MCQ 11 Mark
If $\text{cosec x}+\cot \text{x}=\frac{11}{2},$ then $\tan\text{x}$ is equal to:
- A
$\frac{21}{22}$
- B
$\frac{15}{16}$
- ✓
$\frac{44}{117}$
- D
$\frac{117}{44}$
AnswerCorrect option: C. $\frac{44}{117}$
We have:
$\text{cosec}\text{ x}+\cot\text{x}=\frac{11}{2}\cdots(1)$
$\Rightarrow\frac{1}{\text{cosec}\text{ x}+\cot\text{x}}=\frac{2}{11}$
$\Rightarrow\frac{\text{cosec}^2\text{x}-\cot\text{x}}{\text{cosec }\text{x}-\cot\text{x}}=\frac{2}{11}$
$\Rightarrow\frac{(\text{cosec}\text{ x}+\cot\text{x})(\text{cosec}\text{ x}-\cot\text{x})}{(\text{cosec}\text{ x}+\cot\text{x})}=\frac{2}{11}$
$\therefore\text{cosec}\text{ x}-\cot\text{x}=\frac{2}{11}\cdots(2)$
subtracting (2) from (1):
$\Rightarrow 2\cot\text{x}=\frac{11}{2}-\frac{2}{11}$
$\Rightarrow2\cot\text{x}=\frac{121-4}{22}$
$\Rightarrow2\cot\text{x}=\frac{117}{22}$
$\Rightarrow\cot\text{x}=\frac{117}{44}$
$\Rightarrow\frac{1}{\tan\text{x}}=\frac{117}{44}$
$\Rightarrow\tan\text{x}=\frac{44}{117}$
View full question & answer→MCQ 21 Mark
$\sec^2\text{x}=\frac{4\text{xy}}{(\text{x}+\text{y})^2}$ is true if and only if
AnswerCorrect option: B. $\text{x=y, x}\neq0$
We have:
$\sec^2\text{x}=\frac{4\text{xy}}{(\text{x}+\text{y})^2}$
$\Rightarrow\frac{4\text{x}\text{y}}{(\text{x}+\text{y})^2}\geq1 $ $[\therefore\sec^2\text{x}\geq1]$
$\Rightarrow4\text{xy}\geq(\text{x}+\text{y})^2$
$\Rightarrow4\text{xy}\geq\text{x}^2+\text{y}^2+2\text{xy}$
$\Rightarrow2\text{xy}\geq\text{x}^2+\text{y}^2$
$\Rightarrow(\text{x}-\text{y})^2\leq0$
$\Rightarrow(\text{x}-\text{y})\leq0$
$\Rightarrow\text{x}=\text{y}$
For $\text{x}=0,\sec^2\text{x}$ will not be defined,
$\Rightarrow\text{x}\neq 0$
$\therefore\text{x}=\text{y}$
View full question & answer→MCQ 31 Mark
If A lies in second quadrant $3\tan\text{A}+4=0,$ then the value of $2\cot\text{A}-5\cot\text{A}+\sin\text{A}$ is:
- A
$-\frac{53}{10}$
- ✓
$\frac{23}{10}$
- C
$\frac{37}{10}$
- D
$\frac{7}{10}$
AnswerCorrect option: B. $\frac{23}{10}$
It is given that $\frac{\pi}{2}<\text{A}<\pi$
$3\tan\text{A}+4=0$
$\Rightarrow\tan\text{A}=-\frac{4}{3}$
$\Rightarrow\cot\text{A}=-\frac{3}{4}$
Now,
$\sec\text{A}=\pm\sqrt{1+\tan^2\text{A}}$
$=\pm\sqrt{1+\frac{16}{9}}$
$=\pm\sqrt{\frac{25}{9}}=\pm\frac{5}{3}$
Also,
$\sin\text{A}=\pm\sqrt{1-\cos^2\text{A}}$
$=\pm\sqrt{1-\frac{9}{25}}$
$=\pm\sqrt{\frac{16}{25}}=\pm\frac{4}{5}$
$\therefore\sin\text{A}=\frac{4}{5}$ (A lines in 2 nd quadrant)
so,
$2\cot\text{A}-5\cos\text{A}+\sin\text{A}$
$=2\times\Big(-\frac{3}{5}\Big)-5\times\Big(-\frac{3}{5}\Big)+\frac{4}{5}$
$=-\frac{3}{2}+3+\frac{4}{5}$
$=-\frac{15+30+8}{10} $
$=\frac{23}{10}$
Hence,the correct answer is option B.
View full question & answer→MCQ 41 Mark
If $\text{cosec x}+\cot\text{x}=\frac{11}{2},$ then $\tan\text{x}=$
- A
$\frac{21}{22}$
- B
$\frac{15}{16}$
- ✓
$\frac{44}{117}$
- D
$\frac{117}{43}$
AnswerCorrect option: C. $\frac{44}{117}$
We have:
$\text{cosec}\text{ x}+\cot\text{x}=\frac{11}{2}\cdots(1)$$$
$\Rightarrow\frac{1}{\text{cosec}\text{ x}+\cot\text{x}}=\frac{2}{11}$
$\Rightarrow\frac{\text{cosec}^2\text{x}-\cot\text{x}}{\text{cosec}\text{ x}-\cot\text{x}}=\frac{2}{11}$
$\Rightarrow\frac{(\text{cosec}\text{ x}+\cot\text{x})(\text{cosec}\text{ x}-\cot\text{x})}{(\text{cosec}\text{ x}+\cot\text{x})}=\frac{2}{11}$
$\therefore\text{cosec}\text{ x}-\cot\text{x}=\frac{2}{11}\cdots(2)$
subtracting (2) from (1):
$\Rightarrow 2\cot\text{x}=\frac{11}{2}-\frac{2}{11}$
$\Rightarrow2\cot\text{x}=\frac{121-4}{22}$
$\Rightarrow2\cot\text{x}=\frac{117}{22}$
$\Rightarrow\cot\text{x}=\frac{117}{44}$
$\Rightarrow\frac{1}{\tan\text{x}}=\frac{117}{44}$
$\Rightarrow\tan\text{x}=\frac{44}{117}$
View full question & answer→MCQ 51 Mark
The value of $\sin^25^\circ+\sin^210^\circ+\sin^215^\circ+\ ...\ +\sin^285^\circ+\sin^290^\circ$ is:
AnswerWe have:
$\sin^25^\circ+\sin^210^\circ+\sin^215^\circ+...+\sin^285^\circ+\sin^290^\circ$
$=\sin^25^\circ+\sin^210^\circ+\sin^215^\circ+...+\sin^2(90^\circ-10^\circ)+\sin^2(90^\circ-5^\circ)+\sin^290^\circ$
$=\sin^25^\circ+\sin^210^\circ+\sin^215^\circ+...+\cos^210^\circ+\cos^25^\circ+\sin^290^\circ$
$=(\sin^25^\circ+\cos^25^\circ)+(\sin^210^\circ+\cos^210^\circ)+(\sin^215^\circ+\cos^215^\circ)$
$+(\sin^220^\circ+\cos^220^\circ)+(\sin^225^\circ+\cos^225^\circ)+(\sin^230^\circ+\cos^230^\circ)$
$+(\sin^235^\circ+\cos^235^\circ)+(\sin^240^\circ+\cos^240^\circ)+\sin^245^\circ+\sin^290^\circ$
$=1+1+1+1+1+1+1+1+\Big(\frac{1}{\sqrt2}\Big)+(1)^2$ $[\because \sin^2\theta+\cos^2\theta=1]$
$=8+\frac{1}{2}+1$
$=9.5$
View full question & answer→MCQ 61 Mark
If $0<\text{x}<\frac{\pi}{2},$ and if $\frac{\text{y+1}}{1-\text{y}}=\sqrt{\frac{1+\sin\text{x}}{1-\sin\text{x}}},$ then y is equal to:
AnswerCorrect option: B. $\tan\frac{\text{x}}{2}$
We have:
$\frac{\text{y}+1}{1-\text{y}}=\sqrt{\frac{1+\sin\text{x}}{1-\sin\text{x}}}$
$\Rightarrow\frac{\text{y}+1}{1-\text{y}}=\sqrt{\frac{\cos^2\frac{\text{x}}{2}+\sin^2\frac{\text{x}}{2} +2\sin\frac{\text{x}}{2}-\cos\frac{\text{x}}{2}}{\cos^2\frac{\text{x}}{2}+\sin^2\frac{\text{x}}{2}-2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}}$
$\Rightarrow\frac{\text{y}+1}{1-\text{y}}=\sqrt{\frac{\Big(cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\Big)^2}{\Big(\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{2}\Big)^2}}$
$\Rightarrow\frac{\text{y}+1}{1-\text{y}}=\frac{\Big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\Big)}{\Big(\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{2}\Big)}$ $[\because 0< \text{x }<\frac{\pi}{2}\Rightarrow0< \frac{\pi}{2}<\frac{\pi}{4}, 0 \text{ to } \frac{\pi}{4}\cos\text{x}>\sin\text{x}]$
$\Rightarrow\frac{\frac{\cos\frac{\text{x}}{2}}{\cos\frac{\text{x}}{2}} +\frac{\sin\frac{\text{x}}{2}}{\cos\frac{\text{x}}{2}}}{\frac{\cos\frac{\text{x}}{2}}{\cos\frac{\text{x}}{2}}-\frac{\sin\frac{\text{x}}{2}}{\cos\frac{\text{x}}{2}}}$
$\Rightarrow\frac{\text{y}+1}{1-\text{y}}=\frac{1+\tan\frac{\text{x}}{2}}{1-\tan\frac{\text{x}}{2}}$
comparing both the sides:
$\text{y}=\tan\frac{\text{x}}{2}$
View full question & answer→MCQ 71 Mark
If $\frac{\pi}{2}<\text{x}<{\pi},$ and if $=\sqrt{\frac{1-\sin\text{x}}{1+\sin\text{x}}}+\sqrt{\frac{1+\sin\text{x}}{1-\sin\text{x}}},$ is equal to:
- A
$2\sec\text{x}$
- ✓
$-2\sec\text{x}$
- C
$\sec\text{x}$
- D
$-\sec\text{x}$
AnswerCorrect option: B. $-2\sec\text{x}$
$-2\sec\text{x}$
$\sqrt{\frac{1-\sin\text{x}}{1+\sin\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})(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}{1-\sin^2\text{x}}}$
$=\sqrt{\frac{(1-\sin\text{x})^2}{\cos^2\text{x}}}+\sqrt{\frac{(1+\sin\text{x})^2}{\cos^2\text{x}}}$
$=\frac{(1-\sin\text{x})}{-\cos\text{x}}+\frac{(1+\sin\text{x})}{-\cos\text{x}}$ $[\frac{\pi}{2}<\text{x}<\pi,\text{so}\cos\text{x}\text{ will }\text{be }\text{negative}.]$
$-(\sec\text{x}-\tan\text{x})-(\sec\text{x}+\tan\text{x})$
$=-2\sec\text{x}$
View full question & answer→MCQ 81 Mark
$\sin^6\text{A}+\cos^6\text{A}+3\sin^2\text{A}\cos^2\text{A}=$
AnswerWe have:
$\sin^6\text{A}+\cos^2\text{A}+3(\sin^2\text{A})(\cos^2\text{A})$
$=(\sin^2\text{A})^3+(\cos^2\text{A})^3+3(\sin^2\text{A})(\cos^2\text{A})\times1$
$=(\sin^2\text{A})^3+(\cos^2\text{A})^3+3(\sin^2\text{A})(\cos^2\text{A})(\sin^2\text{A}+\cos^2\text{A})$
$=(\sin^2\text{A}+\cos^2\text{A})^3$
$=1^3=1$
View full question & answer→MCQ 91 Mark
If $\frac{3\pi}{4}<\text{a}<\pi,$ then $\sqrt{2\cot\text{a}+\frac{1}{\sin^\text{a}}}$ is equal to:
- A
$1-\cot\text{a}$
- B
$1+\cot\text{a}$
- C
$-1+\cot\text{a}$
- ✓
$-1-\cot\text{a}$
AnswerCorrect option: D. $-1-\cot\text{a}$
We have:
$\sqrt{2\cot\alpha+\frac{1}{\sin\alpha}}$
$=\sqrt{\frac{2\cos\alpha}{\sin\alpha}+\frac{1}{\sin^2\alpha}}$
$=\sqrt{\frac{2\sin\alpha\cos\alpha+1}{\sin\alpha}}$
$=\sqrt{\frac{2\sin\alpha\cos\alpha+\sin^2\alpha+\cos^2\alpha}{\sin\alpha}}$
$=\sqrt{\frac{(\sin\alpha+\cos\alpha)^2}{\sin^2\alpha}}$
$=\sqrt{(1+\cot\alpha)^2}$
$=|1+\cot\alpha|$
$=-(1+\cot\alpha) $ $$ $[\text{ when}\frac{3\pi}{4}<\alpha<\pi,\cot\alpha<-1\Rightarrow\cot\alpha+1<0\big]$
$=-1-\cot\alpha$
View full question & answer→MCQ 101 Mark
If $\tan\theta+\sec\theta=\text{e}^\text{x},$ then $\cos\theta=$
- A
$\frac{\text{e}^\text{x}+\text{e}^{-\text{x}}}{2}$
- ✓
$\frac{2}{\text{e}^\text{x}+\text{e}^{-\text{x}}}$
- C
$\frac{\text{e}^\text{x}-\text{e}^{-\text{x}}}{2}$
- D
$\frac{\text{e}^\text{x}-\text{e}^{-\text{x}}}{\text{e}^\text{x}+\text{e}^{-\text{x}}}$
AnswerCorrect option: B. $\frac{2}{\text{e}^\text{x}+\text{e}^{-\text{x}}}$
We have:
$\tan\theta + \sec\theta=\text{e}^\text{x}$
$\sec\theta + \tan\theta = \text{e}^\text{x}\cdots(1)$
$\Rightarrow\frac{1}{\sec\theta+\tan\theta}=\frac{1}{\text{e}^\text{x}}$
$\Rightarrow\frac{\sec^2\theta-\tan^2\theta}{\sec\theta + \tan\theta}=\frac{1}{\text{e}^\text{x}}$
$\Rightarrow\frac{(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)}{(\sec\theta + \tan\theta)}=\frac{1}{\text{e}^\text{x}}$
$\therefore \sec\theta - \tan\theta=\frac{1}{\text{e}^\text{x}}\cdots(2)$
Adding (1) and (2):
$2\sec\theta = \text{e}^\text{x}+ \frac{1}{\text{e}^\text{x}}$
$\Rightarrow 2\sec\theta = \frac{(\text{e}^\text{x})^2+1}{\text{e}^\text{x}}$
$\Rightarrow \sec\theta = \frac{\text{e}^{2\text{x}}+1}{2\text{e}^\text{x}}$
$\Rightarrow\sec\theta=\frac{1}{2}\times\frac{\text{e}^{2\text{x}}+1}{2\text{e}^\text{x}}$
$\Rightarrow\sec\theta = \frac{1}{2}\times(\text{e}^\text{x}+\text{e}^\text{-x})$
$\Rightarrow\frac{1}{\cos\theta}=\frac{\text{e}^\text{x}+\text{e}^\text{x}}{2}$
$\Rightarrow\cos\theta= \frac{2}{\text{e}^\text{x}+\text{e}^\text{-x}}$
View full question & answer→MCQ 111 Mark
If $\tan\text{x}+\sec\text{x}=\sqrt{3},0<\text{x}<\pi,$ then x is equal to:
- A
$\frac{5\pi}{6}$
- B
$\frac{2\pi}{3}$
- ✓
$\frac{\pi}{6}$
- D
$\frac{\pi}{3}$
AnswerCorrect option: C. $\frac{\pi}{6}$
We have:
$\tan\text{x}+\sec\text{x}=\sqrt{3}$ $[0,<\text{x}<\pi]$
$\Rightarrow\sec\text{x}+\tan\text{x}=\sqrt{3}$
$\Rightarrow\frac{1}{\cos\text{x}}+\frac{\sin\text{x}}{\cos\text{x}}=\sqrt{3}$
$\Rightarrow1+\sin\text{x}=\sqrt{3}\cos\text{x}$
$\Rightarrow(1+\sin\text{x})^2=(\sqrt{3}\cos\text{x})^2$
$\Rightarrow1+\sin^2\text{x}+2\sin\text{x}=3\cos^2\text{x}$
$\Rightarrow1+\sin^2\text{x}+2\sin\text{x}=3(1-\sin^2\text{x})$
$\Rightarrow4\sin^2\text{x}+2\sin\text{x}=2$
$\Rightarrow2\sin^2\text{x}+\sin\text{x}-1=0$
$\Rightarrow\sin\text{x}=-1,\frac{1}{2}$
since $0<\text{x}<\pi,\ \sin\text{x}$ cannot be negative.
$\therefore\sin\text{x}=\frac{1}{2}$
$\therefore\text{x}=\frac{\pi}{6}$
View full question & answer→MCQ 121 Mark
Which of the following is correct?
AnswerCorrect option: B. $\sin1^\circ<\sin1$
We know that, 1 radian is approximately $57^\circ$.
Also, the value of $\sin\text{x}$ is always increasing for $0\leq \text{x}\leq 90^\circ$
$($or $\sin\text{x}$ is an increasing function for $0\leq \text{x}\leq 90^\circ).$
Now, $1^\circ < 57^\circ$
or $1^\circ< 1 \text{ radian}$
$\therefore\sin 1^\circ < \sin1$
Hence, the correct answer is option B.
View full question & answer→MCQ 131 Mark
If $\sec\text{x}+\tan\text{x}=\text{k},\cos\text{x}=$
- A
$\frac{\text{x}^2+1}{2\text{k}}$
- ✓
$\frac{2\text{k}}{\text{x}^2+1}$
- C
$\frac{\text{k}}{\text{x}^2+1}$
- D
$\frac{\text{k}}{\text{x}^2-1}$
AnswerCorrect option: B. $\frac{2\text{k}}{\text{x}^2+1}$
We have:
$\sec\text{x} +\tan\text{x} = \text{k}\cdots(1)$
$\Rightarrow\frac{1}{\sec\text{x} + \tan\text{x}}=\frac{1}{\text{k}}$
$\Rightarrow\frac{\sec^2\text{x}-\tan^2\text{x}}{\sec\text{x}+\tan\text{x}} = \frac{1}{\text{k}}$
$\Rightarrow\frac{(\sec\text{x} + \tan\text{x})(\sec\text{x}-\tan\text{x})}{(\sec\text{x} + \tan\text{x})} = \frac{1}{\text{k}}$
$\therefore\sec\text{x} - \tan\text{x} = \frac{1}{\text{k}}\cdots(2)$
Adding (1) and (2):
$2\sec\text{x}= \text{k} + \frac{1}{\text{k}}$
$\Rightarrow 2\sec\text{x} = \frac{\text{k}^2 + 1}{\text{k}}$
$\Rightarrow \sec\text{x} = \frac{\text{k}^2+1}{2\text{k}}$
$\Rightarrow\frac{1}{\cos\text{x}}= \frac{\text{k}^2 + 1}{2\text{k}}$
$\Rightarrow\cos \text{x} = \frac{2\text{k}}{\text{k}^2 + 1}$
View full question & answer→MCQ 141 Mark
If $\tan\text{x}=-\frac{1}{\sqrt{5}}$ and $\theta$ lies in the IV quadrant, then the value of $\cos\text{x}$ is:
AnswerCorrect option: A. $\frac{\sqrt{5}}{\sqrt{6}}$
In the fourth quadrant, $\cos\text{x}\text{ and }\sec\text{x}$ are positive.
$\cos\text{x}=\frac{1}{\sec\text{x}}$
$=\frac{1}{\sqrt{\sec^2\text{x}}}$
$=\frac{1}{\sqrt{1+\tan^2\text{x}}}$
$=\frac{1}{\sqrt{1+\Big(-\frac{1}{\sqrt{5}}\Big)^2}}$
$=\frac{1}{\sqrt{\frac{6}{5}}}$
$=\frac{\sqrt{5}}{\sqrt{6}}$
View full question & answer→MCQ 151 Mark
Which of the following is incorrect?
AnswerCorrect option: C. $\sec\text{x}=\frac12$
$\sin\text{x} = -\frac{1}{5}$ is correct as $-1\leq \sin\text{x} \leq1$
- $\cos\text{x}=1$ is correct as $-1\leq \cos\text{x}\leq1$
- $\sec\text{x}=\frac{1}{2}$ is correct as $\text{x}\in [(-\infty, -1)\ \cup (1,\infty)]$
- $\tan\text{x} = 20$ is correct as $\tan\text{x}$ can take any real value.
Hence, the correct answer is option C. View full question & answer→MCQ 161 Mark
If $\text{x}=\text{r}\sin\theta\cos\theta,\text{y}=\text{r}\sin\theta$ and $\text{z}=\text{r}\cos\theta,$ then $\text{x}^2+\text{x}^2+\text{z}^2$ is idepandent of
- ✓
$\theta,\phi$
- B
$\text{r},\theta$
- C
$\text{r},\phi$
- D
$\text{r}.$
AnswerCorrect option: A. $\theta,\phi$
We have:
$\text{x}=\text{r}\sin\theta\cos\phi,\text{y}=\text{r}\sin\theta\sin\phi\text{ and }\text{z}=\text{r}\cos\theta,$
$\therefore\text{x}^2+\text{y}^2+\text{z}^2$
$=(\text{r}\sin\theta\cos\phi)^2+(\text{r}\sin\theta\sin\phi)^2+(\text{r}\cos\theta)^2$
$=\text{r}^2\sin^2\theta\cos^2\phi+\text{r}^2\sin^2\theta\sin^2\phi+\text{r}^2\cos^2\theta$
$=\text{r}^2\sin^2\theta(\cos^2\phi+\sin^2\phi)+\text{r}^2\cos^2\theta$
$=\text{r}^2\sin^2\theta\times1+\text{r}^2\cos^2\theta$
$=\text{r}^2\sin^2\theta+\text{r}\cos^2\theta$
$=\text{r}^2(\sin^2\theta+\cos^2\theta)$
$=\text{r}^2\times1$
$=\text{r}^2$
Thus, $\text{x}^2+\text{y}^2+\text{z}^2$ is independent of $\theta\text{ and }\phi$
View full question & answer→MCQ 171 Mark
If $\text{F}(\text{x})=\cos^2\text{x}+\sec^2\text{x},$ then
AnswerCorrect option: D. $\text{F}(\text{x})\geq2$
$\text{f}(\text{x}) = \cos^2\text{x} + \sec^2\text{x}$
$=\cos^2\text{x} +\sec^2\text{x}-2\cos\text{x}\sec\text{x}+2\cos\text{x}\sec\text{x}$
$=(\sec\text{x}-\cos\text{x})^2 +2$
$\therefore\text{f}\text({x})\geq 2 \ \forall \text{ x }$$\Big[(\sec\text{x}-\cos\text{x})^2\geq 0 \ \forall \text{ x}\Big]$
Hence, the correct option is answer D.
View full question & answer→MCQ 181 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 191 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:
$\sec\text{x} = \text{x} +\frac{1}{4\text{x}}$
$\Rightarrow\sec^2\text{x}=\text{x}^2+\frac{1}{16\text{x}^2}+\frac{1}{2}$
$\Rightarrow1+\tan^2\text{x}$
$=1+\text{x}^2+\frac{1}{16\text{x}^2}-\frac{1}{2}$
$\Rightarrow\tan^2\text{x}=\text{x}^2+\frac{1}{16\text{x}^2}-\frac{1}{2}$
$\Rightarrow\tan^2\text{x}=\Big(\text{x}-\frac{1}{4\text{x}}\Big)^2$
$\therefore\tan\text{x}=\pm\Big(\text{x}-\frac{1}{4\text{x}}\Big)$
$\sec\text{x}-\tan\text{x}=\Big(\text{x}+\frac{1}{4\text{x}}\Big)-\Big(\text{x}-\frac{1}{4\text{x}}\Big)\text{or}$
$=\Big(\text{x}+\frac{1}{4\text{x}}\Big)-\Big[-\Big(\text{x}-\frac{1}{4\text{x}}\Big)\Big]$
$=\frac{1}{2\text{x}}\text{ or } 2\text{x}$
View full question & answer→MCQ 201 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 211 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:
$\tan\text{x}=\text{x}-\frac{1}{4\text{x}}$
$\Rightarrow\sec^2\text{x}=1+\tan^2\text{x}$
$\Rightarrow\sec^2\text{x}=1+\Big(\text{x}-\frac{1}{4\text{x}}\Big)^2$
$\Rightarrow\sec^2\text{x}=\text{x}^2+\frac{1}{16\text{x}^2}+\frac{1}{2}$
$\Rightarrow \sec\text{x}^2 =\Big(\text{x}+\frac{1}{4\text{x}}\Big)^2$
$\therefore \sec\text{x}=\pm\Big(\text{x}+\frac{1}{4\text{x}}\Big)$
$\Rightarrow \sec\text{x}-\tan\text{x} =\Big(\text{x}+\frac{1}{4\text{x}}\Big)- \Big(\text{x}-\frac{1}{4\text{x}}\Big)\text{ or} -\Big(\text{x}+\frac{1}{4\text{x}}\Big)\text{or}- \Big(\text{x}-\frac{1}{4\text{x}}\Big)$
$=\frac{1}{2\text{x}}\text{ or} -2\text{x}$
View full question & answer→MCQ 221 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 231 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 241 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 251 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:
$(\tan^2\text{A}+\cot^2\text{A})^2=14^2$
$\tan^4\text{A}+\cot^4\text{A}+2(\tan^2\text{A})(\cot^2\text{A})=196$
$\Rightarrow\tan^4\text{A}+\cot^4\text{A}+2=196$
$\Rightarrow\tan^4\text{A}+\cot^4\text{A}=194$
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