MCQ 11 Mark
tan-1 $\left(\frac{x}{y}\right)$ - tan-1 $\frac{x-y}{x+y}$ का मान है:
- A
$\frac \pi 2$
- B
$\frac \pi 3$
- C
$-\frac{3 \pi}{4}$
- D
$\frac \pi 4$
Answer$\tan ^{-1}\left(\frac{x}{y}\right)$$-\tan ^{-1}\left(\frac{x-y}{x+y}\right)$= $\tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1}$ $\left(\frac{\frac{x}{y}-1}{\frac{x}{y}+1}\right)$
= $\tan ^{-1}\left(\frac{x}{y}\right)$ - $\left(\tan ^{-1} \frac{x}{y}-\tan ^{-1} 1\right)$ [$\because $ tan-1 x - tan-1 y = tan-1 $(\frac{x-y}{1+x y})$]
= tan-1 1 = $\frac{\pi}{4}$
View full question & answer→MCQ 21 Mark
यदि $\sin ^{-1}(1-x)$ $2 \sin ^{-1} x$ = $\frac{\pi}{2} $, तो x का मान बराबर है:
- A
$0, \frac{1}{2}$
- B
$0$
- C
$\frac 1 2$
- D
$1, \frac{1}{2}$
Answerज्ञात है, $\sin ^{-1}(1-x)-2 \sin ^{-1} $ x = $\frac{\pi}{2}$
$\Rightarrow$ $-2 \sin ^{-1} x$ = $\frac{\pi}{2}-\sin ^{-1}(1-x)$ $ \Rightarrow$ $-2 \sin ^{-1} x$ = $\cos ^{-1}(1-x)$ [$\because$ $ \sin ^{-1}(1-x)$ + $\cos ^{-1}(1-x)$ = $\frac{\pi}{2}$]
$\Rightarrow $ $ \cos \left(-2 \sin ^{-1} x\right)$ = 1 - x (दोनों पक्षों में cos x की गुणा करने पर)
$ \Rightarrow$ $ \cos \left(2 \sin ^{-1} x\right)$ = 1 - x [$ \because$ cos (-x) = + cos x]
$\Rightarrow $ $\left[1-2 \sin ^{2}\left(\sin ^{-1} x\right)\right]$ = 1 - x [$\because$ $ \sin ^{2} x$ = $(\sin x)^{2}$]
$\Rightarrow $1 - 2x2 = 1 - x $\Rightarrow$ 2x2 - x = 0
$\Rightarrow $ x(2x - 1) = 0 $ \Rightarrow $ x = 0 या 2x - 1 = 0 $\Rightarrow$ x = 0 या $\frac{1}{2}$
किन्तु x = $ \frac{1}{2}$ ज्ञात समीकरण को सन्तुष्ट नहीं करता है, इसलिए x = 0 सही है।
View full question & answer→MCQ 31 Mark
sin (tan-1 x), |x| < 1 का मान बराबर है:
- A
$\frac{1}{\sqrt{1-x^{2}}}$
- B
$\frac{x}{\sqrt{1+x^{2}}}$
- C
$\frac{1}{\sqrt{1+x^{2}}}$
- D
$\frac{x}{\sqrt{1-x^{2}}}$
Answer(sin tan-1 x) = sin [sin-1 $\frac{x}{\sqrt{1+x^{2}}}$] ($\because$ tan-1 x = sin-1 $\frac{x}{\sqrt{1+x^{2}}}$)
= $\frac{x}{\sqrt{1+x^{2}}}$
View full question & answer→MCQ 41 Mark
$\sin \left[\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right]$ बराबर है
- A
$\frac{1}{3}$
- B
- C
$\frac{1}{4}$
- D
$\frac{1}{2}$
Answer$\sin \left[\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right]$= $\sin \left[\frac{\pi}{3}-\sin ^{-1}\left(-\sin \frac{\pi}{6}\right)\right]$ $\left(\because \sin \frac{\pi}{6}=\frac{1}{2}\right)$
= $\sin \left[\frac{\pi}{3}-\sin ^{-1}\left\{\sin \left(-\frac{\pi}{6}\right)\right\}\right]$ $[\because \sin (-x)=-\sin x]$
= $\sin \left[\frac{\pi}{3}-\left(-\frac{\pi}{6}\right)\right]$ = $\sin \left(\frac{\pi}{3}+\frac{\pi}{6}\right)$ = $\sin \frac{\pi}{2}$ = 1
View full question & answer→MCQ 51 Mark
cos-1 $\left(\cos \frac{7 \pi}{6}\right) $ बराबर है
- A
$\frac{5 \pi}{6}$
- B
$\frac{7 \pi}{6}$
- C
$\frac{ \pi}{3}$
- D
$\frac{ \pi}{6}$
Answer$\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)$ = $\cos ^{-1}\left[\cos \left(2 \pi-\frac{5 \pi}{6}\right)\right]$
जहाँ, $\frac{5 \pi}{6} \in[0, \pi] $ $ \therefore$ cos-1 $\left(\cos \frac{7 \pi}{6}\right)$ = cos-1 $ \left[\cos \left(\frac{5 \pi}{6}\right)\right]=\frac{5 \pi}{6}$ $[\because \cos (2 \pi-\theta)=\cos \theta]$
View full question & answer→MCQ 61 Mark
$\tan ^{-1} \sqrt{3}-\sec ^{-1}(-2) $ का मान बराबर है
- A
$-\frac{\pi}{3}$
- B
$\frac{2 \pi}{3}$
- C
$\frac{\pi}{3}$
- D
$\pi$
Answerमान लीजिए $ \tan ^{-1} \sqrt{3}$ = x $ \Rightarrow$ $ \tan x=\sqrt{3} $ $\Rightarrow$ $ \tan x$ = $\tan \frac{\pi}{3}$
$\Rightarrow$ x = $ \frac{\pi}{3} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) $ (मुख्य अंतराल)
मान लीजिए $ \sec ^{-1}(-2)$ = y $ \Rightarrow$ sec y = - 2 $[\because \sec (\pi-\theta)=-\sec \theta]$
$\Rightarrow$ sec y = sec $ \left(\frac{2 \pi}{3}\right) $
$\Rightarrow$ y = $\frac{2 \pi}{3} \in$ [0, $\pi$] - $\left(\frac{\pi}{2}\right)$ (मुख्य अंतराल)
$\therefore$ $ \tan ^{-1} \sqrt{3}-\sec ^{-1}(-2)$ = x - y = $ \frac{\pi}{3}-\frac{2 \pi}{3}$ = $-\frac{\pi}{3}$
View full question & answer→MCQ 71 Mark
Answerचूँकि sin-1 x का परिसर $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ है, इसलिए $-\frac{\pi}{2} \leq$ y $ \leq \frac{\pi}{2}$
View full question & answer→MCQ 81 Mark
$\cos ^{-1}\left(\cos \frac{8 \pi}{5}\right)=$
- A
$\frac{8 \pi}{5}$
- B
$\frac{12 \pi}{5}$
- ✓
$\frac{2 \pi}{5}$
- D
$\frac{4 \pi}{5}$
AnswerCorrect option: C. $\frac{2 \pi}{5}$
View full question & answer→MCQ 91 Mark
$\sin ^{-1} x+\cot ^{-1} x$ का मान है :
- A
$\pi$
- ✓
$\frac{\pi}{2}$
- C
$-\frac{\pi}{2}$
- D
AnswerCorrect option: B. $\frac{\pi}{2}$
View full question & answer→MCQ 101 Mark
$\cos ^{-1}(2 x-1)=$
- A
$2 \cos ^{-1} x$
- B
$\cos ^{-1} \sqrt{x}$
- ✓
$2 \cos ^{-1} \sqrt{x}$
- D
AnswerCorrect option: C. $2 \cos ^{-1} \sqrt{x}$
View full question & answer→MCQ 111 Mark
$\sin \left(\sin ^{-1} \frac{1}{2}+\cos ^{-1} \frac{1}{2}\right)=$
View full question & answer→MCQ 121 Mark
$\tan \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)$ का मान है :
AnswerCorrect option: A. $\frac{17}{6}$
View full question & answer→MCQ 131 Mark
$\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)$ समान है :
- A
$\frac{1}{2} \cos ^{-1}\left(\frac{3}{5}\right)$
- B
$\frac{1}{2} \sin ^{-1}\left(\frac{3}{2}\right)$
- C
$\frac{1}{2} \tan ^{-1}\left(\frac{3}{5}\right)$
- ✓
$\tan ^{-1}\left(\frac{1}{2}\right)$
AnswerCorrect option: D. $\tan ^{-1}\left(\frac{1}{2}\right)$
View full question & answer→MCQ 141 Mark
$\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)$ का मुख्य मान है :
- A
$\frac{2 \pi}{3}$
- B
$\frac{\pi}{6}$
- C
$\frac{\pi}{4}$
- ✓
$\frac{\pi}{3}$
AnswerCorrect option: D. $\frac{\pi}{3}$
View full question & answer→MCQ 151 Mark
$\tan ^{-1} \sqrt{3}-\cot ^{-1}(-\sqrt{3})=$
- A
$\pi$
- B
$0$
- C
$2 \sqrt{3}$
- ✓
$-\frac{\pi}{2}$
AnswerCorrect option: D. $-\frac{\pi}{2}$
View full question & answer→MCQ 161 Mark
$\tan ^{-1}\left(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right)$ का मान है :
- A
$\frac{6}{17}$
- B
$\frac{7}{16}$
- ✓
$\frac{17}{6}$
- D
AnswerCorrect option: C. $\frac{17}{6}$
View full question & answer→MCQ 171 Mark
$\cos ^{-1}\left(-\frac{1}{2}\right)$ का मुख्य मान है :
- A
$\frac{\pi}{3}$
- B
$\frac{\pi}{6}$
- ✓
$\frac{2 \pi}{3}$
- D
$\frac{3 \pi}{4}$
AnswerCorrect option: C. $\frac{2 \pi}{3}$
View full question & answer→MCQ 181 Mark
$x$ का मान जिसके लिए $\sin \left(\cot ^{-1}(1-x)\right)=\cos \left(\tan ^{-1} x\right)$ :
- A
$\frac{2}{1}$
- B
- C
$0$
- ✓
$\frac{1}{2}$
AnswerCorrect option: D. $\frac{1}{2}$
View full question & answer→MCQ 191 Mark
$\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)$ का मुख्य मान है :
- A
$\frac{2 \pi}{3}$
- B
$\frac{\pi}{3}$
- ✓
$\frac{-\pi}{6}$
- D
$\frac{\pi}{6}$
AnswerCorrect option: C. $\frac{-\pi}{6}$
View full question & answer→MCQ 201 Mark
$\sin \left(\sec ^{-1} x+\operatorname{cosec}^{-1} x\right)=$
- ✓
- B
- C
$\frac{\pi}{2}$
- D
$\frac{\pi}{3}$
View full question & answer→MCQ 211 Mark
$\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}$ तो $x=$ ?
- ✓
$0, \frac{1}{2}$
- B
$1, \frac{1}{2}$
- C
$\frac{1}{2}$
- D
$0$
AnswerCorrect option: A. $0, \frac{1}{2}$
View full question & answer→MCQ 221 Mark
$\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)=$
- A
$\frac{7 \pi}{6}$
- B
$\frac{\pi}{3}$
- C
$\frac{\pi}{6}$
- ✓
$\frac{5 \pi}{6}$
AnswerCorrect option: D. $\frac{5 \pi}{6}$
View full question & answer→MCQ 231 Mark
$\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)=$
- A
$\frac{2 \pi}{3}$
- B
$\frac{\pi}{6}$
- C
$\frac{4 \pi}{3}$
- ✓
$\frac{\pi}{3}$
AnswerCorrect option: D. $\frac{\pi}{3}$
View full question & answer→MCQ 241 Mark
$\tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1}\left(\frac{x-y}{x+y}\right)$ बराबर है :
- A
$\frac{\pi}{3}$
- ✓
$\frac{\pi}{4}$
- C
$\frac{\pi}{2}$
- D
$\frac{-3 \pi}{4}$
AnswerCorrect option: B. $\frac{\pi}{4}$
View full question & answer→MCQ 251 Mark
$\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$ का मुख्य मान है :
- ✓
$\frac{\pi}{4}$
- B
$\frac{3 \pi}{4}$
- C
$\frac{5 \pi}{4}$
- D
AnswerCorrect option: A. $\frac{\pi}{4}$
View full question & answer→MCQ 261 Mark
यदि $\cot ^{-1}\left[(\cos \alpha)^{1 / 2}\right]-\tan ^{-1}\left[(\cot \alpha)^{1 / 2}\right]=x$ तब $\sin x=$
AnswerCorrect option: A. $\tan ^2 \frac{\alpha}{2}$
View full question & answer→MCQ 271 Mark
$2 \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{7}=$
AnswerCorrect option: D. $\frac{\pi}{4}$
View full question & answer→MCQ 281 Mark
$\sin \left(\cot ^{-1} x\right)$ के लिए बीजीय व्यंजक है :
- A
$\frac{1}{1+x^2}$
- ✓
$\frac{1}{\sqrt{1+x^2}}$
- C
$\frac{x}{\sqrt{1+x^2}}$
- D
AnswerCorrect option: B. $\frac{1}{\sqrt{1+x^2}}$
View full question & answer→MCQ 291 Mark
$\tan ^{-1} \frac{2 x}{1-x^2}=$ प्रतिलोम त्रिकोणमितीय फलन :
- A
$2 \sin ^{-1} x$
- B
$\sin ^{-1} 2 x$
- C
$\tan ^{-1} 2 x$
- ✓
$2 \tan ^{-1} x$
AnswerCorrect option: D. $2 \tan ^{-1} x$
View full question & answer→MCQ 301 Mark
$\tan ^{-1}(1)=\ldots .$. ?
- ✓
$\frac{\pi}{4}$
- B
$\frac{\pi}{2}$
- C
$\frac{\pi}{3}$
- D
$\frac{\pi}{8}$
AnswerCorrect option: A. $\frac{\pi}{4}$
View full question & answer→MCQ 311 Mark
$\tan ^{-1} x+\cot ^{-1} x$ का मान है :
- A
$-\pi$
- B
$-\frac{\pi}{2}$
- ✓
$\frac{\pi}{2}$
- D
$\frac{\pi}{4}$
AnswerCorrect option: C. $\frac{\pi}{2}$
View full question & answer→MCQ 321 Mark
यदि $\cot ^{-1} \sqrt{\cos \theta}-\tan ^{-1} \sqrt{\cos \theta}=x$ तब $\sin x=$ :
- ✓
$\frac{1-\cos \theta}{1+\cos \theta}$
- B
$\frac{1+\cos \theta}{1-\cos \theta}$
- C
$\frac{2 \sqrt{\cos \theta}}{1+\cos \theta}$
- D
$\frac{2 \sqrt{\cos \theta}}{1-\cos \theta}$
AnswerCorrect option: A. $\frac{1-\cos \theta}{1+\cos \theta}$
View full question & answer→MCQ 331 Mark
$\sec ^{-1} x+\operatorname{cosec}^{-1} x$ का मान है :
- ✓
$\frac{\pi}{2}$
- B
$\frac{\pi}{4}$
- C
- D
$-\frac{\pi}{2}$
AnswerCorrect option: A. $\frac{\pi}{2}$
View full question & answer→MCQ 341 Mark
यदि $\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}$ तब $x$ का मान है :
- ✓
$0$
- B
$0,-\frac{1}{2}$
- C
$0, \frac{1}{2}$
- D
View full question & answer→MCQ 351 Mark
$\operatorname{cosec}^{-1}(-2)$ का मुख्य मान है :
- A
$-\frac{2 \pi}{3}$
- B
$\frac{\pi}{6}$
- C
$\frac{2 \pi}{3}$
- ✓
$-\frac{\pi}{6}$
AnswerCorrect option: D. $-\frac{\pi}{6}$
View full question & answer→MCQ 361 Mark
$\tan ^{-1}(1)+\cot ^{-1}\left(\frac{1}{2}\right)+\cot ^{-1}\left(\frac{1}{3}\right)$ का मान है :
View full question & answer→MCQ 371 Mark
$\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}=$
- ✓
$\frac{\pi}{4}$
- B
$\frac{\pi}{2}$
- C
$\frac{\pi}{3}$
- D
$\pi$
AnswerCorrect option: A. $\frac{\pi}{4}$
View full question & answer→MCQ 381 Mark
$\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)$ के बराबर है :
- A
$\frac{7 \pi}{6}$
- ✓
$\frac{5 \pi}{6}$
- C
$\frac{\pi}{3}$
- D
$\frac{\pi}{6}$
AnswerCorrect option: B. $\frac{5 \pi}{6}$
View full question & answer→MCQ 391 Mark
$\tan ^{-1} \sqrt{3}-\sec ^{-1}(-z)$ के बराबर है :
- A
$\pi$
- ✓
$-\frac{\pi}{3}$
- C
$\frac{\pi}{3}$
- D
$\frac{2 \pi}{3}$
AnswerCorrect option: B. $-\frac{\pi}{3}$
View full question & answer→MCQ 401 Mark
$\tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)=$
- A
$0$
- B
$\frac{\pi}{4}$
- ✓
$\frac{3 \pi}{4}$
- D
AnswerCorrect option: C. $\frac{3 \pi}{4}$
View full question & answer→MCQ 411 Mark
$\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{2}{11}=\tan ^{-1} a$ तब $a=$ ?
- A
$\frac{1}{4}$
- B
$\frac{1}{2}$
- ✓
$\frac{3}{4}$
- D
AnswerCorrect option: C. $\frac{3}{4}$
View full question & answer→MCQ 421 Mark
$\sin \left[\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right]$ के बराबर है :
- A
$\frac{1}{2}$
- B
$\frac{1}{3}$
- C
$\frac{1}{4}$
- ✓
View full question & answer→MCQ 431 Mark
$\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)$ का मुख्य मान है :
- ✓
$\frac{3 \pi}{4}$
- B
$\frac{5 \pi}{4}$
- C
$-\frac{\pi}{4}$
- D
AnswerCorrect option: A. $\frac{3 \pi}{4}$
View full question & answer→MCQ 441 Mark
$3 \sec ^{-1} \frac{1}{x}-\sin ^{-1}\left(4 x^3-3 x\right)=$
- A
$\pi$
- B
$\frac{\pi}{2}$
- ✓
$\frac{3 \pi}{2}$
- D
AnswerCorrect option: C. $\frac{3 \pi}{2}$
View full question & answer→MCQ 451 Mark
$\tan ^{-1} \frac{4 \sqrt{x}}{2-2 x}$
AnswerCorrect option: A. $2 \tan ^{-1} \sqrt{x}$
View full question & answer→MCQ 461 Mark
$\sin ^{-1} x+\cos ^{-1} x=$
- ✓
$\frac{\pi}{2}$
- B
$\pi$
- C
$\frac{\pi}{4}$
- D
$2 \pi$
AnswerCorrect option: A. $\frac{\pi}{2}$
View full question & answer→MCQ 471 Mark
$\sin ^{-1}\left(\sin \frac{7 \pi}{6}\right)=$
- A
$\frac{7 \pi}{6}$
- B
$\frac{\pi}{6}$
- ✓
$\frac{-\pi}{6}$
- D
$\frac{5 \pi}{6}$
AnswerCorrect option: C. $\frac{-\pi}{6}$
View full question & answer→MCQ 481 Mark
$\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{4}=\ldots \ldots$
- A
$\tan ^{-1} \frac{3}{2}$
- ✓
$\tan ^{-1} \frac{6}{7}$
- C
$\tan ^{-1} \frac{5}{6}$
- D
$\tan ^{-1} \frac{1}{2}$
AnswerCorrect option: B. $\tan ^{-1} \frac{6}{7}$
View full question & answer→MCQ 491 Mark
$2 \tan ^{-1} x=$
- ✓
$\sin ^{-1} \frac{2 x}{1+x^2}$
- B
$\sin ^{-1} \frac{2 x}{1-x^2}$
- C
$\sin ^{-1} \frac{1-x^2}{1+x^2}$
- D
$\sin ^{-1} \frac{1+x^2}{1-x^2}$
AnswerCorrect option: A. $\sin ^{-1} \frac{2 x}{1+x^2}$
View full question & answer→MCQ 501 Mark
$\tan ^{-1} \sqrt{3}-\cot ^{-1}(-\sqrt{3})$ के बराबर है :
- A
$\pi$
- ✓
$-\frac{\pi}{2}$
- C
$0$
- D
$2 \sqrt{3}$
AnswerCorrect option: B. $-\frac{\pi}{2}$
View full question & answer→MCQ 511 Mark
यदि $\sin ^{-1} x+\sin ^{-1} y=\frac{2 \pi}{3}$ तो $\cos ^{-1} x+\cos ^{-1} y=$
- A
$\frac{2 \pi}{3}$
- B
$\frac{\pi}{4}$
- ✓
$\frac{\pi}{3}$
- D
$\frac{\pi}{4}$
AnswerCorrect option: C. $\frac{\pi}{3}$
View full question & answer→MCQ 521 Mark
$4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{239}=$
- A
$\frac{\pi}{4}$
- B
$\frac{\pi}{2}$
- C
$\frac{\pi}{3}$
- ✓
$\frac{\pi}{4}$
AnswerCorrect option: D. $\frac{\pi}{4}$
View full question & answer→MCQ 531 Mark
यदि $|x| \leq 1$ तो $\tan \left(\cos ^{-1} x\right)=$
- ✓
$\frac{\sqrt{1-x^2}}{x}$
- B
$\frac{x}{1+x^2}$
- C
$\frac{\sqrt{1+x^2}}{x}$
- D
$\sqrt{1-x^2}$
AnswerCorrect option: A. $\frac{\sqrt{1-x^2}}{x}$
View full question & answer→MCQ 541 Mark
$\tan ^{-1}(-1)$ का मान है :
- A
$\frac{\pi}{4}$
- B
$\frac{-\pi}{2}$
- C
$\frac{5 \pi}{4}$
- ✓
$\frac{-\pi}{4}$
AnswerCorrect option: D. $\frac{-\pi}{4}$
View full question & answer→MCQ 551 Mark
$\cos ^{-1}\left(4 x^3-3 x\right)$ का सरल रूप है :
- A
$3 \sin ^{-1} x$
- ✓
$3 \cos ^{-1} x$
- C
$\pi-3 \sin ^{-1} x$
- D
AnswerCorrect option: B. $3 \cos ^{-1} x$
View full question & answer→MCQ 561 Mark
$\cos ^{-1} \frac{1-x^2}{1+x^2}=$
- A
$2 \cos ^{-1} x$
- B
$2 \sin ^{-1} x$
- ✓
$2 \tan ^{-1} x$
- D
$\cos ^{-1} 2 x$
AnswerCorrect option: C. $2 \tan ^{-1} x$
View full question & answer→MCQ 571 Mark
$\cos ^{-1}\left(\frac{-1}{2}\right)=$
- ✓
$\frac{2 \pi}{3}$
- B
$\frac{\pi}{3}$
- C
$\frac{-\pi}{3}$
- D
$\frac{-2 \pi}{3}$
AnswerCorrect option: A. $\frac{2 \pi}{3}$
View full question & answer→MCQ 581 Mark
$\cot ^{-1}(-x)=$
- ✓
$-\cot ^{-1} x$
- B
$\cot ^{-1} x$
- C
$\pi+\cot ^{-1} x$
- D
$\pi-\cot ^{-1} x$
AnswerCorrect option: A. $-\cot ^{-1} x$
View full question & answer→MCQ 591 Mark
$\sin ^{-1} x=y$ तब :
- A
$0 \leq y \leq \pi$
- ✓
- C
$0\lt y\lt\pi$
- D
View full question & answer→MCQ 601 Mark
$\cos ^{-1}\left(\cos \frac{5 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{5 \pi}{3}\right)$ का मान है :
- A
$\frac{\pi}{2}$
- B
$\frac{5 \pi}{3}$
- C
$\frac{10 \pi}{3}$
- ✓
$0$
View full question & answer→MCQ 611 Mark
$\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)$ का मान है :
- A
$\frac{\pi}{4}$
- B
$\frac{\pi}{6}$
- ✓
$\frac{2 \pi}{3}$
- D
$\frac{5 \pi}{6}$
AnswerCorrect option: C. $\frac{2 \pi}{3}$
View full question & answer→MCQ 621 Mark
यदि $\cos ^{-1} x-\cos ^{-1}\left(\frac{y}{2}\right)=a$ तब $4 x^2-4 x y \cos \alpha+y^2$ समान है :
- ✓
$4 \sin ^2 \alpha$
- B
$-4 \sin ^2 \alpha$
- C
$2 \sin ^2 \alpha$
- D
AnswerCorrect option: A. $4 \sin ^2 \alpha$
View full question & answer→MCQ 631 Mark
$\sin ^{-1} x=$
- A
$\cot ^{-1} x$
- B
$\frac{1}{\cot ^{-1} x}$
- ✓
- D
View full question & answer→MCQ 641 Mark
$2 \cot ^{-1} 3+\cot ^{-1} 7=$
- A
$\frac{\pi}{2}$
- ✓
$\frac{\pi}{4}$
- C
$\pi$
- D
$\frac{\pi}{6}$
AnswerCorrect option: B. $\frac{\pi}{4}$
View full question & answer→MCQ 651 Mark
$\sec ^2\left(\tan ^{-1} 5\right)+\operatorname{cosec}^2\left(\cot ^{-1} 5\right)$ का मान है :
View full question & answer→MCQ 661 Mark
$\sin ^{-1}\left(\sin \frac{\pi}{3}\right)+\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)$ का मान है :
- A
$\frac{17}{12} \pi$
- B
$\frac{5}{12} \pi$
- ✓
$\frac{\pi}{12}$
- D
$-\frac{7 \pi}{12}$
AnswerCorrect option: C. $\frac{\pi}{12}$
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