MCQ 1011 Mark
The general solution of the equation $7\cos^2\text{x}+3\sin^2\text{x}=4$ is:
- A
$\text{x}=2\text{n}\pi\pm\frac{\pi}{6},\text{n}\in\text{Z}$
- B
$\text{x}=2\text{n}\pi\pm\frac{2\pi}{3},\text{n}\in\text{Z}$
- ✓
$\text{x}=\text{n}\pi\frac{\pi}{3},\text{n}\in\text{Z}$
- D
AnswerCorrect option: C. $\text{x}=\text{n}\pi\frac{\pi}{3},\text{n}\in\text{Z}$
Given:
$7\cos^2\text{x}+3\sin^2\text{x}=4$
$\Rightarrow7\cos^2\text{x}+3\Big(1-\cos^2\text{x}\Big)=4$
$\Rightarrow7\cos^2\text{x}+3-3\cos^2\text{x}=4$
$\Rightarrow4\cos^2\text{x}+3=4$
$\Rightarrow4\Big(1-\cos^2\text{x}\Big)=3$
$\Rightarrow4\sin^2\text{x}=3$
$\Rightarrow\sin^2\text{x}=\frac{3}{4}$
$\Rightarrow\sin\text{x}=\frac{\sqrt{3}}{2}$
$\Rightarrow\sin\text{x}=\sin\frac{\pi}{3}$
$\Rightarrow\text{x}=\text{n}\pi\pm\frac{\pi}{3},\text{n}\in\text{Z}$
View full question & answer→MCQ 1021 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 1031 Mark
In $(0,\ \pi)$ the number of solutions of the equation$\tan\text{x}+\tan2\text{x}+\tan3\text{x}=\tan\text{x}\tan2\text{x}\tan3\text{x}$ is:
AnswerGiven equation:
$\tan\text{x}+\tan2\text{x}=-\tan3\text{x}+\tan\text{x}\tan2\text{x}\tan3\text{x}$
$\Rightarrow\tan\text{x}+\tan2\text{x}=-\tan3\text{x}+\tan\text{x}\tan2\text{x}\tan3\text{x}$
$\Rightarrow\tan\text{x}+\tan2\text{x}=-\tan3\text{x}\Big(1-\tan\text{x}\tan2\text{x}\Big)$
$\Rightarrow\frac{\tan\text{x}+\tan2\text{x}}{1-\tan\text{x}\tan2\text{x}}=-\tan3\text{x}$
$\Rightarrow\tan(\text{x}+2\text{x})=-\tan3\text{x}$
$\Rightarrow\tan3\text{x}=-\tan3\text{x}$
$\Rightarrow2\tan3\text{x}=0$
$\Rightarrow\tan3\text{x}=0$
$\Rightarrow3\text{x}=\text{n}\pi$
$\Rightarrow\text{x}=\frac{\text{n}\pi}{3}$
Now,
$\text{x}=\frac{\pi}{3},\ \text{n}=1$
$\text{x}=\frac{2\pi}{3},\ \text{n}=2$
$\text{x}=\frac{3\pi}{3}=180^\circ,$ which is not possible, as it is not in the interval $(0,\ 2\pi).$
Hence, the number of solutions of the given equation is $2.$
View full question & answer→MCQ 1041 Mark
The maximum value of $\sin^2\Big(\frac{2\pi}{3}+\text{x}+\sin^2\Big(\frac{2\pi}{3}-\text{x}\Big)$ is:
- A
$\frac12$
- ✓
$\frac32$
- C
$\frac14$
- D
$\frac34$
AnswerCorrect option: B. $\frac32$
$\frac{2\pi}{3}=120^\circ$
Let $\text{f(x)}=\sin^2(90+30+\text{x})+\sin^2(90+30-\text{x})$
$=[\cos(30+\text{x})]^2+[\cos(30=\text{x})]^2$ $[\text{Using }\sin(90+\text{A})=\cos\text{A}]$
$=\Big[\frac{\sqrt{3}}{2}\cos\text{x}-\frac12\sin\text{x}\Big]^2+\Big[\frac{\sqrt{3}}{2}\cos\text{x}-\frac12\sin\text{x}\Big]^2$
$=\frac{\sqrt{3}}{2}\cos^2\text{x}-\frac14\sin^2\text{x}-\frac{\sqrt{3}}{2}\cos\text{x}\sin\text{x}+\frac34\cos^2\text{x}+\frac14\sin^2\text{x}+\frac{\sqrt{3}}{2}\cos\text{x}\sin\text{x}$
$=\frac{3}{2}\cos^2\text{x}-\frac12\sin^2\text{x}$
$=\frac{3}{2}(1-\sin^2\text{x})+\frac12\sin^2\text{x}$
$=\frac32-\frac{3}{2}\sin^2\text{x}+\frac12\sin^2\text{x}$
$=\frac32-\sin^2\text{x}$.
For $f(x)$ to be maximum, $\sin^2\text{x}$ must have minimum value, which is $0.$
$\therefore\frac32$ is the maximum value of $f(x).$
View full question & answer→MCQ 1051 Mark
Choose the correct answer. The value of $\cos12^\circ+\cos84^\circ+\cos156^\circ+\cos132^\circ$ is:
- A
$\frac{1}{2}$
- B
$1$
- ✓
$-\frac{1}{2}$
- D
$\frac{1}{8}$
AnswerCorrect option: C. $-\frac{1}{2}$
$\cos12^\circ+\cos84^\circ+\cos156^\circ+\cos132^\circ$
$=(\cos12^\circ+\cos132^\circ)+(\cos84^\circ+\cos156^\circ)$
$=2\cos72^\circ\cos60^\circ+2\cos120^\circ\cos36^\circ$
$=\cos72^\circ-\cos36^\circ$
$=\sin18^\circ-\cos36^\circ$
$=\Big(\frac{\sqrt5-1}{4}\Big)-\Big(\frac{\sqrt5+1}{4}\Big)$
$=\frac{-1}{2}$
View full question & answer→MCQ 1061 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 1071 Mark
If $\sin(\pi\cos\text{x})=\cos(\pi\sin\text{x}),$ then $\sin2\text{x}=$
- A
$\pm\frac34$
- B
$\pm\frac43$
- ✓
$\pm\frac13$
- D
AnswerCorrect option: C. $\pm\frac13$
$\sin(\pi\cos\text{x})=\cos(\pi\sin\text{x})$
As we know that $\sin\text{x}=-\cos\Big(\frac\pi2+\text{x}\Big)$
$\Rightarrow-\cos\Big(\frac\pi2+\pi\cos\text{x}\Big)=\cos(\pi\sin\text{x})$
$\Rightarrow\frac{-\pi}{2}-\pi\cos\text{x}=\pi\sin\text{x}$
$\Rightarrow\pi\sin\text{x}-\pi\cos\text{x}=\frac12$
$\Rightarrow\sin\text{x}-\cos\text{x}=\frac12$
Squaring both sides we get,
$\Rightarrow\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}\cos\text{x}=\frac14$
$\Rightarrow1+\sin2\text{x}=\frac{1}{4}$
$\Rightarrow\sin2\text{x}=\frac13$
$\therefore\sin2\text{x}=\pm\frac13$
View full question & answer→MCQ 1081 Mark
If $\text{x}\sin45^\circ\cos^260^\circ=\frac{\tan^260^\circ\text{cosec }30^\circ}{\sec45^\circ\cot^230^\circ},$ then $x =$
AnswerWe have:
$\text{x}\sin45^\circ\cos^260^\circ=\frac{\tan^260^\circ\text{cosec}30^\circ}{\sec45^\circ\cot^230^\circ}$
$\Rightarrow\text{x}\times\Big(\frac{1}{\sqrt2}\Big)\times\Big(\frac{1}{2}\Big)^2=\frac{(\sqrt3)^2\times(2)}{(\sqrt2)\times(\sqrt3)^2}$
$\Rightarrow\frac{\text{x}}{4\sqrt2}=\frac{6}{3\sqrt2}$
$\Rightarrow\text{x}=\frac{6}{3\sqrt2}\times4\sqrt2$
$\Rightarrow\text{x}=8$
View full question & answer→MCQ 1091 Mark
If $\sin(\text{B+C}-\text{A}),\sin(\text{C+A}-\text{B}),\sin(\text{A+B}-\text{C})$ are in $A.P.$, than $\cot\text{A},\cot\text{B},\cot\text{C}$ are in
- A
$\text{GP}$
- ✓
$\text{HP}$
- C
$\text{AP}$
- D
AnswerCorrect option: B. $\text{HP}$
Given:
$\sin(\text{B+C}-\text{A}),\sin(\text{C+A}-\text{B}),$ and $\sin(\text{A+B}-\text{C})$ are in $A.P.$
$\Rightarrow\ \sin(\text{C+A}-\text{B})-\sin(\text{B+C}-\text{A})\\ \ \ \ =\sin(\text{A+B}-\text{C})-\sin(\text{C+A}-\text{B})$
$\Rightarrow\ 2\sin\Big(\frac{\text{C+A}-\text{B}-\text{B}-\text{C+A}}{2}\Big)\cos\Big(\frac{\text{C+A}-\text{B+B+C}-\text{A}}{2}\Big)\\ \ \ \ =2\sin\Big(\frac{\text{A+B}-\text{C}-\text{C}-\text{A+B}}{2}\Big)\cos\Big(\frac{\text{A+B}-\text{C+C+A}-\text{B}}{2}\Big)$
$\Rightarrow\ \sin(\text{A}-\text{B})\cos\text{C}=\sin(\text{B}-\text{C})\cos\text{A}$
$\Rightarrow\ \sin\text{A}\cos\text{B}\cos\text{C}-\cos\text{A}\sin\text{B}\cos\text{C}\\ \ \ =\sin\text{B}\cos\text{C}\cos\text{A}-\cos\text{B}\sin\text{C}\cos\text{A}$
$\Rightarrow\ 2\sin\text{B}\cos\text{A}\cos\text{C}=\sin\text{A}\cos\text{B}\cos\text{C}+\cos\text{A}\cos\text{B}\sin\text{C}$
Dividing both sides by $\cos\text{A}\cos\text{B}\cos\text{C}:$
$2\tan\text{B}=\tan\text{A}+\tan\text{C}$
$\Rightarrow\ \frac{2}{\cot\text{B}}=\frac{1}{\cot\text{A}}+\frac{1}{\cot\text{C}}$
Hence, $\cot\text{A}, \cot\text{B}$ and $\cot\text{C}$ are in $H.P.$
View full question & answer→MCQ 1101 Mark
The value of$\cos1^\circ\cos2^\circ\cos^\circ...\cos179$ is:
- A
$\frac{1}{\sqrt{2}}$
- ✓
$0$
- C
$1$
- D
$-1$
View full question & answer→MCQ 1111 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 1121 Mark
If $\text{e}^{\sin\text{x}}-\text{e}^{-\sin\text{x}}-4=0,$ then $\text{x}=$
AnswerGiven equation: $\text{e}^{\sin\text{x}}-\text{e}^{-\sin\text{x}}-4=0,$
Let:
$\text{e}^{\sin\text{x}}=\text{y}$
Now,
$\text{y}-\text{y}^{-1}-4=0$
$\Rightarrow\text{y}^2-4\text{y}-1=0$
$\therefore\text{y}=\frac{4\pm\sqrt{16+4}}{2}$
$\Rightarrow\text{y}=\frac{4\pm\sqrt{20}}{2}$
$\Rightarrow\text{y}=\frac{4\pm2\sqrt{5}}{2}=2\pm\sqrt{5}$
And
$\text{y}=\text{e}^{\sin\text{x}}$
$\Rightarrow\text{y}^{\sin\text{x}}={2\pm\sqrt{5}}$
Taking log on both sides, we get:
$\sin\text{x}=\log_\text{e}\big(2\pm\sqrt{5}\big)$
$\Rightarrow\sin\text{x}=\log_{e}\big(2+\sqrt{5}\big)$ or $\sin\text{x}=\log_\text{e}\big(2-\sqrt{5}\big)$
$\Rightarrow\sin\text{x}=\log_{e}\big(4.24\big)$ or $\sin\text{x}=\log_\text{e}\big(-0.24\big)$
$\log(4.24)>1$ and $\sin\text{x}$ cannot be greater than $1.$
In the other case, the $\log$ of negative term occurs, which is not defined.
View full question & answer→MCQ 1131 Mark
A solution of the equation $\cos^2\text{x}+\sin\text{x}+1=0,$ 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. $\Big(\frac{5\pi}{4},\ \frac{7\pi}{4}\Big)$
Given:
$\cos^2\text{x}+\sin\text{x}+1=0$
$\Rightarrow(1-\sin^2\text{x})+\sin\text{x}+1=0$
$\Rightarrow1-\sin^2\text{x}+\sin\text{x}+1=0$
$\Rightarrow\sin^2\text{x}-\sin\text{x}-2=0$
$\Rightarrow\sin^2\text{x}-2\sin\text{x}+\sin\text{x}-2=0$
$\Rightarrow\sin\text{x}(\sin\text{x}-2)+1(\sin\text{x}-2)=0$
$\Rightarrow(\sin\text{x}-2)(\sin\text{x}+1)=0$
$\Rightarrow\sin\text{x}-2=0$ or $\sin\text{x}+1=0$
$\Rightarrow\sin\text{x}=2$ or $\sin\text{x}=-1$
$\sin\text{x}=2$ is not possible.
$\Rightarrow\sin\text{x}=-1$
$\therefore\sin\text{x}=\sin\frac{3\pi}{2}$
$\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}\frac{3\pi}{2},\ \text{n}\in\text{Z}$
The values of $x$ lies in the third and fourth quadrants.
Hence, $x$ lies in $\Big(\frac{5\pi}{4},\ \frac{7\pi}{4}\Big).$
View full question & answer→MCQ 1141 Mark
Choose the correct answer. The minimum value of $3\cos\text{x}+4\sin\text{x}+8$ is:
AnswerThe given expression is $3\cos\text{x}+4\sin\text{x}+8$
Let $\text{y}=3\cos\text{x}+4\sin\text{x}+8$
$\Rightarrow\text{y}-8=3\cos\text{x}+4\sin\text{x}$
Minimum value of $\text{y}-8=-\sqrt{(3)^2+(4)^2}$
$\Rightarrow\text{y}-8=-\sqrt{9+16}=-5$
$\Rightarrow\text{y}=8-5=3$
so, the minimum value of the given expression is $3.$
Hence, the correct option is $(d).$
View full question & answer→MCQ 1151 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 1161 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 1171 Mark
If $\sec\text{x}=\text{x}+\frac{1}{4\text{x}},$ then $\sec\text{x}+\tan\text{x}=$
- A
$\text{x},\frac{1}{\text{x}}$
- ✓
$2\text{x},\frac{1}{2\text{x}}$
- C
$-2\text{x},\frac{1}{2\text{x}}$
- D
$-\frac{1}{\text{x}},\text{x}$
AnswerCorrect option: B. $2\text{x},\frac{1}{2\text{x}}$
We have:
$\sec\text{x} = \text{x} +\frac{1}{4\text{x}}$
$\Rightarrow\sec^2\text{x}=\text{x}^2+\frac{1}{16\text{x}^2}+\frac{1}{2}$
$\Rightarrow1+\tan^2\text{x}$
$=1+\text{x}^2+\frac{1}{16\text{x}^2}-\frac{1}{2}$
$\Rightarrow\tan^2\text{x}=\text{x}^2+\frac{1}{16\text{x}^2}-\frac{1}{2}$
$\Rightarrow\tan^2\text{x}=\Big(\text{x}-\frac{1}{4\text{x}}\Big)^2$
$\therefore\tan\text{x}=\pm\Big(\text{x}-\frac{1}{4\text{x}}\Big)$
$\sec\text{x}-\tan\text{x}=\Big(\text{x}+\frac{1}{4\text{x}}\Big)-\Big(\text{x}-\frac{1}{4\text{x}}\Big)\text{or}$
$=\Big(\text{x}+\frac{1}{4\text{x}}\Big)-\Big[-\Big(\text{x}-\frac{1}{4\text{x}}\Big)\Big]$
$=\frac{1}{2\text{x}}$ or $2\text{x}$
View full question & answer→MCQ 1181 Mark
$\text{If}\ \sin2\theta+\sin2\phi=\frac{1}{2}\ \text{and}\ \cos2\theta \ +\cos2\phi=\frac{3}{2},\text{then}\cos^2(\theta-\phi)=$
- A
$\frac{3}{8}$
- ✓
$\frac{5}{8}$
- C
$\frac{3}{4}$
- D
$\frac{5}{4}$
AnswerCorrect option: B. $\frac{5}{8}$
Given:
$\sin2\theta+\sin2\phi=\frac{1}{2}....(\text{i})$
and
$\cos2\theta+\cos2\phi=\frac{3}{2}....(\text{ii})$
Squaring and adding $(i)$ and $(ii),$ we get:
$(\sin2\theta+\sin2\phi)^2+(\cos2\theta+\cos2\phi)^2=\frac{1}{4}+\frac{9}{4}$
$\Rightarrow\ \Big[2\sin\Big(\frac{2\theta+2\phi}{2}\Big)\cos\Big(\frac{2\theta-2\phi}{2}\Big)\Big]^2+\Big[2\cos\Big(\frac{2\theta+2\phi}{2}\Big)\cos\Big(\frac{2\theta-2\phi}{2}\Big)\Big]^2=\frac{5}{2}$
$\Rightarrow\ 4\sin^2(\theta+\phi)\cos^2(\theta-\phi)+4\cos^2(\theta+\phi)\cos^2(\theta-\phi)=\frac{5}{2}$
$\Rightarrow\ 4\cos^2(\theta-\phi)[\sin^2(\theta+\phi)+\cos^2(\theta+\phi)]=\frac{5}{2}$
$\Rightarrow\ 4\cos^2(\theta-\phi)=\frac{5}{2}$
$\Rightarrow\ \cos^2(\theta-\phi)=\frac{5}{8}$
View full question & answer→MCQ 1191 Mark
Choose the correct answer.
The value of $\sin\frac{\pi}{18}+\sin\frac{\pi}{9}+\sin\frac{2\pi}{9}+\sin\frac{5\pi}{18}$ is given by:
- ✓
$\sin\frac{7\pi}{18}+\sin\frac{4\pi}{9}$
- B
$1$
- C
$\cos\frac{\pi}{6}+\cos\frac{3\pi}{7}$
- D
$\cos\frac{\pi}{9}+\sin\frac{\pi}{9}$
AnswerCorrect option: A. $\sin\frac{7\pi}{18}+\sin\frac{4\pi}{9}$
The given expression is $\sin \frac{\pi}{18}+\sin\frac{\pi}{9}+\sin\frac{2\pi}{9}+\sin\frac{5\pi}{18}$
$=\Big(\sin\frac{5\pi}{18}+\sin\frac{\pi}{18}\Big)+\Big(\sin\frac{2\pi}{9}+\sin\frac{\pi}{9}\Big)$
$=2\sin\bigg(\frac{\frac{5\pi}{18}+\frac{\pi}{18}}{2}\bigg)\cdot\cos\bigg(\frac{\frac{5\pi}{18}-\frac{\pi}{8}}{2}\bigg)+2\sin\bigg(\frac{\frac{2\pi}{9}+\frac{\pi}{9}}{2}\bigg)\cdot\cos\bigg(\frac{\frac{2\pi}{9}-\frac{\pi}{9}}{2}\bigg)$
$=2\sin\frac{\pi}{6}\cdot\cos\frac{\pi}{9}+2\sin\frac{\pi}{6}\cdot\cos\frac{\pi}{18}$
$=2\times\frac{1}{2}\cos\frac{\pi}{9}+2\times\frac{1}{2}\cos\frac{\pi}{18}=\cos\frac{\pi}{9}+\cos\frac{\pi}{18}$
$=\sin\Big(\frac{\pi}{2}-\frac{\pi}{9}\Big)+\sin\Big(\frac{\pi}{2}-\frac{\pi}{18}\Big)=\sin\frac{7\pi}{18}+\sin\frac{8\pi}{18}$
$=\sin\frac{7\pi}{18}+\sin\frac{4\pi}{9}.$
Hence, the correct option is $(a).$
View full question & answer→MCQ 1201 Mark
At $3 : 40,$ the hour and minute hands of a clock are inclined at:
- A
$\frac{2\pi^{\text{c}}}{3}$
- B
$\frac{7\pi^{\text{c}}}{12}$
- ✓
$\frac{13\pi^{\text{c}}}{18}$
- D
$\frac{13\pi^{\text{c}}}{4}$
AnswerCorrect option: C. $\frac{13\pi^{\text{c}}}{18}$
We know that the hour of a clock completes one rotation in $12$ hours $= 360^\circ$
Angle traced by the hour hand in $12$ hours $= 360^\circ$
Now,
Angle traced by the hour hand in $8$ hours $30$ minutes, i.e.
We also know that the minute hand of a clock completes one rotation in $60$ minutes.
Angle traced by the minute hade in $30$ minutes $=\Big(\frac{360}{60}\times40^{\circ}\Big)=240^{\circ}$
Required angle between the two hands of the clock $= 240^\circ - 110^\circ = 130^\circ$
Value of the angle $($in radians$)$ between the two hands of the clock $\Big(130\times\frac{\pi}{180}\Big)^{\text{c}}=\frac{13\pi^{\text{c}}}{18}$
View full question & answer→MCQ 1211 Mark
If $\tan\alpha=\frac{\text{x}}{\text{x}+1}$ and $\tan\beta=\frac{1}{2\text{x}+1},$ then $\alpha+\beta$ is equal to:
- A
$\frac\pi2$
- B
$\frac\pi3$
- C
$\frac\pi6$
- ✓
$\frac\pi4$
AnswerCorrect option: D. $\frac\pi4$
It is given that $\tan\alpha=\frac{\text{x}}{\text{x}+1}$ and $\tan\beta=\frac{1}{2\text{x}+1}.$
$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$
$=\frac{\frac{\text{x}}{\text{x}+1}+\frac{1}{2\text{x+1}}}{1-\frac{\text{x}}{\text{x}+1}\times\frac{1}{2\text{x+1}}}$
$=\frac{\frac{\text{x}(2\text{x}+1)+(\text{x}+1)}{(\text{x}+1)(2\text{x}+1)}}{\frac{(\text{x}+1)(2\text{x}+1)-\text{x}}{(\text{x}+1)(2\text{x}+1)}}$
$=\frac{2\text{x}^2+\text{x}+\text{x}+1}{2\text{x}^2+3\text{x}+1+\text{x}}$
$=\frac{2\text{x}^2+2\text{x}+1}{2\text{x}^2+2\text{x}+1}$
$=1$
$\therefore\alpha+\beta=\frac\pi4\ (\tan\frac\pi4=1)$
Hence, the correct answer is option $D.$
View full question & answer→MCQ 1221 Mark
$\sin^2\frac{\pi}{18}+\sin^2\frac{\pi}{9}+\sin^2\frac{7\pi}{18}+\sin^2\frac{4\pi}{9}=$
AnswerWe have:
$\sin^2\frac{\pi}{18}+\sin^2\frac{\pi}{9}+\sin^2\frac{7\pi}{2}+\sin^2\frac{4\pi}{9}$
$=\sin^2\frac{\pi}{18}+\sin^2\frac{2\pi}{8}+\sin^2\frac{7\pi}{18}+\sin^2\frac{8\pi}{2}$
$=\sin^2\frac{\pi}{18}+\sin^2\frac{2\pi}{8}+\sin^2\Big(\frac{7\pi}{18}\Big)+\sin^2\Big(\frac{8\pi}{2}\Big)$
$=\sin^2\frac{\pi}{18}+\sin^2\frac{2\pi}{18}+\sin^2\Big(\frac{\pi}{2}-\frac{\pi}{18}\Big)+\sin^2\Big(\frac{\pi}{2}-\frac{\pi}{18}\Big)$
$=\sin^2\frac{\pi}{18}+\sin^2\frac{2\pi}{18}+\cos^2\frac{2\pi}{18}+\cos^2\frac{\pi}{18}$
$=\sin^2\frac{\pi}{18}+\sin^2\frac{\pi}{18}+\cos^2\frac{2\pi}{18}+\cos^2\frac{2\pi}{18}$
$=1+1$
$=2$
View full question & answer→MCQ 1231 Mark
Choose the correct answer. $\cos2\theta\cos2\phi+\sin^2(\theta-\phi)-\sin^2(\theta+\phi)$ is equal to:
- A
$\sin2(\theta+\phi)$
- ✓
$\cos2(\theta+\phi)$
- C
$\sin2(\theta-\phi)$
- D
$\cos2(\theta-\phi)$
AnswerCorrect option: B. $\cos2(\theta+\phi)$
$\cos2\theta\cos2\phi+\sin^2(\theta-\phi)-\sin^2(\theta+\phi)$
$=\cos2\theta\cos2\phi+\sin(\theta-\phi+\theta+\phi)\sin(\theta-\phi-\theta-\phi)$
$=\cos2\theta\cos2\phi-\sin2\theta\sin2\phi$
$=\cos(2\theta+2\phi)$
$=\cos2(\theta+\phi)$
View full question & answer→MCQ 1241 Mark
Choose the correct answer. Which of the following is correct?
AnswerCorrect option: B. $\sin1^\circ<\sin1$
we know that if $\theta$ increase then the value of $\sin\theta$ also increase
so, $\sin1^\circ<\sin1\Big[\because1\text{radian}=\frac{\pi}{180}\sin1\Big]$
Hence the correct option is $(b).$
View full question & answer→MCQ 1251 Mark
$\text{If }\text{A},\text{B},\text{C}\text{ are in }\text{A}.\text{P.},\text{than}\frac{\sin\text{A}-\sin\text{C}}{\cos\text{C}-\cos\text{A}}=$
- A
$\tan\text{B}$
- ✓
$\cot\text{B}$
- C
$\tan2\text{B}$
- D
AnswerCorrect option: B. $\cot\text{B}$
Since $A,B$ and $C$ are in $A.P,$
$\text{B}-\text{A}=\text{C}-\text{B}$
$\text{Or },2\text{B}=\text{A}+\text{C}$
$\frac{\sin\text{A}-\sin\text{C}}{\cos\text{C}-\cos\text{A}}$
$=\ \frac{2\sin\Big(\frac{\text{A}-\text{C}}{2}\Big)\cos\Big(\frac{\text{A+C}}{2}\Big)}{-2\sin\Big(\frac{\text{C+A}}{2}\Big)\sin\Big(\frac{\text{C}-\text{A}}{2}\Big)}$ $\Big[\because\ \sin\text{A}-\sin\text{B}=2\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\cos\Big(\frac{\text{A+B}}{2}\Big)\\\text{ and }\cos\text{A}-\cos\text{B}=-2\sin\Big(\frac{\text{A+B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\Big]$
$=\ \frac{\sin\Big(\frac{\text{A}-\text{C}}{2}\Big)\cos\Big(\frac{\text{A+C}}{2}\Big)}{-\sin\Big(\frac{\text{A+C}}{2}\Big)\sin\Big(\frac{\text{C}-\text{A}}{2}\Big)}$
$=\ \frac{\sin\Big(\frac{\text{A}-\text{C}}{2}\Big)\cos\Big(\frac{\text{A+C}}{2}\Big)}{\sin\Big(\frac{\text{A+C}}{2}\Big)\sin\Big(\frac{\text{C}-\text{A}}{2}\Big)}$
$=\ \frac{\cos\Big(\frac{\text{A+C}}{2}\Big)}{\sin\Big(\frac{\text{A+C}}{2}\Big)}$
$=\ \frac{\cos\text{B}}{\sin\text{B}}$
$=\ \cot\text{B}$
View full question & answer→MCQ 1261 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 1271 Mark
$\text{If }\cos\text{A}=\text{m}\cos\text{B},\text{ than }\cot\frac{\text{A+B}}{2}\cot\frac{\text{B}-\text{A}}{2}=$
- A
$\frac{\text{m}-1}{\text{m}+1}$
- B
$\frac{\text{m}+2}{\text{m}-2}$
- ✓
$\frac{\text{m}+1}{\text{m-1}}$
- D
AnswerCorrect option: C. $\frac{\text{m}+1}{\text{m-1}}$
Given:
$\cos\text{A}=\text{m}\cos\text{B}$
$\Rightarrow\ \frac{\cos\text{A}}{\cos\text{B}}=\frac{\text{m}}{1}$
$\Rightarrow\ \frac{\cos\text{A}+\cos\text{B}}{\cos\text{A}-\cos\text{B}}=\frac{\text{m}+1}{\text{m}-1}$
$\Rightarrow\ \frac{2\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\cos\Big(\frac{\text{A+B}}{2}\Big)}{-2\sin\Big(\frac{\text{B+A}}{2}\Big)\sin\Big(\frac{\text{B}-\text{A}}{2}\Big)}=\frac{\text{m}+1}{\text{m}-1}$ $\Big[\because\ \cos\text{A}+\cos\text{B}=2\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\cos\Big(\frac{\text{A+B}}{2}\Big)\\\text{ and } \cos\text{A}-\cos\text{B}=2\sin\Big(\frac{\text{A}+\text{B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\Big]$
$\Rightarrow\ \frac{\cos\Big(\frac{\text{B}-\text{A}}{2}\Big)\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)}{\sin\Big(\frac{\text{A+B}}{2}\Big)\sin\Big(\frac{\text{B}-\text{A}}{2}\Big)}=\frac{\text{m}+1}{\text{m}-1}$
$\Rightarrow\ \cot\Big(\frac{\text{A+B}}{2}\Big)\cot\Big(\frac{\text{B}-\text{A}}{2}\Big)=\frac{\text{m}+1}{\text{m}-1}$
View full question & answer→MCQ 1281 Mark
Let $A$ and $B$ denote the statements : $\text{A}:\cos\text{a}+\cos\text{b}+\cos\text{g}=0$
$\text{B}:\sin\text{a}+\sin\text{b}+\sin\text{g}=0$
If $\cos(\beta-\text{y})+\cos(\text{y}-\alpha)+\cos(\alpha-\beta)=\frac{-3}{2}$ then:
- ✓
$A$ is false and $B$ is true.
- B
Both $A$ and $B$ are true.
- C
Both $A$ and $B$ are false.
- D
$A$ is true and $B$ is false.
AnswerCorrect option: A. $A$ is false and $B$ is true.
View full question & answer→MCQ 1291 Mark
If $\cos\text{P}=\frac{1}{7}$ then $\cos\text{Q}=\frac{13}{14},$ where $P$ and $Q$ both are acute angles. Then, the value of $P - Q$ is:
- A
$\frac{\pi}{6}$
- ✓
$\frac\pi3$
- C
$\frac\pi4$
- D
$\frac{5\pi}{12}$
AnswerCorrect option: B. $\frac\pi3$
$\cos\text{P}=\frac17,\cos\text{Q}=\frac{13}{14}$
$\therefore\sin\text{P}=\sqrt{1-\frac{1}{49}}=\frac{4\sqrt{3}}{7}$ and $\sin\text{Q}=\sqrt{1-\frac{169}{196}}=\frac{3\sqrt{3}}{14}$
Hence, $\tan\text{P}=4\sqrt{3},\tan\text{Q}=\frac{3\sqrt{3}}{13}$
$\cos(\text{P - Q})=\cos\text{P}\cos\text{Q}+\sin\text{P}\sin\text{Q}$
$=\frac{1}{7}\times\frac{13}{14}+\frac{4\sqrt{3}}{7}\times\frac{3\sqrt{3}}{14}$
$=\frac{13+36}{98}$
$=\frac{49}{98}$
$\therefore\cos(\text{P - Q})=\frac12$
$\Rightarrow\text{P - Q}=\cos^{-1}\frac12$
$\Rightarrow\text{P - Q}=60^\circ$
Hence, the correct answer is option $B.$
View full question & answer→MCQ 1301 Mark
$\tan3\text{A}-\tan2\text{A}-\tan\text{A}$ is equal to:
- ✓
$\tan3\text{A}\tan2\text{A}\tan\text{A}$
- B
$-\tan3\text{A}-\tan2\text{A}\tan\text{A}$
- C
$\tan\text{A}\tan2\text{A}\tan2\text{A}\tan3\text{A}-\tan3\text{A}\tan\text{A}$
- D
AnswerCorrect option: A. $\tan3\text{A}\tan2\text{A}\tan\text{A}$
$3\text{A}=2\text{A}+\text{A}$
$\Rightarrow\tan3\text{A}=\tan(2\text{A}+\text{A})$
$\Rightarrow\tan3\text{A}=\tan(2\text{A}+\text{A})=\frac{\tan2\text{A}+\tan\text{A}}{1-\tan2\text{A}\tan\text{A}}$
$\Rightarrow\tan3\text{A}-\tan3\text{A}\tan2\text{A}\tan\text{A}=\tan2\text{A}+\tan\text{A}$
$\Rightarrow\tan3\text{A}-\tan2\text{A}-\tan\text{A}=\tan3\text{A}\tan2\text{A}\tan\text{A}$
View full question & answer→MCQ 1311 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 1321 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 1331 Mark
The smallest positive angle which satisfies the equation $2\sin^2\text{x}+\sqrt{3}\cos\text{x}+1=0$ is:
- ✓
$\frac{5\pi}{6}$
- B
$\frac{2\pi}{3}$
- C
$\frac{\pi}{3}$
- D
$\frac{\pi}{6}$
AnswerCorrect option: A. $\frac{5\pi}{6}$
Given:
$2\sin^2\text{x}+\sqrt{3}\cos\text{x}+1=0$
$\Rightarrow2(1-\cos^2\text{x})+\sqrt{3}\cos\text{x}+1=0$
$\Rightarrow2-2\cos^2\text{x}+\sqrt{3}\cos\text{x}+1=0$
$\Rightarrow2\cos^2\text{x}-\sqrt{3}\cos\text{x}-3=0$
$\Rightarrow2\cos^2\text{x}-2\sqrt{3}\cos\text{x}+\sqrt{3}\cos\text{x}-3=0$
$\Rightarrow2\cos\text{x}\Big(\cos\text{x}-\sqrt{3}\Big)+\sqrt{3}\Big(\cos\text{x}-\sqrt{3}\Big)=0$
$\Rightarrow\Big(2\cos\text{x}+\sqrt{3}\Big)\Big(\cos\text{x}-\sqrt{3}\Big)=0$
$\therefore\cos\text{x}+\sqrt{3}=0$ or, $\cos\text{x}=\sqrt{3}$ is not possible.
$\Rightarrow\cos\text{x}=\cos\Big(\frac{\pi}{6}\Big)$
$\Rightarrow\text{x}=2\text{n}\pi\pm\frac{5\pi}{6},\ \text{n}\in\text{Z}$
For $n = 0$, the value of $x$ is $\pm\frac{5\pi}{6}.$
Hence, the smallest positive angle is $\frac{5\pi}{6}.$
View full question & answer→MCQ 1341 Mark
Choose the correct answer. The value of $\cos^248^\circ-\sin^212^\circ$ is:
AnswerCorrect option: A. $\frac{\sqrt{5}+1}{8}$
Given expression is $\cos^248^\circ-\sin^212^\circ$
$\cos^248^\circ-\sin^212^\circ=\cos(48^\circ+12^\circ).\cos(48^\circ-12^\circ)$
$[\therefore\cos^2\text{A}-\sin^2\text{B}=\cos(\text{A+B}).\cos(\text{A}-\text{B})]$
$=\cos60^\circ.\cos36^\circ=\frac{1}{2}\times\frac{\sqrt5+1}{4}=\frac{\sqrt5+1}{8}$
Hence,the correct option is $(a).$
View full question & answer→MCQ 1351 Mark
If $\tan\text{A}=\frac{\text{a}}{\text{a}+1}$ and $\text{B}=\frac{1}{2\text{a}+1},$ then the value of $A + B$ is:
- A
$0$
- B
$\frac{\pi}{2}$
- C
$\frac\pi3$
- ✓
$\frac\pi4$
AnswerCorrect option: D. $\frac\pi4$
$\tan(\text{A+B})=\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}$
$=\frac{\frac{\text{a}}{\text{a}+1}+\frac{1}{2\text{a}+1}}{1-\frac{\text{a}}{(\text{a}+1)(2\text{a}+1)}}$
$=\frac{2\text{a}^2+\text{a}+\text{a}+1}{2\text{a}^2+3\text{a}+1-\text{a}}$
$=\frac{2\text{a}^2+2\text{a}+1}{2\text{a}^2+2\text{a}+1}$
$=1$
$\therefore \text{ A+B}=\tan^{-1}(1)=\frac\pi4.$
View full question & answer→MCQ 1361 Mark
The number of solution of $\tan\text{x}+\sec\text{x}=2\cos\text{x}$ in $[0, 2 p]$ is:
View full question & answer→MCQ 1371 Mark
If $\tan\text{x}=\text{x}+\frac{1}{4\text{x}},$ then $\sec\text{x}+\tan\text{x}=$
- ✓
$-2\text{x},\frac{1}{2\text{x}}$
- B
$-\frac{1}{2\text{x}},2\text{x}$
- C
$2\text{x}$
- D
$2\text{x},\frac{1}{\text{x}2}$
AnswerCorrect option: A. $-2\text{x},\frac{1}{2\text{x}}$
We have:
$\tan\text{x}=\text{x}-\frac{1}{4\text{x}}$
$\Rightarrow\sec^2\text{x}=1+\tan^2\text{x}$
$\Rightarrow\sec^2\text{x}=1+\Big(\text{x}-\frac{1}{4\text{x}}\Big)^2$
$\Rightarrow\sec^2\text{x}=\text{x}^2+\frac{1}{16\text{x}^2}+\frac{1}{2}$
$\Rightarrow \sec\text{x}^2 =\Big(\text{x}+\frac{1}{4\text{x}}\Big)^2$
$\therefore \sec\text{x}=\pm\Big(\text{x}+\frac{1}{4\text{x}}\Big)$
$\Rightarrow \sec\text{x}-\tan\text{x} =\Big(\text{x}+\frac{1}{4\text{x}}\Big)- \Big(\text{x}-\frac{1}{4\text{x}}\Big)\text{ or} -\Big(\text{x}+\frac{1}{4\text{x}}\Big)\text{or}- \Big(\text{x}-\frac{1}{4\text{x}}\Big)$
$=\frac{1}{2\text{x}}$ or $-2\text{x}$
View full question & answer→MCQ 1381 Mark
Choose the correct answer.If $\tan\alpha=\frac{\text{m}}{\text{m}+1},\tan\beta=\frac{1}{2\text{m}+1},$ then $\alpha+\beta$ is equal to:
- A
$\frac{\pi}{2}$
- B
$\frac{\pi}{3}$
- C
$\frac{\pi}{6}$
- ✓
$\frac{\pi}{4}$
AnswerCorrect option: D. $\frac{\pi}{4}$
Given that, $\tan\alpha=\frac{\text{m}}{\text{m}+1}$ and $\tan\beta=\frac{1}{2\text{m}+1}$
Now, $\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\cdot\tan\beta}=\frac{\frac{\text{m}}{\text{m}+1}+\frac{1}{2\text{m}+1}}{1-\Big(\frac{\text{m}}{\text{m}+1}\Big)\Big(\frac{1}{2\text{m}+1}\Big)}$
$=\frac{\text{m}(2\text{m}+1)+\text{m}+1}{(\text{m}+1)(2\text{m}+1)-\text{m}}=\frac{2\text{m}^2+2\text{m}+1}{2\text{m}^2+3\text{m}+1-\text{m}}=1$
$\therefore\alpha+\beta=\frac{\pi}{4}$
View full question & answer→MCQ 1391 Mark
Choose the correct answer. If $\alpha+\beta=\frac{\pi}{4},$ then the value of $(1+\tan\alpha)(1+\tan\beta)$ is:
AnswerGiven that, $\alpha+\beta=\frac{\pi}{4}$
$\Rightarrow\tan(\alpha+\beta)=\tan\frac{\pi}{4}$
$\Rightarrow\frac{\tan\alpha+\tan\beta}{1-tan\alpha\tan\beta}=1$
$\Rightarrow\tan\alpha+\tan\beta=1-\tan\alpha\tan\beta$
$\Rightarrow\tan\alpha+\tan\beta+\tan\alpha\tan\beta=1$
On adding $1$ both sides, we get,
$\Rightarrow1+\tan\alpha+\tan\beta+\tan\alpha\tan\beta=1+1$
$\Rightarrow1(1+\tan\alpha)+\tan\beta(1+\tan\alpha)=2$
$\Rightarrow(1+\tan\alpha)(1+\tan\beta)=2$
Hence, the correct option is $(b)$
View full question & answer→MCQ 1401 Mark
If $\cos\text{x}=-\frac{1}{2}$ and $0<\text{x}<2\pi,$ then the solutions are:
- A
$\text{x}=\frac{\pi}{3},\ \frac{4\pi}{3}$
- ✓
$\text{x}=\frac{2\pi}{3},\ \frac{4\pi}{3}$
- C
$\text{x}=\frac{2\pi}{3},\ \frac{7\pi}{3}$
- D
$\theta=\frac{2\pi}{3},\ \frac{5\pi}{3}$
AnswerCorrect option: B. $\text{x}=\frac{2\pi}{3},\ \frac{4\pi}{3}$
Given equation:
$\cos\text{x}=-\frac{1}{2}$
$\Rightarrow\cos\text{x}=\cos\frac{2\pi}{3}$
$\Rightarrow\text{x}=\frac{2\pi}{3}$
or
$\cos\text{x}=\cos\frac{4\pi}{3}$
$\Rightarrow\text{x}=\frac{4\pi}{3}$
so, both $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$ line in $0<\text{}\text{x}<2\pi.$
View full question & answer→MCQ 1411 Mark
Let $a, b$ be such that $\pi<\alpha-\beta<3\pi \text{ D}$
If $\sin\alpha+\sin\beta=-\frac{21}{65}$ and, $\cos\alpha+\cos\beta=-\frac{27}{65} .$ then the value is $\alpha-\frac{\beta}{2}$
- A
$\frac{-6}{65}$
- B
$\frac{3}{\sqrt{130}}$
- C
$\frac{6}{65}$
- ✓
$\frac{-3}{130}$
AnswerCorrect option: D. $\frac{-3}{130}$
View full question & answer→MCQ 1421 Mark
In a $\triangle\text{ABC},$ if $a = 2, \angle\text{B}=60^{\circ}$ and $\angle\text{C}=75^{\circ},$ then $b =$
- A
$\sqrt{3}$
- ✓
$\sqrt{6}$
- C
$\sqrt{9}$
- D
$1+\sqrt{2}$
AnswerCorrect option: B. $\sqrt{6}$
It is given that $\text{a}=2,\angle\text{B}=60^{\circ}$ and $\angle\text{C}=75^{\circ}.$
In $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^{\circ} ($Angle sum property$)$
$\Rightarrow\angle\text{A}+60^{\circ}+75^{\circ}=180^{\circ}$
$\Rightarrow\angle\text{A}=180^{\circ}-135^{\circ}=45^{\circ}$
Using sine rule, we get
$\frac{2}{\sin45^{\circ}}=\frac{\text{b}}{\sin60^{\circ}}$ $\Big(\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\sin\text{B}}=\frac{\text{c}}{\sin\text{C}}\Big)$
$\Rightarrow\text{b}=\frac{2\times\frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}}}=\sqrt{6}$
Hence, the correct answer is option $(b).$
View full question & answer→MCQ 1431 Mark
If $\frac{\pi}{2}<\text{x}<\frac{3\pi}{2},$ then $\sqrt{\frac{1-\sin\text{x}}{1+\sin\text{x}}}$ is equal to:
- A
$\sec\text{x}-\tan\text{x}$
- B
$\sec\text{x}+\tan\text{x}$
- ✓
$\tan\text{x}-\sec\text{x}$
- D
AnswerCorrect option: C. $\tan\text{x}-\sec\text{x}$
$\tan\text{x}-\sec\text{x}$
$\sqrt{\frac{1-\sin\text{x}}{1+\sin\text{x}}}$
$=\sqrt{\frac{(1-\sin\text{x})(1-\sin\text{x})}{(1+\sin\text{x})(1-\sin\text{x})}}$
$=\sqrt{\frac{(1-\sin\text{x})^2}{1-\sin^2\text{x}}}$
$=\sqrt{\frac{(1-\sin\text{x})^2}{\cos^2\text{x}}}$
$=\frac{(1-\sin\text{x})}{-\cos\text{x}}$ $[\text{as},\frac{\pi}{2}<\text{x}<\frac{3\pi}{2},\text{so}\cos\theta \text{ will}\text{ be}\text{ negative}]$
$=-(\sec\text{x}-\tan\text{x})$
$=-\sec\text{x} +\tan\text{x}$
View full question & answer→MCQ 1441 Mark
What is the value of $\sin\Big(\frac{5\pi}{12}\Big)?$
AnswerCorrect option: B. $\sqrt{6}+\frac{\sqrt{2}}{4}$
View full question & answer→MCQ 1451 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 1461 Mark
A value of $x$ satisfying $\cos\text{x}+\sqrt{3}\sin\text{x}=2$ is:
- A
$\frac{5\pi}{3}$
- B
$\frac{4\pi}{3}$
- C
$\frac{2\pi}{3}$
- ✓
$\frac{\pi}{3}$
AnswerCorrect option: D. $\frac{\pi}{3}$
Given equation:
$\cot\text{x}+\sqrt{3}\sin\text{x}=2\ .....(1)$
Thus, the equation is of the form $\text{a}\cos\text{x}+\text{b}\sin\text{x}=\text{c},$ where $\text{a}=1,\ \text{b}=\sqrt{3}$ and $\text{c}=3$
Let:
$\text{a}=\text{r}\cos\alpha$ and $\text{b}=\text{r}\sin\alpha$
$1=\text{r}\cos\alpha$ and $\sqrt{3}=\text{r}\sin\alpha$
$\Rightarrow\text{r}\sqrt{\text{a}^2+\text{b}^2}=\sqrt{\Big(\sqrt{3}\Big)^2+1^2}=2$ and
$\tan\alpha=\frac{\text{b}}{\text{a}}$
$\Rightarrow\tan\alpha=\frac{\sqrt{3}}{1}$
$\Rightarrow\tan\alpha\tan\frac{\pi}{3}$
$\Rightarrow\alpha=\frac{\pi}{3}$
On putting $\text{a}=1=\text{r}\cos\alpha$ and $\text{b}=\sqrt{3}=\text{r}\sin\alpha$ in equation $(1)$ we get:
$\text{r}\cos\alpha\cos\text{x}+\text{r}\sin\alpha\sin\text{x}=2$
$\Rightarrow\text{r}\cos(\text{x}-\alpha)=2$
$\Rightarrow\text{r}\cos(\text{x}-\frac{\pi}{3})=2$
$\Rightarrow\text{2}\cos(\text{x}-\frac{\pi}{3})=2$
$\Rightarrow\text{}\cos(\text{x}-\frac{\pi}{3})=1$
$\Rightarrow\text{}\cos(\text{x}-\frac{\pi}{3})=\cos0$
$\Rightarrow\text{x}-\frac{\pi}{3}=0$
$\Rightarrow\text{x}=\frac{\pi}{3}$
View full question & answer→MCQ 1471 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 1481 Mark
Choose the correct answer. The value of $\cos1^\circ\cos2^\circ\cos3^\circ...\cos179^\circ$ is:
- A
$\frac{1}{\sqrt{2}}$
- ✓
$0$
- C
$1$
- D
$-1$
Answersince $\cos90^\circ=0,$ we have
$\cos1^\circ\cos2^\circ\cos3^\circ\dots\cos90^\circ\dots\cos179^\circ=0$
View full question & answer→MCQ 1491 Mark
Choose the correct answer. If $\text{f(x)}=\cos^2\text{x}+\sec^2\text{x},$ then:
- A
$\text{f(x)}<1$
- B
$\text{f(x)}=1$
- C
$2<\text{f(x)}<1$
- ✓
$\text{f(x)}\geq2$
AnswerCorrect option: D. $\text{f(x)}\geq2$
Given that; $\text{f(x)}=\cos^2\text{x}+\sec^2\text{x}$
We know that $\text{AM}\geq\text{GM}$
$\Rightarrow\frac{\cos^2\text{x}+\sec^2\text{x}}{2}\geq\sqrt{\cos^2\text{x}.\sec^2\text{x}}$
$\Rightarrow\frac{\cos^2\text{x}+\sec^2\text{x}}{2}\geq1$ $\big[\text{since}\sec\theta=\frac{1}{\cos\theta}\big]$
$\Rightarrow\cos^2\text{x}+\sec^2\text{x}\geq2$
$\Rightarrow\text{f(x)}\geq2$
Hence, the correct option is $(d)$
View full question & answer→MCQ 1501 Mark
If $\sqrt{3}\cos\text{x}+\sin\text{x}=\sqrt{2},$ then general value of $\theta$ is:
- A
$\text{n}\pi+(-1)^{\text{n}}\frac{\pi}{4},\ \text{n}\in\text{Z}$
- B
$(-1)^\text{n}\frac{\pi}{4}-\frac{\pi}{3},\ \text{n}\in\text{Z}$
- C
$\text{n}\pi+\frac{\pi}{4}-\frac{\pi}{3},\ \text{n}\in\text{Z}$
- ✓
$\text{n}\pi+(-1)^\text{n}\frac{\pi}{4}-\frac{\pi}{3},\ \text{n}\in\text{Z}$
AnswerCorrect option: D. $\text{n}\pi+(-1)^\text{n}\frac{\pi}{4}-\frac{\pi}{3},\ \text{n}\in\text{Z}$
Given equation:
$\sqrt{3}\cos\text{x}+\sin\text{x}\sqrt{2}\ .....(1)$
Thus, is of the form $\text{a}\cos\text{x}+\text{b}\sin\text{x}=\text{c},$ where $\text{a}=\sqrt{3},\ \text{b}=1$ and $\text{c}=\sqrt2$
Let:
$\text{a}=\text{r}\sin\alpha$ and $\text{b}=\text{r}\cos\alpha$
Now,
$\text{r}=\sqrt{\text{a}^2+\text{b}^2}=\sqrt{\big(\sqrt{3}\big)^2+1^2}=2$
And
$\tan\alpha=\frac{\sqrt{3}}{1}$
$\Rightarrow\tan\alpha=\frac{\sqrt{3}}{1}$
$\Rightarrow\tan\alpha=\tan\frac{\pi}{3}$
$\Rightarrow\alpha=\frac{\pi}{3}$
Putting $\text{a}=\sqrt{3}=\text{r}\sin\alpha$ and $\text{b}=1=\text{r}\cos\alpha$ in equation $(i)$, we get:
$\text{r}\cos\text{x}\sin\alpha+\text{r}\sin\text{x}\cos\alpha=\sqrt{2}$
$\Rightarrow\text{r}\sin(\text{x}+\alpha)=\sqrt{2}$
$\Rightarrow2\sin(\text{x}+\alpha)=\sqrt{2}$
$\Rightarrow\sin\big(\text{x}+\frac{\pi}{3}\big)=\frac{1}{\sqrt{2}}$
$\Rightarrow\sin\big(\text{x}+\frac{\pi}{3}\big)=\cos\frac{\pi}{4}$
$\Rightarrow\text{x}\frac{\pi}{3}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{4},\ \text{n}\in\text{Z}$
$\Rightarrow\text{x}\text{n}\pi+(-1)^\text{n}\frac{\pi}{4}-\frac{\pi}{3},\ \ \text{n}\in\text{Z}$
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