{2}\big]$ of the equation $3\text{x}\tan5\text{x}=\sin7\text{x}$ is:
AnswerGiven:
$\Rightarrow\cos(5\text{x}-2\text{x})\tan5\text{x}=\sin(5\text{x}+2\text{x})$
$\Rightarrow\tan5\text{x}=\frac{\sin(5\text{x}+2\text{x})}{\cos(5\text{x}-2\text{x})}$
$\Rightarrow\tan5\text{x}=\frac{\sin5\text{x}\cos2\text{x}+\cos5\text{x}\sin2\text{x}}{\cos5\text{x}\cos2\text{x}+\sin5\text{x}\cos2\text{x}}$
$\Rightarrow\frac{\sin5\text{x}}{\cos5\text{x}}=\frac{\sin5\text{x}\cos2\text{x}+\cos5\text{x}\sin2\text{x}}{\cos5\text{x}\cos2\text{x}+\sin5\text{x}\cos2\text{x}}$
$\Rightarrow\sin5\text{x}\cos5\text{x}+\sin^25\text{x}\sin2\text{x}\\=\sin5\text{x}\cos5\text{x}\cos2\text{x}+\cos^25\text{x}\sin2\text{x}$
$\Rightarrow\sin^25\text{x}\sin2\text{x}=\cos^25\text{x}\sin2\text{x}$
$\Rightarrow\Big(\sin^25\text{x}\cos^25\text{x}\Big)\sin2\text{x}=0$
$\Rightarrow\Big(\sin5\text{x}-\cos5\text{x}\Big)\Big(\sin5\text{x}+\cos5\text{x}\Big)\sin2\text{x}=0$
$\Rightarrow\sin5\text{x}-\cos5\text{x}=0,\ \sin5\text{x}+\cos5\text{x}=0$ or $\sin2\text{x}=0$
$\Rightarrow\frac{\sin5\text{x}}{\cos5\text{x}}=1,\ \frac{\sin5\text{x}}{\cos5\text{x}}=-1$ or $\sin2\text{x}=0$
Now,
$\Rightarrow\tan5\text{x}=\tan\frac{\pi}{4}$
$\Rightarrow5\text{x}=\text{n}\pi+\frac{\pi}{4},\ \text{n}\in\text{Z}$
$\Rightarrow\text{x}=\frac{\text{n}\pi}{5}+\frac{\pi}{20},\text{n}\in\text{Z}$
$F$ or $n = 0, 1$ and $2$, the values of x are $\frac{\pi}{20},\ \frac{\pi}{4}$ and $\frac{9\pi}{20},$ respectively.
Or,
$\tan5\text{x}=1$
$\Rightarrow\tan5\text{x}=\tan\frac{3\pi}{4}$
$\Rightarrow5\pi=\text{n}\pi+\frac{3\pi}{4},\ \text{n}\in\text{Z}$
$\Rightarrowx\text{x}=\frac{\text{n}\pi}{5}+\frac{3\pi}{20},\ \text{n}\in\text{Z}$
For $n = 0$ and $1,$ the values of $x$ are $\frac{3\pi}{20}$ and $\frac{7\pi}{20},$ respectively.
And,
$\sin2\text{x}=\sin0$
$\Rightarrow\sin2\text{x}=\sin0$
$\Rightarrow2\text{x}=\text{n}\pi,\ \text{n}\in\text{Z}$
$\Rightarrow\text{x}=\frac{\text{n}\pi}{2},\ \text{n}\in\text{Z}$
For $n = 0,$ the value of $x$ is $0.$
Also, for the odd multiple of $\frac{\pi}{2,} \tan x$ is not defined.
Hence, there are six solutions.
View full question & answer→MCQ 1731 Mark
Choose the correct answer. If $\tan\text{A}=\frac{1}{2},\tan\text{B}=\frac{1}{3},$ then $\tan(2\text{A + B})$ is equal to:
AnswerGiven that, $\tan\text{A}=\frac{1}{2}$ and $\tan\text{B}=\frac{1}{3}$
$\tan2\text{A}=\frac{2\tan\text{A}}{1-\tan^2\text{A}}=\frac{2\times\frac{1}{2}}{1-\Big(\frac{1}{2}\Big)^2}$
$=\frac{1}{1-\frac{1}{4}}=\frac{1}{\frac{3}{4}}=\frac{4}{3}$
so, $\tan2\text{A}=\frac{4}{3}$ and $\tan\text{B}=\frac{1}{3}$
$\tan(\text{2A+B})=\frac{\tan2\text{A}+\tan\text{B}}{1-\tan\text{A}.\tan\text{B}}=\frac{\frac{4}{3}+\frac{1}{3}}{1-\frac{4}{3}\times\frac{1}{3}}$
$=\frac{\frac{5}{3}}{\frac{9-4}{9}}=\frac{5}{3}\times\frac{9}{5}=3$
Hence, the correct option is $(c).$
View full question & answer→MCQ 1741 Mark
$\cos40^\circ+\cos80^\circ+\cos160^\circ+\cos240=$
- A
$0$
- B
$1$
- C
$\frac{1}{2}$
- ✓
$-\frac{1}{2}$
AnswerCorrect option: D. $-\frac{1}{2}$
$\cos40^\circ+\cos80^\circ+\cos160^\circ+\cos240^\circ$
$=\ 2\cos\Big(\frac{40^\circ+80^\circ}{2}\Big)\cos\Big(\frac{40^\circ-80^\circ}{2}\Big)+\cos160^\circ-\cos(180^\circ+60^\circ)$
$\big[\because\ \cos\text{A}+\cos\text{B}=2\big]$
$=\ 2\cos60^\circ\cos(-20^\circ)+\cos160^\circ-\frac{1}{2}$
$=\ 2\times\frac{1}{2}\cos20^\circ+\cos160^\circ-\frac{1}{2}$
$=\ -\cos(180-20)^\circ+\cos160^\circ-\frac{1}{2}$
$=\ -\frac{1}{2}$
View full question & answer→MCQ 1751 Mark
Choose the correct answer. Which of the following is not correct?
AnswerCorrect option: C. $\sec\theta=\frac{1}{2}$
$\sin\theta=-\frac{1}{5}$ is correct.
$\because-1\leq\sin\theta\leq1$
so $(a)$ is correct.
$\cos\theta=1$ is correct.
$\because\cos0^\circ=1$
so $(b)$ is correct.
$\sec\theta=-\frac{1}{2}$
$\Rightarrow\cos\theta=2$ is not correct.
$\because-1\leq\cos\theta\leq1$
Hence, $(c)$ is not correct.
View full question & answer→MCQ 1761 Mark
If $\tan\theta=\frac12$ and $\tan\phi=\frac13,$ then the value of $\theta+\phi$ is:
- A
$\frac\pi6$
- B
$\pi$
- C
$0$
- ✓
$\frac\pi4$
AnswerCorrect option: D. $\frac\pi4$
It is given that $\tan\theta=\frac12$ and $\tan\phi=\frac13.$
Now,
$\tan(\theta+\phi)=\frac{\tan+\tan\phi}{1-\tan\theta\tan\phi}$
$=\frac{\frac12+\frac13}{1-\frac12\times\frac13}$
$=\frac{\frac{5}{6}}{\frac56}$
$=1$
$\therefore\theta+\phi=\frac\pi4$ $\Big(\tan\frac\pi4=1\Big)$
Hence, the correct answer is option $D.$
View full question & answer→MCQ 1771 Mark
If $x$ is an acute angle and $\text{x}=\frac{1}{\sqrt{7}},$ than the value of $\frac{\text{cosec}^2\text{x}-\sec^2\text{x}}{\text{cosec}^2\text{x}+\sec^2\text{x}}$ is:
- ✓
$\frac{3}{4}$
- B
$\frac{1}{2}$
- C
$2$
- D
$\frac{5}{4}$
AnswerCorrect option: A. $\frac{3}{4}$
We have:
$\tan\text{x}=\frac{1}{\sqrt{7}}$
$\therefore \tan^2\text{x}=\frac{1}{7}$
Now, dividing the numerator and the denominator of $\frac{\text{cosec}^2\text{x}-\sec^2\text{x}}{\text{cosec}^2\text{x}+\sec^2\text{x}{}}$ by $\text{cosec}^2\text{x}=\frac{1-\tan^2\text{x}}{1+\tan^2\text{x}}$
$=\frac{1-\frac{1}{7}}{1+\frac{1}{7}}$
$=\frac{6}{8}$
$=\frac{3}{4}$
View full question & answer→MCQ 1781 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→MCQ 1791 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:
- A
$\sin2\text{A}$
- ✓
$\cos2\text{A}$
- C
$\cos3\text{A}$
- D
$\sin3\text{A}$
AnswerCorrect option: B. $\cos2\text{A}$
$\cos(36^\circ-\text{A})\cos(36^\circ+\text{A})+\cos(54^\circ+\text{A})\cos(54^\circ-\text{A})$
$=\cos(36^\circ-\text{A})\cos(36^\circ+\text{A})+\sin[90^\circ-(54^\circ+\text{A})]\sin[90^\circ-(54^\circ-\text{A})]$
$=\cos(36^\circ-\text{A})\cos(36^\circ+\text{A})+\sin(36^\circ-\text{A})\sin(36^\circ-\text{A})$
$=\cos(36^\circ-\text{A}-36^\circ+\text{A})$$\big[\text{Using}\cos(\text{A - B) formula} \big]$
$=\cos2\text{A}$
View full question & answer→MCQ 1801 Mark
If $\tan\text{A}+\cot\text{A}=4,$ then $\tan^4\text{A}+\cot^4\text{A}$ is equal to:
AnswerWe have:
$\tan\text{A}+\cot\text{A}=4$
squaring both the sides:
$(\tan\text{A}+\cot\text{A})^2=4^2$
$\Rightarrow\tan^2\text{A}+\cot^2\text{A}+2(\tan\text{A})(\cot\text{A})=16$
$\Rightarrow\tan^2\text{A}+\cot^2\text{A}+2=16$
$\Rightarrow\tan^2\text{A}+\cot\text{A}=14$
squaring both the sides again:
$(\tan^2\text{A}+\cot^2\text{A})^2=14^2$
$\tan^4\text{A}+\cot^4\text{A}+2(\tan^2\text{A})(\cot^2\text{A})=196$
$\Rightarrow\tan^4\text{A}+\cot^4\text{A}+2=196$
$\Rightarrow\tan^4\text{A}+\cot^4\text{A}=194$
View full question & answer→MCQ 1811 Mark
$\sin47^\circ+\sin61^\circ-\sin11^\circ-\sin25^\circ$ is equal to
- A
$\sin36^\circ$
- B
$\cos36^\circ$
- C
$\sin7^\circ$
- ✓
$\cos7^\circ$
AnswerCorrect option: D. $\cos7^\circ$
$\sin47^\circ+\sin61^\circ-\sin11^\circ-\sin25^\circ$
$=\ \sin47^\circ-\sin25^\circ+\sin61^\circ-\sin11^\circ$
$=\ 2\sin\Big(\frac{47^\circ-25^\circ}{2}\Big)\cos\Big(\frac{47^\circ+25^\circ}{2}\Big) +2\sin\Big(\frac{61^\circ-11^\circ}{2}\Big)\cos\Big(\frac{61^\circ+11^\circ}{2}\Big)$
$=\ 2\sin11^\circ\cos36^\circ+2\sin25^\circ\cos36^\circ$
$=\ 2\cos36^\circ(\sin11^\circ+\sin25^\circ)$
$=\ 2\cos36^\circ\Big\{2\sin\Big(\frac{11^\circ+25^\circ}{2}\Big)\cos\Big(\frac{11^\circ-25^\circ}{2}\Big)\Big\}$
$=\ 4\cos36^\circ\sin18^\circ\cos7^\circ$
$=\ 4\times\Big(\frac{\sqrt5-1}{4}\Big)\Big(\frac{\sqrt5+1}{4}\Big)\cos7^\circ$ $\Big[\cos36^\circ=\frac{\sqrt5+1}{4}\text{ and }\sin18^\circ=\frac{\sqrt5-1}{4}\Big]$
$=\ \frac{5-1}{4}\cos7^\circ$
$=\ \cos7^\circ$
View full question & answer→MCQ 1821 Mark
If $\tan\Big(\frac\pi4+\text{x}\Big)+\tan\Big(\frac\pi4-\text{x}\Big)=\text{a},$ then $\tan^2\Big(\frac\pi4+\text{x}\Big)+\tan^2\Big(\frac\pi4-\text{x}\Big)=$
- A
$a^2 + 1$
- B
$a^2 + 2$
- ✓
$a^2 - 2$
- D
AnswerCorrect option: C. $a^2 - 2$
Given:
$\tan\Big(\frac\pi4+\text{x}\Big)+\tan\Big(\frac\pi4-\text{x}\Big)=\text{a}$
$\Rightarrow\Big[\tan\Big(\frac\pi4+\text{x}\Big)+\tan\Big(\frac\pi4-\text{x}\Big)\Big]^2=\text{a}^2$
$\Rightarrow\tan^2\Big(\frac\pi4+\text{x}\Big)+\tan^2\Big(\frac\pi4-\text{x}\Big)+2\tan\Big(\frac\pi4-\text{x}\Big)\tan\Big(\frac\pi4-\text{x}\Big)=\text{a}^2$
$\Rightarrow\tan^2\Big(\frac\pi4+\text{x}\Big)+\tan^2\Big(\frac\pi4-\text{x}\Big)\text{a}^2-2\tan\Big(\frac\pi4-\text{x}\Big)\tan\Big(\frac\pi4-\text{x}\Big)$
$\Rightarrow\tan^2\Big(\frac\pi4+\text{x}\Big)+\tan^2\Big(\frac\pi4-\text{x}\Big)\text{a}^2-2\Big[\frac{\tan45^\circ-\tan\text{x}}{1+\tan45^\circ\tan\text{x}}\times\frac{\tan45^\circ+\tan\text{x}}{1+\tan45^\circ\tan\text{x}}\Big]$
$\Rightarrow\tan^2\Big(\frac\pi4+\text{x}\Big)+\tan^2\Big(\frac\pi4-\text{x}\Big)\text{a}^2-2\Big[\frac{1^\circ-\tan\text{x}}{1+\tan\text{x}}\times\frac{1^\circ+\tan\text{x}}{1-\tan\text{x}}\Big]$
$\Rightarrow\tan^2\Big(\frac\pi4+\text{x}\Big)+\tan^2\Big(\frac\pi4-\text{x}\Big)\text{a}^2-2\Big(\frac{1-\tan^2\text{x}}{1-\tan^2\text{x}}\Big)$
$\Rightarrow\tan^2\Big(\frac\pi4+\text{x}\Big)+\tan^2\Big(\frac\pi4-\text{x}\Big)\text{a}^2-2$
View full question & answer→MCQ 1831 Mark
If $\text{A+B+C}=\pi,$ then $\sec\text{A}(\cos\text{B}\cos\text{C}-\sin^2\text{B}\sin\text{C})$ is equal to:
Answer$\pi=180^\circ$
$\sec\text{A}(\cos\text{B}\cos\text{C}-\sin^2\text{B}\sin\text{C})=\frac{\cos\text{B}\cos(\pi-(\text{A+B}))-\sin\text{B}\sin(\pi(\text{A+B}))}{\cos\text{A}}$
We know that, $\cos(\pi-\theta)=-\cos\theta$ and $\sin(\pi-\theta)=\sin\theta,$
$\therefore\sec\text{A}(\cos\text{B}\cos\text{C}-\sin\text{B}\sin\text{C})=\frac{\cos\text{B}\cos(\text{A+B})-\sin\text{B}\sin(\text{A+B})}{\cos\text{A}}$
Now, using the identities $\cos(\text{A+B})=\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}$ and $\sin(\text{A+B})=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B},$ we get
$\sec\text{A}(\cos\text{B}\cos\text{C}-\sin\text{B}\sin\text{C})=\frac{-\cos\text{A}\cos\text{B}^2+\cos\text{B}\sin\text{A}\sin\text{B}-\sin\text{B}\sin\text{A}\cos\text{B}-\sin^2\cos\text{A}}{\cos\text{A}}$
$\Rightarrow\sec\text{A}(\cos\text{B}\cos\text{C}-\sin\text{B}\sin\text{C})=\frac{-\cos\text{A}(\cos^2\text{B}+\sin^2\text{B})}{\cos\text{A}}$
$\Rightarrow\sec\text{A}(\cos\text{B}\cos\text{C}-\sin\text{B}\sin\text{C})=\frac{-\cos\text{A}}{\cos\text{A}}=-1$
View full question & answer→MCQ 1841 Mark
What is the value of $\cot (-870^\circ )?$
- ✓
$3$
- B
$\frac{1}{\sqrt{3}}$
- C
$-\sqrt{3}$
- D
$\frac{-1}{\sqrt{3}}$
View full question & answer→