MCQ 511 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 521 Mark
Choose the correct answer. The value of $\frac{1-\tan^215^\circ}{1+\tan^215^\circ}$ is:
- A
$1$
- B
$\sqrt{3}$
- ✓
$\frac{\sqrt{3}}{2}$
- D
$2$
AnswerCorrect option: C. $\frac{\sqrt{3}}{2}$
Given that, $\frac{1-\tan^215^\circ}{1+\tan^215^\circ}$
Let $\theta=15^\circ\therefore2\theta=30^\circ$
$\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}$
$\Rightarrow\cos30^\circ=\frac{1-\tan^215^\circ}{1+\tan^215^\circ}$
$\Rightarrow\frac{\sqrt3}{2}=\frac{1-\tan^215^\circ}{1+\tan^215^\circ}$
Hence, the correct option is $(c).$
View full question & answer→MCQ 531 Mark
$\sin163^\circ\cos347^\circ+\sin73^\circ\sin167^\circ=$
AnswerCorrect option: B. $\frac{1}{2}$
$\sin163^\circ\cos347^\circ+\sin73^\circ\sin167^\circ$
$=\ \sin(180^\circ-17^\circ)\cos(360^\circ-13^\circ) +\sin(90^\circ-17^\circ)\sin(180^\circ+13^\circ)$
$=\ \sin17^\circ\cos13^\circ+\cos17^\circ\sin13^\circ$
$=\ \sin(17^\circ+13^\circ)$ $[\sin(\text{A+B})=\sin\text{A}\cos\text{B}+\sin\text{B}\cos\text{A}]$
$=\ \sin30^\circ$
$=\ \frac{1}{2}$
View full question & answer→MCQ 541 Mark
The value of $\tan20^\circ +2\tan50^\circ-\tan70^\circ$ is:
- A
$1$
- ✓
$0$
- C
$\tan50^\circ$
- D
$\text{None of these}$
View full question & answer→MCQ 551 Mark
If $\sin\text{x}+\sin\text{y}=\sqrt3(\cos\text{y}-\cos\text{x}),$ than $\sin3\text{x}+\sin3\text{y}=$
- A
$2\sin3\text{x}$
- ✓
$0$
- C
$1$
- D
AnswerWe have,
$\sin\text{x}+\sin\text{y}=\sqrt3(\cos\text{y}-\cos\text{x})$
$\Rightarrow\ 2\sin\Big(\frac{\text{x+y}}{2}\Big)\cos\Big(\frac{\text{x}-\text{y}}{2}\Big)=2\sqrt3\sin\Big(\frac{\text{x+y}}{2}\Big)\sin\Big(\frac{\text{x}-\text{y}}{2}\Big)$
$\Rightarrow\ 2\sin\Big(\frac{\text{x+y}}{2}\Big)\cos\Big(\frac{\text{x}-\text{y}}{2}\Big)-2\sqrt3\sin\Big(\frac{\text{x+y}}{2}\Big)\sin\Big(\frac{\text{x}-\text{y}}{2}\Big)=0$
$\Rightarrow\ 2\sin\Big(\frac{\text{x+y}}{2}\Big)\Big[\cos\Big(\frac{\text{x}-\text{y}}{2}\Big)-\sqrt3\sin\Big(\frac{\text{x}-\text{y}}{2}\Big)\Big]=0$
$\Rightarrow\ \sin\Big(\frac{\text{x+y}}{2}\Big)\Big[\cos\Big(\frac{\text{x}-\text{y}}{2}\Big)-\sqrt3\sin\Big(\frac{\text{x}-\text{y}}{2}\Big)\Big]=0$
$\Rightarrow\ \sin\frac{\text{x+y}}{2}=0$ Or, $\cos\Big(\frac{\text{x}-\text{y}}{2}\Big)-\sqrt3\sin\Big(\frac{\text{x}-\text{y}}{2}\Big)=0$
$\Rightarrow\ \frac{\text{x+y}}{2}=0$ Or, $\tan\Big(\frac{\text{x}-\text{y}}{2}\Big)=\frac{1}{\sqrt3}=\tan\frac{\pi}{6}$
$\Rightarrow\ \text{x}=-\text{y}$ Or $, \frac{\text{x}-\text{y}}{2}=\frac{\pi}{6}$
$\Rightarrow\ \text{x}=-\text{y}$ Or $, \text{x}-\text{y}=\frac{\pi}{3}$
Case $1:$
When $\text{x}=-\text{y}$
In this case,
$\sin3\text{x}+\sin3\text{y}=\sin(-3\text{y})+\sin3\text{y}=-\sin3\text{y}+\sin3\text{y}=0$
Case $2:$
When $\text{x}-\text{y}=\frac{\pi}{3}$
Or, $3\text{x}=\pi+3\text{y}$
So, $\sin3\text{x}+\sin3\text{y}=\sin(\pi+3\text{y})+\sin3\text{y}$
$=\ -\sin3\text{y}+\sin3\text{y}$
$=\ 0$
View full question & answer→MCQ 561 Mark
Choose the correct answer. If $\sin\theta=\frac{-4}{5}$ and $\theta$ lies in third quadrant then the value of $\cos\frac{\theta}{2}$ is:
- A
$\frac{1}{5}$
- B
$-\frac{1}{\sqrt{10}}$
- ✓
$-\frac{1}{\sqrt{5}}$
- D
$\frac{1}{\sqrt{10}}$
AnswerCorrect option: C. $-\frac{1}{\sqrt{5}}$
Given that, $\sin\theta=-\frac{4}{5},\theta$ lies in third quadrant
$\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\Big(\frac{-4}{5}\Big)^2}$
$=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{9}{25}}=\pm\frac{3}{5}$
$\therefore\cos\theta=-\frac{3}{5},\theta$ lies in third quadrant
$\cos\theta=2\cos^2\frac{\theta}{2}-1\Big[\because\pi<\theta\frac{3\pi}{2},\therefore\frac{\pi}{2}<\frac{\theta}{2}<\frac{3\pi}{4}\Big]$
$\Rightarrow\frac{-3}{5}=2\cos^2\frac{\theta}{2}-1$
$\Rightarrow2\cos^2\frac{\theta}{2}=1-\frac{3}{5}=\frac{2}{5}$
$\Rightarrow\cos^2\frac{\theta}{2}=\frac{2}{5\times2}=\frac{1}{5}$
$\Rightarrow\cos\frac{\theta}{2}=\pm\frac{1}{\sqrt5}$
$\Rightarrow\cos\frac{\theta}{2}=-\frac{1}{\sqrt5}\Big[\because\frac{\pi}{2}<\frac{\theta}{2}<\frac{3\pi}{4}\Big]$
Hence, the correct option is $(c).$
View full question & answer→MCQ 571 Mark
In a $\triangle\text{ABC},$ if $(c + a + b)(a + b − c) = ab,$ then the measure of angle $C$ is:
- A
$\frac{\pi}{3}$
- B
$\frac{\pi}{6}$
- ✓
$\frac{2\pi}{3}$
- D
$\frac{\pi}{2}$
AnswerCorrect option: C. $\frac{2\pi}{3}$
Given: $\text{(c + a + b) (a + b − c) = ab}$
$\Rightarrow(\text{a + b})^2-\text{c}^2=\text{ab}$
$\Rightarrow\text{a}^2+\text{b}^2+2\text{ab}-\text{c}^2=\text{ab}$
$\Rightarrow\text{a}^2+\text{b}^2-\text{c}^2=-\text{ab}$
$\Rightarrow\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}=-\frac{1}{2}$
$\Rightarrow\cos\text{C}=-\frac{1}{2}=\cos\frac{2\pi}{3} ($Using cosine rule$)$
$\Rightarrow\text{C}=\frac{2\pi}{3}$
Thus, the measure of angle $C$ is $\frac{2\pi}{3}.$
Hence, the correct answer is option $(c).$
View full question & answer→MCQ 581 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 $1^\circ< 1$ radian
$\therefore\sin 1^\circ < \sin1$
Hence, the correct answer is option $B.$
View full question & answer→MCQ 591 Mark
In the sides of a triangle are in the ratio $1:\sqrt{3}:2,$ then the measure of its greatest angle is:
- A
$\frac{\pi}{6}$
- B
$\frac{\pi}{3}$
- ✓
$\frac{\pi}{2}$
- D
$\frac{2\pi}{3}$
AnswerCorrect option: C. $\frac{\pi}{2}$
Let $\triangle\text{ABC}$ be the given triangle such that its sides are in the ratio $1:\sqrt{3}:2.$
$\therefore\text{a = k,b}=\sqrt{3}\text{k,c}=2\text{k}$
Now, $\text{a}^2+\text{b}^2=\text{k}^2+3\text{k}^2=4\text{k}^2=\text{c}^2$
So, $\triangle\text{ABC}$ is a right triangle right angled at $C.$
$\therefore\text{C}=90^{\circ}$
Using sine rule, we have
$\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\sin\text{B}}=\frac{\text{c}}{\sin\text{C}}$
$\Rightarrow\frac{\text{k}}{\sin\text{A}}=\frac{\sqrt{3}\text{k}}{\sin\text{B}}=\frac{2\text{k}}{\sin90^{\circ}}$
$\Rightarrow\sin\text{A}=\frac{1}{2}$ and $\sin\text{B}=\frac{\sqrt{3}}{2}$
$\Rightarrow\text{A}=30^{\circ}$ and $\text{B}=60^{\circ}$
Thus, the measure of its greatest angle is $\frac{\pi}{2}$
Hence, the correct answer is option $(c).$
View full question & answer→MCQ 601 Mark
The radius of the circle whose arc of length $15\pi$ makes an angle of $\frac{3\pi}{4}$ radian at the centre is:
AnswerCorrect option: B. $20\text{cm}$
$\theta=\frac{\text{Arc}}{\text{Radius}}$
$\Rightarrow\frac{3\pi}{4}=\frac{15\pi}{\text{Radius}}$
$\Rightarrow\frac{60}{3}$
$\Rightarrow20\text{cm}$
View full question & answer→MCQ 611 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 621 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 631 Mark
If a is any real number, the number of roots of $\cot\text{x}-\tan\text{x}=\text{a}$ in the first quadrant is (are):
Answer(C)
Solution:
Given:
$\cot\text{x}-\tan\text{x}=\text{a}$
$\Rightarrow\frac{1}{\tan\text{x}}-\tan\text{x}=\text{a}$
$\Rightarrow1-\tan^2\text{x}=\text{a}\tan\text{x}$
$\Rightarrow\tan^2\text{x}+\text{a}\tan\text{x}-1=0$
It is a quadratic equation.
If $\tan\text{x}=\text{z},$ then the equation becomes
$\text{z}^2+\text{az}-1=0$
$\Rightarrow\text{z}=\frac{-\text{a}\pm\sqrt{\text{a}^2+4}}{2}$
$\Rightarrow\tan\text{x}=\frac{-\text{a}\pm\sqrt{\text{a}^2+4}}{2}$
$\Rightarrow\text{x}=\tan^{-1}\Big(\frac{-\text{a}\pm\sqrt{\text{a}^2+4}}{2}\Big)$
There are two roots of the given equation, but we need to find the number of roots in the first quadrant.
There is exactly one root of the equation, that is, $\text{x}=\tan^{-1}\Big(\frac{-\text{a}+\sqrt{\text{a}^2+4}}{2}\Big).$
View full question & answer→MCQ 641 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 651 Mark
If the arac of the same langth in two circles subtend angles $65^\circ$ and $110^\circ$ at the center, the ratio of the circle is:
- ✓
$22 : 13$
- B
$11 : 13$
- C
$22 : 15$
- D
$21 : 13$
AnswerCorrect option: A. $22 : 13$
Let the angle subtended at the by the arec and radii of the first second circle $\theta_{1}$ and $\text{r}_{1}$ and $\theta_{2}$ and $\text{r}_{2}.$
We have,
$\theta_{1}=65^{\circ}=\Big(65\times\frac{\pi}{180}\Big)\ \text{radian}$
$\theta_{2}=65^{\circ}=\Big(110\times\frac{\pi}{180}\Big)\ \text{radian}$
$\theta_{1}=\frac{1}{\text{r}_{1}}$
$\Rightarrow \text{r}_{1}=\frac{1}{\big(65\times\frac{\pi}{180}\big)}$
$\Rightarrow \text{r}_{2}=\frac{1}{\big(110\times\frac{\pi}{180}\big)}$
$\frac{\text{r}_{1}}{\text{r}_{2}}=\frac{\frac{l}{\big(65\times\frac{\pi}{180}\big)}}{\frac{i}{\big(110\times\frac{\pi}{180}\big)}} $
$=\frac{110}{65}=\frac{22}{13}$
$\text{r}_{1}:\text{r}_{2}=22:13$
View full question & answer→MCQ 661 Mark
If $\cot(\alpha+\beta)=0,$ then $\sin(\alpha+2\beta)$ is equal to:
- ✓
$\sin\alpha$
- B
$\cos2\beta$
- C
$\cos\alpha$
- D
$\sin2\alpha$
AnswerCorrect option: A. $\sin\alpha$
Given:
$\cot(\alpha+\beta)=0$
$\Rightarrow\frac{\cos(\alpha+\beta)}{\sin(\alpha+\beta)}=0$
$\Rightarrow\cos(\alpha+\beta)=0$
$\Rightarrow\alpha+\beta=\frac\pi2$
$\therefore\sin(\alpha+2\beta)=\sin(\alpha+\alpha+\beta)$
$=\sin\alpha$
View full question & answer→MCQ 671 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 681 Mark
Choose the correct answer. If $\tan\theta=\frac{\text{a}}{\text{b}},$ then $\text{b}\cos2\theta+\text{a}\sin2\theta$ is equal to:
AnswerGiven that, $\tan\theta=\frac{\text{a}}{\text{b}}$
$\text{b}\cos2\theta+\text{a}\sin2\theta=\text{b}\Big[\frac{1-\tan^2\theta}{1+\tan^2\theta}\Big]+\text{a}\Big[\frac{2\tan\theta}{1+\tan^2\theta}\Big]$
$=\text{b}\Bigg[\frac{1-\frac{\text{a}^2}{\text{b}^2}}{1+\frac{\text{a}^2}{\text{b}^2}}\Bigg]+\text{a}\Bigg[\frac{\frac{2\text{a}}{\text{b}}}{1+\frac{\text{a}^2}{\text{b}^2}}\Bigg]$
$=\text{b}\Big[\frac{\text{b}^2-\text{a}^2}{\text{b}^2+\text{a}^2}\Big]+\Bigg[\frac{\frac{2\text{a}^2}{\text{b}}}{\frac{\text{b}^2+\text{a}^2}{\text{b}^2}}\Bigg]$
$=\frac{\text{b}^3+\text{a}^2\text{b}}{\text{b}^2+\text{a}^2}+\frac{2\text{a}^2\text{b}}{\text{b}^2+\text{a}^2}=\frac{\text{b}^3-\text{a}^2\text{b}+2\text{a}^2\text{b}}{\text{b}^2+\text{a}^2}$
$=\frac{\text{b}^3+\text{a}^2\text{b}}{\text{b}^2+\text{a}^2}=\frac{\text{b}(\text{b}^2+\text{a}^2)}{\text{b}^2+\text{a}^2}$
$=\text{b}$
Hence, the correct option is $(b).$
View full question & answer→MCQ 691 Mark
The equation $3\cos\text{x}+4\sin\text{x}=6$ has $....$ solution.
AnswerGiven equation:
$3\cos\text{x}+4\sin\text{x}=6\ .....(1)$
Thus, the equation is of the form $\text{a}\cos\text{x}+\text{b}\sin\text{x}=\text{c},$ where and $\text{c}=6.$
Let:
$\text{a}=3=\text{r}\cos\alpha$ and $\text{b}=4=\text{r}\sin\alpha$
Now,
$\tan\alpha=\frac{\text{b}}{\text{a}}=\frac{4}{3}$
$\Rightarrow\alpha=\tan^{-1}\Big(\frac{4}{3}\Big)$
Also,
$\text{r}=\sqrt{\text{a}^2+\text{b}^2}=\sqrt{9+16}=\sqrt{25}=5$
On putting $\text{a}=3=\text{r}\cos\alpha$ and $\text{b}=4=\text{r}\sin\alpha$ in equation $(i),$ we get:
$\Rightarrow\text{r}\cos(\theta-\alpha)=6$
$\Rightarrow\text{5}\cos(\theta-\alpha)=6$
$\Rightarrow\text{}\cos(\theta-\alpha)=\frac{6}{5}$
From here, we cannot find the value of $\theta.$
View full question & answer→MCQ 701 Mark
If $\sin\text{cosec q}=2,$ then $\sin2\text{q}+\text{cosec 2}\text{q}$ is equal to:
View full question & answer→MCQ 711 Mark
$\cos35^\circ+\cos85^\circ+\cos155^\circ=$
- ✓
$0$
- B
$\frac{1}{\sqrt3}$
- C
$\frac{1}{\sqrt2}$
- D
$\cos275^\circ$
Answer$\cos35^\circ+\cos85^\circ+\cos155^\circ$
$=\ 2\cos\Big(\frac{35^\circ+85^\circ}{2}\Big)\cos\Big(\frac{35^\circ-85^\circ}{2}\Big)+\cos155^\circ$ $\Big[\because\ \cos\text{A}+\cos\text{B}=2\cos\Big(\frac{\text{A+B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\Big]$
$=\ 2\cos60^\circ\cos(-25^\circ)+\cos155^\circ$
$=\ 2\times\frac{1}{2}\cos25^\circ+\cos155^\circ$
$=\ \cos25^\circ+\cos155^\circ$
$=\ 2\cos\Big(\frac{25^\circ+155^\circ}{2}\Big)\cos\Big(\frac{25^\circ-155^\circ}{2}\Big)$
$=\ 2\cos90^\circ\cos65^\circ$
$=\ 0$
View full question & answer→MCQ 721 Mark
If $\text{OP}$ makes $4$ revolutions in on second the angular velocity in radians per seconds is:
- A
$\pi$
- B
$2\pi$
- C
$4\pi$
- ✓
$8\pi$
AnswerCorrect option: D. $8\pi$
$\text{Angular velocity}=\frac{\text{Distance}}{\text{Time}}$
$=\frac{4\times2\pi}{1}$
$=8\pi\ \text{radians}$
View full question & answer→MCQ 731 Mark
If $\tan\text{x}=-\frac{1}{\sqrt{5}}$ and $\theta$ lies in the $\text{IV}$ quadrant, then the value of $\cos\text{x}$ is:
AnswerCorrect option: A. $\frac{\sqrt{5}}{\sqrt{6}}$
In the fourth quadrant, $\cos\text{x}$ 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 741 Mark
Choose the correct answer. The value of $\tan3\text{A}-\tan2\text{A}-\tan\text{A}$ is equal to:
- ✓
$\tan3\text{A}\tan2\text{A}\tan\text{A}$
- B
$-\tan3\text{A}\tan2\text{A}\tan\text{A}$
- C
$\tan\text{A}\tan2\text{A}-\tan2\text{A}\tan3\text{A}\tan\text{A}$
- D
$\text{None of these}$
AnswerCorrect option: A. $\tan3\text{A}\tan2\text{A}\tan\text{A}$
$3\text{A}=\text{A}+2\text{A}$
$\Rightarrow\tan3\text{A}=\tan(\text{A+2A})$
$\Rightarrow\tan3\text{A}=\tan\text{A}+\tan\frac{2\text{A}}{1}-\tan\text{A}.\tan2\text{A}$
$\Rightarrow\tan\text{A}+\tan2\text{A}=\tan3\text{A}-\tan3\text{A}.\tan2\text{A}.\tan\text{A}$
$\Rightarrow\tan3\text{A}-\tan2\text{A}-\tan\text{A}=\tan3\text{A}.\tan2\text{A}.\tan\text{A}$
View full question & answer→MCQ 751 Mark
If in a $\Delta\text{ABC},\tan\text{A}+\tan\text{B}+\tan\text{C=0,}$ then $\cot\text{A}\cot\text{B}\cot\text{C}=$
Answer$\text{ABC}$ 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 761 Mark
If $\text{A+B+C}=\pi,$ then $\frac{\tan\text{A}+\tan\text{B}+\tan\text{C}}{\tan\text{A}\tan\text{B}\tan\text{C}}$ is equal to:
Answer$\pi=180^\circ$
Using $\tan(180^\circ-\text{A})=-\tan\text{A},$ we get:
$\text{C}=\pi-(\text{A+B})$
Now, $\frac{\tan\text{A}+\tan\text{B}+\tan\text{C}}{\tan\text{A}\tan\text{B}\tan\text{C}}$
$=\frac{\tan\text{A}+\tan\text{B}-\tan[\pi-\text{(A+B)}]}{\tan\text{A}\tan\text{B}\tan[\pi-\text{(A+B)}]}$
$=\frac{\tan\text{A}+\tan\text{B}-\tan\text{(A+B)}}{-\tan\text{A}\tan\text{B}\tan\text{(A+B)}}$
$=\frac{\tan\text{A}+\tan\text{B}-\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}}{-\tan\text{A}\tan\text{B}\times\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}}$
$=\frac{\tan\text{A}+\tan\text{B}-\tan^2\text{A}\tan\text{B}-\tan\text{A}\tan^2\text{B}-\tan\text{A}-\tan\text{B}}{-\tan^2\text{A}\tan\text{B}-\tan\text{A}\tan^2\text{B}}$
$=\frac{-\tan^2\text{A}\tan\text{B}-\tan\text{A}\tan^2\text{B}}{-\tan^2\text{A}\tan\text{B}-\tan\text{A}\tan^2\text{B}}$
$=1$
View full question & answer→MCQ 771 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 781 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 791 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 $\theta$ and $\phi$
View full question & answer→MCQ 801 Mark
If $D, G$ and $R$ denote respectively the number of degrees, grades and radians in an angle, then:
- A
$\frac{\text{D}}{100}=\frac{\text{G}}{90}=\frac{2\text{R}}{\pi}$
- B
$\frac{\text{D}}{90}=\frac{\text{G}}{100}=\frac{\text{R}}{\pi}$
- ✓
$\frac{\text{D}}{100}=\frac{\text{G}}{100}=\frac{2\text{R}}{\pi}$
- D
$\frac{\text{D}}{90}=\frac{\text{G}}{100}=\frac{\text{R}}{\pi}$
AnswerCorrect option: C. $\frac{\text{D}}{100}=\frac{\text{G}}{100}=\frac{2\text{R}}{\pi}$
It is the relation between degree, grade and radian.
View full question & answer→MCQ 811 Mark
In any $\triangle\text{ABC},\text{a}(\text{b}\cos\text{C}-\text{c}\cos\text{B})=$
- A
$a^2$
- ✓
$b^2- c^2$
- C
$0$
- D
$b^2+ c^2$
AnswerCorrect option: B. $b^2- c^2$
Using cosine rule, we have
$\text{a(b}\cos\text{C}-\text{c}\cos\text{B})$
$=\text{ab}\Big(\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}\Big)-\text{ca}\Big(\frac{\text{c}^2+\text{a}^2-\text{b}^2}{2\text{ca}}\Big)$
$=\frac{\text{a}^2+\text{b}^2-\text{c}^2-\text{c}^2-\text{a}^2+\text{b}^2}{2}$
$=\frac{2\text{b}^2-2\text{c}^2}{2}$
$=\text{b}^2-\text{c}^2$
Hence, the correct answer is option $(b).$
View full question & answer→MCQ 821 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 831 Mark
If $\tan(\text{A}-\text{B})=1,\sec(\text{A+B})=\frac{2}{\sqrt{3}},$ then the smallest positive value of $B$ is:
- A
$\frac{25\pi}{24}$
- ✓
$\frac{19\pi}{24}$
- C
$\frac{13\pi}{24}$
- D
$\frac{11\pi}{24}$
AnswerCorrect option: B. $\frac{19\pi}{24}$
Given:
$\tan(\text{A - B})=1$ and $\sec(\text{A+B})=\frac{2}{\sqrt{3}}$
$\Rightarrow\text{A - B}=\frac{\pi}{4}\cdots(1)$ and $\text{A + B}=\frac{\pi}{4}\cdots(2)$
Adding these equations we get:
$2\text{A}=\frac{\pi}{4}+\frac\pi6$
$\Rightarrow\text{A}=\frac{5\pi}{24}$
$\Rightarrow$ Smallest possible value of $\text{B}=\pi-\frac{5\pi}{24}=\frac{19\pi}{24}.$
View full question & answer→MCQ 841 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 851 Mark
Choose the correct answer. The value of $\sin\frac{\pi}{10}\sin\frac{13\pi}{10}$ is:
- A
$\frac{1}{2}$
- B
$-\frac{1}{2}$
- ✓
$-\frac{1}{4}$
- D
$1$
AnswerCorrect option: C. $-\frac{1}{4}$
$\sin\frac{\pi}{10}\sin\frac{13\pi}{10}=\sin\frac{\pi}{10}\sin\Big(\pi+\frac{3\pi}{10}\Big)=-\sin\frac{\pi}{10}\sin\frac{3\pi}{10}$
$=-\sin18^\circ\sin54^\circ=-\sin18^\circ\cos36^\circ$
$=-\Big(\frac{\sqrt5-1}{4}\Big)\Big(\frac{\sqrt5+1}{4}\Big)$
$=\frac{1-5}{16}$
$=-\frac{1}{4}$
View full question & answer→MCQ 861 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 871 Mark
The value of $\sin^2\frac{5\pi}{12}-\sin^2\frac{\pi}{12}$ is:
- A
$\frac12$
- ✓
$\frac{\sqrt{3}}{2}$
- C
$1$
- D
$0$
AnswerCorrect option: B. $\frac{\sqrt{3}}{2}$
$\frac{5\pi}{12}=75^\circ,\frac{\pi}{12}=15^\circ$
$\sin^275^\circ-\sin^215^\circ$
$=\sin^275^\circ-\cos^275^\circ$ $[\sin(90^\circ-\theta)=\cos\theta]$.
Now, $\sin75^\circ=\sin(45^\circ+30^\circ)$
$=\sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ$
$=\frac{1}{\sqrt{2}}\times\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times\frac12$
$=\frac{\sqrt{3}+1}{2\sqrt{2}}$
$\cos75^\circ=\cos(45^\circ+30^\circ)$
$=\cos45^\circ\cos30^\circ-\sin45^\circ\sin30^\circ$
$=\frac{1}{\sqrt{2}}\times\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\times\frac12$
$=\frac{\sqrt{3}-1}{2\sqrt{2}}$
Hence,
$\sin^275^\circ-\cos^275^\circ=\Big(\frac{\sqrt{3}+1}{2\sqrt{2}}\Big)^2-\Big(\frac{\sqrt{3}-1}{2\sqrt{2}}\Big)^2$
$=\frac{3+1+2\sqrt{3}-3-1+2\sqrt{3}}{8}$
$=\frac{4\sqrt{3}}{8}$
$=\frac{\sqrt{3}}{2}$
View full question & answer→MCQ 881 Mark
If $4\sin^2\text{x}=1$ then the values of $x$ are:
- A
$2\text{n}\pi\pm\frac{\pi}{3},\text{n}\in\text{Z}$
- B
$\text{n}\pi\pm\frac{\pi}{3},\text{n}\in\text{Z}$
- ✓
$\text{n}\pi\pm\frac{\pi}{6},\text{n}\in\text{Z}$
- D
$2\text{n}\pi\pm\frac{\pi}{6},\text{n}\in\text{Z}$
AnswerCorrect option: C. $\text{n}\pi\pm\frac{\pi}{6},\text{n}\in\text{Z}$
Given:
$\Rightarrow\sin^2\text{x}=1$
$\Rightarrow\sin^2\text{x}=\frac{1}{4}$
$\Rightarrow\sin\text{x}=\frac{1}{2}$ or $\sin\text{x}=-\frac{1}{2}$
$\Rightarrow\sin\text{x}=\sin\frac{\pi}{6}$ or $\sin\text{x}=\sin\Big(-\frac{\pi}{6}\Big)$
$\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{6},\text{n}\in\text{Z}$ or $\text{n}\pi+(-1)^\text{n}\Big(-\frac{\pi}{6}\Big),\text{n}\in\text{Z}$
$\Rightarrow\text{x}=\text{n}\pi\pm\frac{\pi}{6},\text{n}\in\text{Z}$
View full question & answer→MCQ 891 Mark
Choose the correct answer.If $\sin\theta+\text{cosec}\theta=2,$ then $\sin^2\theta+\text{cosec}^2\theta$ is equal to:
Answer$\sin\theta+\text{cosec}\theta=2$
$\Rightarrow\sin\theta+\frac{1}{\sin\theta}=2$
$\Rightarrow\sin^2\theta+1=2\sin\theta$
$\Rightarrow(\sin\theta-1)^2=0$
$\Rightarrow\sin\theta=1$
$\therefore\sin^2\theta+\text{cosec}^2\theta=1+1=2$
View full question & answer→MCQ 901 Mark
If $\tan20^\circ+\tan40^\circ+\sqrt{3}\tan20\circ\tan40^\circ$ is equal to:
- A
$\frac{\sqrt{3}}{4}$
- B
$\frac{\sqrt{3}}{2}$
- ✓
$\sqrt{3}$
- D
$1$
AnswerCorrect option: C. $\sqrt{3}$
$\tan20^\circ+\tan40^\circ+\sqrt{3}\tan20^\circ\tan40^\circ$
$=\tan60^\circ(1-\tan20^\circ\tan40^\circ)+\tan60^\circ\tan20^\circ\tan40^\circ$
$=\tan60^\circ-\tan60^\circ\tan20^\circ\tan40^\circ+\tan60^\circ\tan20^\circ\tan40^\circ$
$=\tan60^\circ$
$=\sqrt{3}$
View full question & answer→MCQ 911 Mark
Choose the correct answer. If $\tan\theta=\frac{1}{2}$ and $\tan\phi=\frac{1}{3},$ then the value of $\theta+\phi$ is:
- A
$\frac{\pi}{6}$
- B
$\pi$
- C
$0$
- ✓
$\frac{\pi}{4}$
AnswerCorrect option: D. $\frac{\pi}{4}$
We have, $\tan\theta=\frac{1}{2}$ and $\tan\phi=\frac{1}{3}$
$\tan(\theta+\phi)=\frac{\tan\theta+\tan\phi}{1-\tan\theta\cdot\tan\phi}=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\cdot\frac{1}{3}}=\frac{\frac{5}{6}}{1-\frac{1}{6}}=1$
$\therefore\theta+\phi=\frac{\pi}{4}$
View full question & answer→MCQ 921 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 931 Mark
If $\tan\text{px}-\tan\text{qx}=0,$ then the values of $\theta$ form a series in:
AnswerGiven:
$\tan\text{px}-\tan\text{qx}=0$
$\Rightarrow\tan\text{px}=\tan\text{qx}$
$\Rightarrow\frac{\sin\text{px}}{\cos\text{px}}=\frac{\sin\text{qx}}{\cos\text{qx}}$
$\Rightarrow\sin\text{px}\cos\text{qx}=\sin\text{qx}\cos\text{px}$
$\Rightarrow\frac{1}{2}\Big[\sin\Big(\frac{\text{p+q}}{2}\Big)\text{x}+\sin\Big(\frac{\text{p}-\text{q}}{2}\Big)\text{x}\Big]\\=\frac{1}{2}\Big[\sin\Big(\frac{\text{q+p}}{2}\Big)\text{x}+\sin\Big(\frac{\text{q}-\text{p}}{2}\text{x}\Big)\Big]$
Now,
$\Rightarrow\sin\text{A}\cos\text{B}=\frac{1}{2}\Big[\Big(\frac{\text{A+B}}{2}\Big)+\Big(\frac{\text{A}-\text{B}}{2}\Big)\sin\Big]$
$\Rightarrow\sin\Big(\frac{\text{p}-\text{q}}{2}\Big)\text{x}=\sin\Big(\frac{\text{q}-\text{p}}{2}\Big)\text{x}$
$\Rightarrow\sin\Big(\frac{\text{p}-\text{q}}{2}\Big)\text{x}=-\sin\Big(\frac{\text{q}-\text{p}}{2}\Big)\text{x}$
$\Rightarrow2\sin\Big(\frac{\text{p}-\text{q}}{2}\Big)\text{x}=0$
$\Rightarrow\sin\Big(\frac{\text{p}-\text{q}}{2}\Big)\text{x}=0$
$\Rightarrow\Big(\frac{\text{p}-\text{q}}{2}\Big)\text{x}=\text{n}\pi,\text{n}\in\text{Z}$
Now, on putting the value of $n,$ we get:
$\text{n}=1,\text{x}=\frac{2\pi}{(\text{p}-\text{q})}=\text{a}_1$
$\text{n}=2,\text{x}=\frac{4\pi}{(\text{p}-\text{q})}=\text{a}_2$
$\text{n}=3,\text{x}=\frac{6\pi}{(\text{p}-\text{q})}=\text{a}_3$
$\text{n}=4,\text{x}=\frac{4\pi}{(\text{p}-\text{q})}=\text{a}_4$
And so on.
Also,
$\text{d}=\text{a}_2-\text{a}_1=\frac{4\pi}{(\text{p}-\text{q})}-\frac{2\pi}{(\text{p}-\text{q})}=\frac{2\pi}{(\text{p}-\text{q})}$
$\text{d}=\text{a}_3-\text{a}_2=\frac{6\pi}{(\text{p}-\text{q})}-\frac{4\pi}{(\text{p}-\text{q})}=\frac{2\pi}{(\text{p}-\text{q})}$
$\text{d}=\text{a}_4-\text{a}_3=\frac{8\pi}{(\text{p}-\text{q})}-\frac{6\pi}{(\text{p}-\text{q})}=\frac{2\pi}{(\text{p}-\text{q})}$
And so on.
Thus, $xx$ forms a series in $AP.$
View full question & answer→MCQ 941 Mark
Choose the correct answer. Number of solutions of the equation $\tan\text{x}+\sec\text{x}=2\cos\text{x}$ lying in the interval $[0,2\pi]$ is:
AnswerGiven equation is $\tan\text{x}+\sec\text{x}=2\cos\text{x}$
$\Rightarrow\frac{\sin\text{x}}{\cos\text{x}}+\frac{1}{\cos\text{x}}=2\cos\text{x}$
$\Rightarrow1+\sin\text{x}=2\cos^2\text{x}$
$\Rightarrow2\cos^2\text{x}-\sin\text{x}-1=0$
$\Rightarrow2(1-\sin^2\text{x})-\sin\text{x}-1=0$
$\Rightarrow2-2\sin^2\text{x}-\sin\text{x}-1=0$
$\Rightarrow-2\sin^2\text{x}-\sin\text{x}+1=0$
$\Rightarrow2\sin^2\text{x}+\sin\text{x}-1=0$
since, the equation is a quadratic equation in $\sin\text{x}$ so it will have $2$ solutions.
Hence, the correct option is $(c)$
View full question & answer→MCQ 951 Mark
Choose the correct answer. The value of $\tan75^\circ-\cot75^\circ$ is equal to:
- ✓
$2\sqrt{3}$
- B
$2+\sqrt{3}$
- C
$2-\sqrt{3}$
- D
$1$
AnswerCorrect option: A. $2\sqrt{3}$
$\tan75^\circ-\cot75^\circ=\frac{\sin75^\circ}{\cos75^\circ}-\frac{\cos75^\circ}{\sin75^\circ}\\=\frac{2(\sin^275^\circ-\cos^275^\circ)}{2\sin75^\circ\cos75^\circ}$
$=\frac{-2\cos150^\circ}{\sin150^\circ}$
$=-2\cot150^\circ$
$=-2\cot(180^\circ-30^\circ)$
$=2\cot30^\circ$
$=2\sqrt3$
View full question & answer→MCQ 961 Mark
The value of $\cos^\text{2}48^\circ-\sin^\text{2}48^\circ$ is:
AnswerCorrect option: A. $\sqrt{5}+\frac{1}{8}$
View full question & answer→MCQ 971 Mark
The number of values of $x$ in $[0,\ 2\pi]$ that satisfy the equation $\sin^2\text{x}-\cos\text{x}=\frac{1}{4}$
Answer$\sin^2\text{x}-\cos\text{x}=\frac{1}{4}$
$\Rightarrow(1-\cos^2\text{x})-\cos\text{x}=\frac{1}{4}$
$\Rightarrow4-4\cos^2\text{x}-4\cos\text{x}=1$
$\Rightarrow4\cos^2-4\cos\text{x}=1$
$\Rightarrow4\cos^2\text{x}+4\cos\text{x}-3=0$
$\Rightarrow4\cos^2\text{x}+6\cos\text{x}-2\cos\text{x}-3=0$
$\Rightarrow2\cos\text{x}(2\cos\text{x}+3)-1(2\cos\text{x}-1)=0$
$\Rightarrow(2\cos\text{x}+3)(2\cos\text{x}-1)=0$
$\Rightarrow2\cos\text{x}+3=0$ or $2\cos\text{x}-1=0$
$\Rightarrow\cos\text{x}=-\frac{3}{2}$ is not possible.
$\therefore\cos\text{x}=\frac{1}{2}$
$\Rightarrow\cos\text{x}=\cos\frac{\pi}{3}$
$\Rightarrow\text{x}=2\text{n}\pi\pm\frac{\pi}{3}$
Now for $n = 0$ and $1$, the values of $x$ are $\frac{\pi}{3},\ \frac{5\pi}{3}$ and $\frac{7\pi}{3},$ but $\frac{7\pi}{3}$ is not in $[0,\ 2\pi]$
Hence, there are two solutions in $[0,\ 2\pi].$
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The value of $\sin50^\circ-\sin70^\circ+\sin10^\circ$is equal to
- A
$1$
- ✓
$0$
- C
$\frac{1}{2}$
- D
$2$
Answer$\sin50^\circ-\sin70^\circ+\sin10^\circ$
$=\ 2\sin\Big(\frac{50^\circ-70^\circ}{2}\Big)\cos\Big(\frac{50^\circ+70^\circ}{2}\Big)+\sin10^\circ$
$\Big[\because\ \sin\text{A}-\sin\text{B}=2\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\Big]$
$=\ 2\sin(-10^\circ)\cos60^\circ+\sin10^\circ$
$=\ 2\times\frac{1}{2}\sin(-10^\circ)+\sin10^\circ$
$=\ -\sin10^\circ+\sin10^\circ$
$=\ 0$
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The number of values of $x$ in the interval $[0,\ 5\pi]$ satisfying the equation $3\sin^2\text{x}-7\sin\text{x}+2=0$ is:
AnswerGiven:
$3\sin^2\text{x}-7\sin\text{x}+2=0$
$\Rightarrow3\sin^2\text{x}-6\sin\text{x}-\sin\text{x}+2=0$
$\Rightarrow3\sin\text{x}(\sin\text{x}-2)-1(\sin\text{x}-2)=0$
$\Rightarrow(3\sin\text{x}-1)(\sin\text{x}-2)=0$
$\Rightarrow3\sin\text{x}-1=0$ or $\sin\text{x}-2=0$
Now, $\sin\text{x}=2$ is not possible, as the value of sin $x\sin x$ lies between $-1$ and $1.$
$\Rightarrow\sin\text{x}=\frac{1}{4}$
Also, $\sin x$ is positive only in first two quadrants.
Therefore, sin $x$ is positive twice in the interval $[0,\ \pi].$
Hence, it is positive six times in the interval $[0,\ 5\pi]$ viz $[0,\ \pi],$ $2\pi,\ 3\pi$ and $[4\pi,\ 5\pi].$
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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 $D.$
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