MCQ 11 Mark
The value of $\sin50^\circ-\sin70^\circ+\sin10^\circ$ is equal to:
- A
$1$
- ✓
$0$
- C
$\frac{1}{2}$
- D
$2$
Answer$\sin50^\circ-\sin70^\circ+\sin10^\circ$
$=2\cos\Big(\frac{50^\circ+70^\circ}{2}\Big)\sin\Big(\frac{50^\circ-70^\circ}{2}\Big)+\sin10^\circ$
$=-2\cos60^\circ\sin10^\circ+\sin10^\circ$
$=-2\cdot\frac{1}{2}\sin10^\circ+\sin10^\circ$
$=0$
View full question & answer→MCQ 21 Mark
If $\tan\theta=3$ and $\theta$ lies in third quadrant, then the value of $\sin\theta$ is:
- A
$\frac{1}{\sqrt{10}}$
- B
$-\frac{1}{\sqrt{10}}$
- ✓
$\frac{-3}{\sqrt{10}}$
- D
$\frac{3}{\sqrt{10}}$
AnswerCorrect option: C. $\frac{-3}{\sqrt{10}}$
$\tan\theta=3,\theta$ lies in third quadrant, it is positive.
$\tan\theta=\frac{\text{P}}{\text{B}}=\frac{3}{1}$
Then, Hypotenuse $=\sqrt{(3)^2+(1)^2}=\sqrt{9+1}=\sqrt10$
$\therefore\sin\theta=\frac{3}{\sqrt{10}}$ where $\theta$ lies in third quadrant
Hence the correct option is $(c).$ View full question & answer→MCQ 31 Mark
If A lies in the second quadrant and $3\tan\text{A}+4=0,$ then the value of $2\cot\text{A}-5\cos\text{A}+\sin\text{A}$ is equal to:
- A
$\frac{-53}{10}$
- ✓
$\frac{23}{10}$
- C
$\frac{37}{10}$
- D
$\frac{7}{10}$
AnswerCorrect option: B. $\frac{23}{10}$
Given that, $3\tan\text{A}+4=0,$
$A$ lies in second quadrant
$\therefore\tan\text{A}=\frac{-4}{3}$
$\cos\text{A}=\frac{-3}{5} [A$ lies in second quadrant$]$
and $\sin\text{A}=\frac{4}{5}$ and $\cot\text{A}=\frac{-3}{4}$
$\therefore2\cot\text{A}-5\cos\text{A}+\sin\text{A}=2\Big(\frac{-3}{4}\Big)-5\Big(\frac{-3}{5}\Big)+\frac{4}{5}\\=\frac{-3}{2}+3+\frac{4}{5}=\frac{-15+30+8}{10}=\frac{23}{10}$
Hence, the correct option is $(b).$
View full question & answer→MCQ 41 Mark
The value of $\tan1^\circ\tan2^\circ\tan3^\circ...\tan89^\circ$ is:
Answer$\tan1^\circ\tan2^\circ\tan^\circ\dots\tan89^\circ$
$=[\tan1^\circ\tan2^\circ\dots\tan44^\circ]\\\tan45^\circ[\tan(90^\circ-44^\circ)\tan(90^\circ-43^\circ)\dots\tan(90^\circ-1^\circ)]$
$=[\tan1^\circ\tan2^\circ\dots\tan44^\circ][\cot44^\circ\cot43^\circ\dots\cot1^\circ]$
$=1-1\dots1-1$
$=1$
View full question & answer→MCQ 51 Mark
The value of $\sin(45^\circ+\theta)-\cos(45^\circ-\theta)$ is:
- A
$2\cos\theta$
- B
$2\sin\theta$
- C
$1$
- ✓
$0$
AnswerGiven expression is $(\sin45^\circ+\theta)-\cos(45^\circ-\theta)$
$\sin(45^\circ+\theta)=\sin45^\circ\cos\theta+\cos45^\circ\sin\theta$
$=\frac{1}{\sqrt2}\cos\theta+\frac{1}{\sqrt2}\sin\theta$
$\cos(45^\circ-\theta)=\cos45^\circ\cos\theta+\sin45^\circ\sin\theta$
$=\frac{1}{\sqrt2}\cos\theta+\frac{1}{\sqrt2}\sin\theta$
$\sin(45^\circ+\theta)-\cos(45^\circ-\theta)$
$=\frac{1}{\sqrt2}\cos\theta+\frac{1}{\sqrt2}\sin\theta-\frac{1}{\sqrt2}\cos\theta-\frac{1}{\sqrt2}\sin\theta
= 0.$
Hence, the correct option is $(d). $
View full question & answer→MCQ 61 Mark
The value of $\cot\Big(\frac{\pi}{4}+\theta\Big)\cot\Big(\frac{\pi}{4}-\theta\Big)$ is:
Answer$\cot\Big(\frac{\pi}{4}+\theta\Big)\cdot\cot\Big(\frac{\pi}{4}-\theta\Big)=\frac{\cot\frac{\pi}{4}\cot\theta-1}{\cot\theta+\cot\frac{\pi}{4}}\times\frac{\cot\frac{\pi}{4}\cot\theta+1}{\cot\theta-\cot\frac{\pi}{4}}$
$=\frac{1.\cot\theta-1}{\cot\theta+1}\times\frac{1.\cot\theta+1}{\cot-1}$
$=\frac{\cot\theta-1}{\cot\theta+1}\times\frac{\cot\theta+1}{\cot\theta-1}$
$=1$
Hence, the correct option is $(c).$
View full question & answer→MCQ 71 Mark
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 81 Mark
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 91 Mark
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 101 Mark
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 111 Mark
The value of $\sin\frac{\pi}{10}\sin\frac{13\pi}{10}$ is : $[$Hint: Use $\sin18^\circ=\frac{\sqrt{5}-1}{4}$ and $\cos36^\circ=\frac{\sqrt{5}+1}{4}]$
- 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 121 Mark
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 131 Mark
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 141 Mark
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 151 Mark
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 161 Mark
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 171 Mark
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 181 Mark
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 191 Mark
$\cos2\theta\cos2\phi+\sin^2(\theta-\phi)-\sin^2(\theta+\phi)$ is equal to:
$[$Hint: Use $\sin^2\text{A}-\sin^2\text{B}=\sin(\text{A+B})\sin(\text{A}-\text{B})]$
- 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 201 Mark
Which of the following is correct?
$[$Hint: $1\text{radian} =\frac{180^\circ}{\pi}=57^\circ30'\text{approx}]$
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 211 Mark
The value of $\cos^248^\circ-\sin^212^\circ$ is:
$[$Hint: Use $\cos^2\text{A}-\sin^2\text{B}=\cos(\text{A + B})\cos(\text{A}-\text{B})]$
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 221 Mark
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 231 Mark
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 241 Mark
The value of $\cos1^\circ\cos2^\circ\cos3^\circ...\cos179^\circ$ is:
Answersince $\cos90^\circ=0,$ we have
$\cos1^\circ\cos2^\circ\cos3^\circ\dots\cos90^\circ\dots\cos179^\circ=0$
View full question & answer→MCQ 251 Mark
If $\text{f(x)}=\cos^2\text{x}+\sec^2\text{x},$ then :
$[$Hint: $\text{A.M}\geq\text{G.M.}]$
- 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 261 Mark
If $\sin\theta+\cos\theta=1,$ then the value of $\sin2\theta$ is equal to:
- A
$1$
- B
$\frac{1}{2}$
- ✓
$0$
- D
$-1$
AnswerGiven that, $\sin\theta+\cos\theta=1$
Squaring both sides, we get,
$\Rightarrow(\sin\theta+\cos\theta)^2=(1)^2$
$\Rightarrow\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=1$
$\Rightarrow1+\sin2\theta=1$
$\Rightarrow\sin2\theta=1-1=0$
Hence, the correct option is $(c).$
View full question & answer→MCQ 271 Mark
If for real values of $x, \cos\theta=\text{x}+\frac{1}{\text{x}},$ then:
- A
$\theta$ is an acute angle
- B
$\theta$ is right angle
- C
$\theta$ is an obtuse angle
- ✓
No value of $\theta$ is possible
AnswerCorrect option: D. No value of $\theta$ is possible
given that, $\cos\theta=\text{x}+\frac{1}{\text{x}}$
$\Rightarrow\cos\theta=\frac{\text{x}^2+1}{\text{x}}$
$\Rightarrow\text{x}^2+1=\text{x}\cos\theta$
$\Rightarrow\text{x}^2-\text{x}\cos\theta+1=0$
For real value of $\text{x},\text{b}^2-4\text{a}\text{c}\geq0$
$\Rightarrow(-\cos\theta)^2-4\times1\times1\geq0$
$\Rightarrow\cos^2\theta-4\geq0$
$\Rightarrow\cos^2\theta\geq4$
$\Rightarrow\cos\theta\geq\pm2[-1\leq\cos\theta\leq1]$
so, the value of $\theta$ is not possible.
Hence, the correct options $(d).$
View full question & answer→MCQ 281 Mark
If $\tan\alpha=\frac{1}{7},\tan\beta=\frac{1}{3},$ then $\cos2\alpha$ is equal to:
- A
$\sin2\beta$
- ✓
$\sin4\beta$
- C
$\sin3\beta$
- D
$\cos2\beta$
AnswerCorrect option: B. $\sin4\beta$
Given that, $\tan\alpha=\frac{1}{7}$ and $\tan\beta=\frac{1}{3}$
$\cos2\alpha=\frac{1-\tan^2\alpha}{1+\tan^2\alpha}=\frac{1-\Big(\frac{1}{7}\Big)^2}{1+\Big(\frac{1}{7}\Big)^2}=\frac{1-\frac{1}{49}}{1+\frac{1}{49}}$
$=\frac{48}{50}=\frac{24}{25}$
Now $\tan2\beta=\frac{2\tan\beta}{1-\tan^2\beta}=\frac{2\times\frac{1}{3}}{1-\frac{1}{9}}=\frac{\frac{2}{3}}{\frac{8}{9}}=\frac{2}{3}\times\frac{9}{8}=\frac{3}{4}$
$\therefore\tan^2\beta=\frac{3}{4}$
$\sin4\beta=\frac{2\tan2\beta}{1+\tan^22\beta}$
$=\frac{2\times\frac{3}{4}}{1+\Big(\frac{3}{4}\Big)^2}$
$=\frac{\frac{3}{2}}{1+\frac{9}{16}}$
$=\frac{3}{2}\times\frac{16}{25}$
$=\frac{24}{25}$
$\cos2\alpha=\sin4\beta=\frac{24}{25}$
Hence, the correct option is $(b).$
View full question & answer→MCQ 291 Mark
If $\tan\text{A}=\frac{1}{2},\tan\text{B}=\frac{1}{3},$ then $\tan(2\text{A + B})$ is equal to:
AnswerGiven that, $\tan\text{A}=\frac{1}{2}$ and $\tan\text{B}=\frac{1}{3}$
$\tan2\text{A}=\frac{2\tan\text{A}}{1-\tan^2\text{A}}=\frac{2\times\frac{1}{2}}{1-\Big(\frac{1}{2}\Big)^2}$
$=\frac{1}{1-\frac{1}{4}}=\frac{1}{\frac{3}{4}}=\frac{4}{3}$
so, $\tan2\text{A}=\frac{4}{3}$ and $\tan\text{B}=\frac{1}{3}$
$\tan(\text{2A+B})=\frac{\tan2\text{A}+\tan\text{B}}{1-\tan\text{A}.\tan\text{B}}=\frac{\frac{4}{3}+\frac{1}{3}}{1-\frac{4}{3}\times\frac{1}{3}}$
$=\frac{\frac{5}{3}}{\frac{9-4}{9}}=\frac{5}{3}\times\frac{9}{5}=3$
Hence, the correct option is $(c).$
View full question & answer→MCQ 301 Mark
Which of the following is not correct?
AnswerCorrect option: C. $\sec\theta=\frac{1}{2}$
$\sin\theta=-\frac{1}{5}$ is correct.
$\because-1\leq\sin\theta\leq1$
so $(a)$ is correct.
$\cos\theta=1$ is correct.
$\because\cos0^\circ=1$
so $(b)$ is correct.
$\sec\theta=-\frac{1}{2}$
$\Rightarrow\cos\theta=2$ is not correct.
$\because-1\leq\cos\theta\leq1$
Hence, $(c)$ is not correct.
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