Topic Quiz • 61 questions • 40 seconds per question.
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What is the value of \(\sin^{-1}1\)?
B. \(\frac{\pi}{2}\)
\(\sin\frac{\pi}{2}=1\) और \(\frac{\pi}{2}\) मुख्य परिसर में है। इसलिए उत्तर \(\frac{\pi}{2}\) है। / Since \(\sin\frac{\pi}{2}=1\) and \(\frac{\pi}{2}\) is in the principal range, the answer is \(\frac{\pi}{2}\).
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What is the principal value of \(\cot^{-1}0\)?
C. \(\frac{\pi}{2}\)
\(\cot\frac{\pi}{2}=0\) और \(\frac{\pi}{2}\) परिसर (\left\(0,\pi\right\)) में है। इसलिए मान \(\frac{\pi}{2}\) है। / Since \(\cot\frac{\pi}{2}=0\) and \(\frac{\pi}{2}\) lies in (\left\(0,\pi\right\)), the value is \(\frac{\pi}{2}\).
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If \(y=\sin^{-1}x\), what does it mean?
A. \(\sin y=x\)
\(\sin^{-1}x\) वह कोण देता है जिसका \(\sin\) मान (x) हो। इसलिए \(y=\sin^{-1}x\) का अर्थ \(\sin y=x\) है। / The expression \(\sin^{-1}x\) gives the angle whose \(\sin\) value is (x). So \(y=\sin^{-1}x\) means \(\sin y=x\).
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What happens when (x=2) is used in \(\sin^{-1}x\)?
B. परिभाषित नहीं होगाIt is not defined
\(\sin^{-1}x\) का प्रांत \(\left[-1,1\right]\) है और \(2\notin\left[-1,1\right]\)। इसलिए यह परिभाषित नहीं है। / The domain of \(\sin^{-1}x\) is \(\left[-1,1\right]\) and \(2\notin\left[-1,1\right]\). Hence it is not defined.
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Which of the following statements is correct about \(\tan^{-1}5\)?
B. यह परिभाषित है — इसका मान किसी सीमित कोण के समान है जो \((-\tfrac{\pi}{2},\tfrac{\pi}{2})\) में आता हैIt is defined — equal to a finite angle in \((-\tfrac{\pi}{2},\tfrac{\pi}{2})\)
विलोम फलन \(\tan^{-1}x\) सभी वास्तविक संख्याओं \(x\in\mathbb{R}\) के लिए परिभाषित होता है और इसका प्रधान मान (principal value) दायरा \((-\tfrac{\pi}{2},\tfrac{\pi}{2})\) में लिया जाता है। इसलिए \(\tan^{-1}5\) निश्चित रूप से परिभाषित है और किसी सीमित वास्तविक कोण के बराबर है जो उपर्युक्त खुला अन्तराल में आता है। विकल्प \(\tfrac{\pi}{2}\) गलत है क्योंकि \(\tan\theta\) के लिए \(\theta=\tfrac{\pi}{2}\) पर अनियतता (vertical asymptote) होती है और arctan उस सीमा बिंदु को कभी ग्रहण नहीं करता। विकल्प \(\pi\) भी गलत है क्योंकि arctan का प्रधान मान \(\pi\) नहीं हो सकता। परीक्षा सुझाव: किसी भी वास्तविक संख्या के लिए \(\tan^{-1}x\) की डोमेन और रेंज याद रखें — domain = all real numbers, range = \((-\tfrac{\pi}{2},\tfrac{\pi}{2})\)। / The inverse tangent function \(\tan^{-1}x\) is defined for every real number \(x\in\mathbb{R}\) and its principal value lies in the interval \((-\tfrac{\pi}{2},\tfrac{\pi}{2})\). Hence \(\tan^{-1}5\) is defined and equals a finite angle inside that open interval. The choice \(\tfrac{\pi}{2}\) is incorrect because \(\tan\theta\) has a vertical asymptote at \(\theta=\tfrac{\pi}{2}\) and arctan never attains that endpoint; \(\pi\) is also impossible since arctan’s principal value cannot be \(\pi\). Exam tip: memorize domain = all real numbers and range = \((-\tfrac{\pi}{2},\tfrac{\pi}{2})\) for \(\tan^{-1}x\).
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What type of function is \(\sin^{-1}x\)?
A. व्युत्क्रम त्रिकोणमितीय फलनInverse trigonometric function
\(\sin^{-1}x\) \(\sin x\) का उपयुक्त शाखा पर व्युत्क्रम है। इसलिए यह व्युत्क्रम त्रिकोणमितीय फलन है। / The function \(\sin^{-1}x\) is the inverse of \(\sin x\) on a suitable branch. Hence it is an inverse trigonometric function.
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What is the correct meaning of the exponent (-1) in \(\sin^{-1}x\)?
B. यह \(\sin x\) का व्युत्क्रम फलन हैIt is the inverse function of \(\sin x\)
\(\sin^{-1}x\) का अर्थ \(\sin x\) का व्युत्क्रम फलन है, प्रतिलोम \(\frac{1}{\sin x}\) नहीं। यह एक महत्वपूर्ण गलती है। / The notation \(\sin^{-1}x\) means the inverse function of \(\sin x\), not the reciprocal \(\frac{1}{\sin x}\). This is an important mistake to avoid.
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What is the value of \(\sin^{-1}x+\cos^{-1}x\) when \(x\in\left[-1,1\right]\)?
A. \(\frac{\pi}{2}\)
\(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\) सभी \(x\in\left[-1,1\right]\) के लिए सही है। परीक्षा में यह मूल पहचान याद रखें। / The identity \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\) is true for all \(x\in\left[-1,1\right]\). Remember this basic identity for exams.
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What is the value of \(\tan^{-1}x+\cot^{-1}x\)?
A. \(\frac{\pi}{2}\)
\(\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}\) सभी वास्तविक (x) के लिए लिया जाता है। यह बहुत उपयोगी प्रारंभिक पहचान है। / The identity \(\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}\) is used for all real (x). This is a very useful introductory identity.
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How does \(\cos^{-1}x\) behave on its domain \(\left[-1,1\right]\)?
B. घटता हुआDecreasing
\(\cos^{-1}x\) \(\left[-1,1\right]\) पर घटता हुआ फलन है। (x) बढ़ने पर मुख्य कोण घटता है। / The function \(\cos^{-1}x\) is decreasing on \(\left[-1,1\right]\). As (x) increases, the principal angle decreases.
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Which statement is correct about \(\tan^{-1}x\)?
A. यह \(\mathbb{R}\) पर बढ़ता हैIt is increasing on \(\mathbb{R}\)
\(\tan^{-1}x\) सभी वास्तविक संख्याओं पर परिभाषित और बढ़ता हुआ है। इसका परिसर (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)) है। / The function \(\tan^{-1}x\) is defined and increasing for all real numbers. Its range is (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)).
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How does \(\cot^{-1}x\) behave on its domain \(\mathbb{R}\)?
B. घटता हुआDecreasing
\(\cot^{-1}x\) \(\mathbb{R}\) पर घटता हुआ फलन है। इसका मुख्य मान परिसर (\left\(0,\pi\right\)) है। / The function \(\cot^{-1}x\) is decreasing on \(\mathbb{R}\). Its principal value range is (\left\(0,\pi\right\)).
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Which statement is correct about \(\sec^{-1}0\)?
B. यह परिभाषित नहीं हैIt is not defined
\(\sec^{-1}x\) का प्रांत (\left\(-\infty,-1\right]\cup\left[1,\infty\right\)) है। (0) इस प्रांत में नहीं है। / The domain of \(\sec^{-1}x\) is (\left\(-\infty,-1\right]\cup\left[1,\infty\right\)). The number (0) is not in this domain.
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What is the horizontal limiting value of \(\tan^{-1}x\) when (x) is very large and positive?
A. \(\frac{\pi}{2}\)
जब \(x\to\infty\), तब \(\tan^{-1}x\to\frac{\pi}{2}\)। \(\frac{\pi}{2}\) सीमा है, पर वास्तविक मान नहीं बनता। / As \(x\to\infty\), \(\tan^{-1}x\to\frac{\pi}{2}\). The value \(\frac{\pi}{2}\) is a limit, not an attained value.
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What is the horizontal limiting value of \(\tan^{-1}x\) when (x) is very large and negative?
A. -\(\frac{\pi}{2}\)
जब \(x\to-\infty\), तब \(\tan^{-1}x\to-\frac{\pi}{2}\)। यह सीमा है, वास्तविक प्राप्त मान नहीं। / As \(x\to-\infty\), \(\tan^{-1}x\to-\frac{\pi}{2}\). This is a limiting value, not an attained value.
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What is the domain of the function \(\sin^{-1}x\)?
A. अंतराल ([-1,1])Interval ([-1,1])
\(\sin x\) का मान केवल ([-1,1]) में आता है इसलिए \(\sin^{-1}x\) का डोमेन ([-1,1]) है। परीक्षा में पहले डोमेन जरूर जांचें। / The value of \(\sin x\) lies only in ([-1,1]) so the domain of \(\sin^{-1}x\) is ([-1,1]). In exams check the domain first.
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What is the domain of \(\cos^{-1}x\)?
A. अंतराल ([-1,1])Interval ([-1,1])
\(\cos x\) का मान भी ([-1,1]) में रहता है इसलिए \(\cos^{-1}x\) का डोमेन ([-1,1]) है। मान रखने से पहले (x) की सीमा देखें। / The value of \(\cos x\) also lies in ([-1,1]) so the domain of \(\cos^{-1}x\) is ([-1,1]). Check the range of (x) before substitution.
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What is the principal value range of \(\tan^{-1}x\)?
A. अंतराल (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\))Interval (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\))
\(\tan^{-1}x\) का प्रधान मान (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)) में होता है। सिरों \(\pm\frac{\pi}{2}\) को शामिल नहीं किया जाता। / The principal value of \(\tan^{-1}x\) lies in (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)). The endpoints \(\pm\frac{\pi}{2}\) are not included.
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What is the value of \(\cos^{-1}1\)?
A. (\0)
क्योंकि \(\cos 0=1\) और \(0\in[0,\pi]\) है इसलिए \(\cos^{-1}1=0\)। \(\cos^{-1}x\) में उत्तर ऋणात्मक नहीं होता। / Because \(\cos 0=1\) and \(0\in[0,\pi]\) so \(\cos^{-1}1=0\). For \(\cos^{-1}x\) the answer is not negative.
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What is the principal value of \(\tan^{-1}1\)?
A. \(\frac{\pi}{4}\)
क्योंकि \(\tan \frac{\pi}{4}=1\) और \(\frac{\pi}{4}\) प्रधान सीमा में है। मान याद रखने से मूल प्रश्न जल्दी हल होता है। / Because \(\tan \frac{\pi}{4}=1\) and \(\frac{\pi}{4}\) lies in the principal range. Remembering basic values helps solve quickly.
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What is (\tan\(\tan^{-1}x\)) equal to when \(x\in\mathbb{R}\)?
A. \(\x\)
\(\tan^{-1}x\) का डोमेन \(\mathbb{R}\) है और (\tan\(\tan^{-1}x\)=x)। यह सबसे अधिक पूछी जाने वाली पहचान है। / The domain of \(\tan^{-1}x\) is \(\mathbb{R}\) and (\tan\(\tan^{-1}x\)=x). This is a very common identity.
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When can (\sin^{-1}\(\sin x\)=x) be directly true?
A. जब \(x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\)When \(x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\)
\(\sin^{-1}x\) की प्रधान सीमा \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) है इसलिए इसी में (\sin^{-1}\(\sin x\)=x) सीधे सही है। सीमा के बिना यह गलती बन सकती है। / The principal range of \(\sin^{-1}x\) is \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) so (\sin^{-1}\(\sin x\)=x) is directly true there. Without range it can be a mistake.
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When can (\cos^{-1}\(\cos x\)=x) be directly true?
A. जब \(x\in[0,\pi]\)When \(x\in[0,\pi]\)
\(\cos^{-1}x\) की प्रधान सीमा \([0,\pi]\) है। इसलिए (\cos^{-1}\(\cos x\)=x) इसी सीमा में सीधे लागू होता है। / The principal range of \(\cos^{-1}x\) is \([0,\pi]\). Therefore (\cos^{-1}\(\cos x\)=x) applies directly only in this range.
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When can (\tan^{-1}\(\tan x\)=x) be directly true?
A. जब (x\in\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\))When (x\in\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\))
\(\tan^{-1}x\) की प्रधान सीमा (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)) है। इसलिए वही अंतराल सही चयन है। / The principal range of \(\tan^{-1}x\) is (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)). Therefore that interval is the correct choice.
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What is the value of \(\sin^{-1}x+\cos^{-1}x\) when \(x\in[-1,1]\)?
A. \(\frac{\pi}{2}\)
हर \(x\in[-1,1]\) के लिए \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\) होता है। यह पहचान अक्सर सीधे उपयोग होती है। / For every \(x\in[-1,1]\), \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\). This identity is often used directly.
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If \(\sin^{-1}x=\frac{\pi}{6}\), what is the value of (x)?
A. \(\frac{1}{2}\)
दोनों तरफ \(\sin\) लेने पर \(x=\sin\frac{\pi}{6}=\frac{1}{2}\) मिलता है। उल्टे फलन में कोण से मूल मान निकालें। / Taking \(\sin\) on both sides gives \(x=\sin\frac{\pi}{6}=\frac{1}{2}\). In inverse functions convert the angle back to the original value.
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If \(\cos^{-1}x=\frac{\pi}{3}\), what is the value of (x)?
A. \(\frac{1}{2}\)
दोनों तरफ \(\cos\) लेने पर \(x=\cos\frac{\pi}{3}=\frac{1}{2}\) है। कोण प्रधान सीमा में है इसलिए उत्तर मान्य है। / Taking \(\cos\) on both sides gives \(x=\cos\frac{\pi}{3}=\frac{1}{2}\). The angle lies in the principal range so the answer is valid.
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If \(\tan^{-1}x=\frac{\pi}{4}\), what is the value of (x)?
A. (\1)
दोनों तरफ \(\tan\) लेने पर \(x=\tan\frac{\pi}{4}=1\) है। \(\tan^{-1}\) वाले प्रश्न में मान सीधे टैन से मिलता है। / Taking \(\tan\) on both sides gives \(x=\tan\frac{\pi}{4}=1\). In \(\tan^{-1}\) questions the value comes directly from tangent.
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What is the domain of \(\cot^{-1}x\)?
A. समस्त वास्तविक संख्याएं \(\mathbb{R}\)All real numbers \(\mathbb{R}\)
\(\cot x\) सभी वास्तविक मान ले सकता है इसलिए \(\cot^{-1}x\) का डोमेन \(\mathbb{R}\) है। इसे \(\sin^{-1}\) के डोमेन से न मिलाएं। / Since \(\cot x\) can take all real values the domain of \(\cot^{-1}x\) is \(\mathbb{R}\). Do not mix it with the domain of \(\sin^{-1}x\).
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What is the usual principal value range of \(\cot^{-1}x\)?
A. अंतराल (\(0,\pi\))Interval (\(0,\pi\))
\(\cot^{-1}x\) का प्रधान मान प्रायः (\(0,\pi\)) में लिया जाता है। इस कारण \(\cot^{-1}0=\frac{\pi}{2}\) होता है। / The principal value of \(\cot^{-1}x\) is usually taken in (\(0,\pi\)). Hence \(\cot^{-1}0=\frac{\pi}{2}\).
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Which expression is not defined?
A. \(\sin^{-1}2\)
\(\sin^{-1}x\) केवल \(x\in[-1,1]\) के लिए परिभाषित है और \(2\notin[-1,1]\)। डोमेन आधारित प्रश्नों में विकल्प तुरंत जांचें। / The expression \(\sin^{-1}x\) is defined only for \(x\in[-1,1]\), and \(2\notin[-1,1]\). In domain questions check each option quickly.
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What condition on (x) makes \(\sin^{-1}x\) real?
A. \(-1\le x\le1\)
\(\sin^{-1}x\) वास्तविक तभी है जब \(x\in[-1,1]\)। इसे असमानता के रूप में \(-1\le x\le1\) लिखते हैं। / The value \(\sin^{-1}x\) is real only when \(x\in[-1,1]\). This is written as \(-1\le x\le1\).
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If \(\cos^{-1}x=0\), what is the value of (x)?
A. (\1)
यदि \(\cos^{-1}x=0\) है तो \(x=\cos0=1\)। उल्टे त्रिकोणमितीय समीकरण में दोनों ओर संबंधित त्रिकोणमितीय फलन लगाएं। / If \(\cos^{-1}x=0\), then \(x=\cos0=1\). In an inverse trigonometric equation apply the corresponding trigonometric function on both sides.
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If \(\sin^{-1}x=\frac{\pi}{2}\), what is the value of (x)?
A. (\1)
दोनों तरफ \(\sin\) लेने पर \(x=\sin\frac{\pi}{2}=1\) मिलता है। यह \(\sin^{-1}x\) का अधिकतम प्रधान मान है। / Taking \(\sin\) on both sides gives \(x=\sin\frac{\pi}{2}=1\). This is the maximum principal value of \(\sin^{-1}x\).
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A. \(\sin^{-1}x\) का डोमेन ([-1,1]) हैThe domain of \(\sin^{-1}x\) is ([-1,1])
\(\sin^{-1}x\) के लिए (x) का मान ([-1,1]) में होना चाहिए। बाकी कथन डोमेन या सीमा में गलत हैं। / For \(\sin^{-1}x\), the value of (x) must lie in ([-1,1]). The other statements are wrong in domain or range.
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A. \(\sin^{-1}x\)
(\sin^{-1}(-x)=-\sin^{-1}x) इसलिए \(\sin^{-1}x\) विषम है। \(\tan^{-1}x\) भी विषम होता है पर विकल्प में नहीं है। / Since (\sin^{-1}(-x)=-\sin^{-1}x), \(\sin^{-1}x\) is odd. The function \(\tan^{-1}x\) is also odd but is not in the options.
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What is (\cos^{-1}(-x)) equal to when \(x\in[-1,1]\)?
A. \(\pi-\cos^{-1}x\)
पहचान (\cos^{-1}(-x)=\pi-\cos^{-1}x) होती है। इसे \(\cos^{-1}x\) को विषम फलन मानकर हल न करें। / The identity is (\cos^{-1}(-x)=\pi-\cos^{-1}x). Do not solve it by treating \(\cos^{-1}x\) as an odd function.
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What is (\sin^{-1}(-x)) equal to when \(x\in[-1,1]\)?
A. \(-\sin^{-1}x\)
\(\sin^{-1}x\) विषम फलन है इसलिए (\sin^{-1}(-x)=-\sin^{-1}x)। चिन्ह बदलने वाले प्रश्नों में यह पहचान उपयोगी है। / The function \(\sin^{-1}x\) is odd, so (\sin^{-1}(-x)=-\sin^{-1}x). This identity is useful in sign-change questions.
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What is (\tan^{-1}(-x)) equal to when \(x\in\mathbb{R}\)?
A. \(-\tan^{-1}x\)
\(\tan^{-1}x\) विषम फलन है इसलिए (\tan^{-1}(-x)=-\tan^{-1}x)। उत्तर हमेशा प्रधान सीमा में रखें। / The function \(\tan^{-1}x\) is odd, so (\tan^{-1}(-x)=-\tan^{-1}x). Keep the answer in the principal range.
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What is the value of (\tan^{-1}\left(\tan\left\(-\frac{\pi}{6}\right\)\right))?
A. \(-\frac{\pi}{6}\)
\(-\frac{\pi}{6}\) प्रधान सीमा (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)) में है इसलिए मान वही है। \(\tan^{-1}\) में सीमा खुली होती है। / The angle \(-\frac{\pi}{6}\) lies in the principal range (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)), so the value is the same. For \(\tan^{-1}\), the interval is open.
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If \(y=\sin^{-1}x\), what is the correct relation between (x) and (y)?
A. \(\sin y=x\) और \(y\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\)\(\sin y=x\) and \(y\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\)
\(y=\sin^{-1}x\) का अर्थ है \(\sin y=x\) और (y) प्रधान सीमा में है। उल्टे फलन को परिभाषा से समझना सबसे सुरक्षित तरीका है। / The statement \(y=\sin^{-1}x\) means \(\sin y=x\) and (y) lies in the principal range. Understanding inverse functions by definition is the safest method.
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If \(y=\tan^{-1}x\), which relation is correct?
A. \(\tan y=x\) और (y\in\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\))\(\tan y=x\) and (y\in\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\))
\(y=\tan^{-1}x\) का मतलब \(\tan y=x\) होता है। यहां (y) प्रधान सीमा (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)) में रहता है। / The expression \(y=\tan^{-1}x\) means \(\tan y=x\). Here (y) stays in the principal range (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)).
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What is the principal value of \(\tan^{-1}\sqrt{3}\)?
A. \(\frac{\pi}{3}\)
क्योंकि \(\tan\frac{\pi}{3}=\sqrt{3}\) है और \(\frac{\pi}{3}\) प्रधान सीमा में है। \(\tan^{-1}\) में कोण (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)) में लें। / Because \(\tan\frac{\pi}{3}=\sqrt{3}\) and \(\frac{\pi}{3}\) lies in the principal range. For \(\tan^{-1}\), take the angle in (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)).
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What is the principal value of (\tan^{-1}\left\(\frac{1}{\sqrt{3}}\right\))?
A. \(\frac{\pi}{6}\)
क्योंकि \(\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}\) है। मूल त्रिकोणमितीय मानों से उल्टे मान जल्दी मिलते हैं। / Because \(\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}\). Inverse values are found quickly from basic trigonometric values.
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What is the value of (\cos^{-1}\left\(\cos\frac{\pi}{4}\right\))?
A. \(\frac{\pi}{4}\)
\(\frac{\pi}{4}\in[0,\pi]\) है इसलिए (\cos^{-1}\left\(\cos\frac{\pi}{4}\right\)=\frac{\pi}{4})। प्रधान सीमा में कोण हो तो वही उत्तर आता है। / Since \(\frac{\pi}{4}\in[0,\pi]\), (\cos^{-1}\left\(\cos\frac{\pi}{4}\right\)=\frac{\pi}{4}). If the angle lies in the principal range, it remains unchanged.
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What is the value of (\tan^{-1}\left\(\tan\frac{\pi}{3}\right\))?
A. \(\frac{\pi}{3}\)
\(\frac{\pi}{3}\) प्रधान सीमा (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)) में आता है। इसलिए व्यंजक का मान \(\frac{\pi}{3}\) है। / The angle \(\frac{\pi}{3}\) lies in the principal range (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)). Therefore the value of the expression is \(\frac{\pi}{3}\).
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What is the value of \(\operatorname{cosec}^{-1}2\)?
A. \(\frac{\pi}{6}\)
क्योंकि \(\operatorname{cosec}\frac{\pi}{6}=2\) है। \(\operatorname{cosec}^{-1}x\) में (x) का परिमाण (1) से कम नहीं हो सकता। / Because \(\operatorname{cosec}\frac{\pi}{6}=2\). In \(\operatorname{cosec}^{-1}x\), the magnitude of (x) cannot be less than (1).
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What is the principal value of \(\cot^{-1}\sqrt{3}\)?
A. \(\frac{\pi}{6}\)
क्योंकि \(\cot\frac{\pi}{6}=\sqrt{3}\) है और (\frac{\pi}{6}\in\(0,\pi\)) है। \(\cot^{-1}\) के प्रधान मान के लिए (\(0,\pi\)) याद रखें। / Because \(\cot\frac{\pi}{6}=\sqrt{3}\) and (\frac{\pi}{6}\in\(0,\pi\)). Remember (\(0,\pi\)) for the principal value of \(\cot^{-1}\).
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Which expression does not give a real number?
A. (\cos^{-1}\left\(\frac{7}{6}\right\))
\(\cos^{-1}x\) केवल \(x\in[-1,1]\) पर वास्तविक है और \(\frac{7}{6}>1\) है। इसलिए यह वास्तविक मान नहीं देगा। / The expression \(\cos^{-1}x\) is real only for \(x\in[-1,1]\), and \(\frac{7}{6}>1\). Hence it will not give a real value.
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Which value is not possible from the range of \(\sin^{-1}x\)?
A. \(\frac{2\pi}{3}\)
\(\sin^{-1}x\) की सीमा \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) है। \(\frac{2\pi}{3}\) इस सीमा के बाहर है। / The range of \(\sin^{-1}x\) is \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). The value \(\frac{2\pi}{3}\) lies outside this range.
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Which value is not possible from the range of \(\cos^{-1}x\)?
A. \(-\frac{\pi}{3}\)
\(\cos^{-1}x\) की सीमा \([0,\pi]\) है इसलिए ऋणात्मक मान संभव नहीं है। ऐसे प्रश्नों में केवल आउटपुट रेंज देखें। / The range of \(\cos^{-1}x\) is \([0,\pi]\), so a negative value is not possible. In such questions check only the output range.
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Which value is not possible from the range of \(\tan^{-1}x\)?
A. \(\frac{\pi}{2}\)
\(\tan^{-1}x\) की सीमा (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)) है जिसमें \(\frac{\pi}{2}\) शामिल नहीं है। खुले अंतराल पर ध्यान दें। / The range of \(\tan^{-1}x\) is (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)), which does not include \(\frac{\pi}{2}\). Pay attention to open intervals.
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What is the value of (\sin\left\(\cos^{-1}\frac{3}{5}\right\))?
A. \(\frac{4}{5}\)
मान लें \(\theta=\cos^{-1}\frac{3}{5}\), तब \(\cos\theta=\frac{3}{5}\) और \(\theta\in[0,\pi]\)। यहां \(\sin\theta=\frac{4}{5}\) होगा। / Let \(\theta=\cos^{-1}\frac{3}{5}\), then \(\cos\theta=\frac{3}{5}\) and \(\theta\in[0,\pi]\). Here \(\sin\theta=\frac{4}{5}\).
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What is the value of (\sin^{-1}\left\(\frac{1}{\sqrt{2}}\right\))?
A. \(\frac{\pi}{4}\)
क्योंकि \(\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}\) है। मानक कोणों के मान परीक्षा में बहुत उपयोगी होते हैं। / Because \(\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}\). Standard angle values are very useful in exams.
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What is the value of (\cos^{-1}\left\(-\frac{1}{\sqrt{2}}\right\))?
A. \(\frac{3\pi}{4}\)
क्योंकि \(\cos\frac{3\pi}{4}=-\frac{1}{\sqrt{2}}\) और \(\frac{3\pi}{4}\in[0,\pi]\) है। \(\cos^{-1}\) में प्रधान मान ऋणात्मक नहीं लेते। / Because \(\cos\frac{3\pi}{4}=-\frac{1}{\sqrt{2}}\) and \(\frac{3\pi}{4}\in[0,\pi]\). For \(\cos^{-1}\), do not take a negative principal value.
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What is the value of (\cos\left\(\sin^{-1}\frac{5}{13}\right\))?
A. \(\frac{12}{13}\)
मान लें \(\theta=\sin^{-1}\frac{5}{13}\), तब \(\sin\theta=\frac{5}{13}\) और \(\theta\) प्रधान सीमा में है। इसलिए \(\cos\theta=\frac{12}{13}\) होगा। / Let \(\theta=\sin^{-1}\frac{5}{13}\), then \(\sin\theta=\frac{5}{13}\) and \(\theta\) is in the principal range. Therefore \(\cos\theta=\frac{12}{13}\).
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What is the value of (\tan\left\(\cos^{-1}\frac{4}{5}\right\))?
A. \(\frac{3}{4}\)
मान लें \(\theta=\cos^{-1}\frac{4}{5}\), तब \(\cos\theta=\frac{4}{5}\) और \(\sin\theta=\frac{3}{5}\)। इसलिए \(\tan\theta=\frac{3}{4}\) होगा। / Let \(\theta=\cos^{-1}\frac{4}{5}\), then \(\cos\theta=\frac{4}{5}\) and \(\sin\theta=\frac{3}{5}\). Hence \(\tan\theta=\frac{3}{4}\).
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For (\sin^{-1}(2x)) to be real, in which interval should (x) lie?
A. \(\left[-\frac{1}{2},\frac{1}{2}\right]\)
(\sin^{-1}(2x)) के लिए \(-1\le 2x\le1\) होना चाहिए। इससे \(x\in\left[-\frac{1}{2},\frac{1}{2}\right]\) मिलता है। / For (\sin^{-1}(2x)), we need \(-1\le 2x\le1\). This gives \(x\in\left[-\frac{1}{2},\frac{1}{2}\right]\).
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For (\cos^{-1}\left\(\frac{x}{3}\right\)) to be real, in which interval should (x) lie?
A. ([-3,3])
(\cos^{-1}\left\(\frac{x}{3}\right\)) के लिए \(-1\le\frac{x}{3}\le1\) होना चाहिए। इसलिए \(-3\le x\le3\) है। / For (\cos^{-1}\left\(\frac{x}{3}\right\)), we need \(-1\le\frac{x}{3}\le1\). Therefore \(-3\le x\le3\).
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If (x>0), what is the value of \(\tan^{-1}x+\cot^{-1}x\)?
A. \(\frac{\pi}{2}\)
जब (x>0) हो तो \(\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}\) होता है। \(\cot^{-1}x\) की प्रधान सीमा का ध्यान रखें। / When (x>0), \(\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}\). Keep the principal range of \(\cot^{-1}x\) in mind.
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What is the principal value of (\operatorname{cosec}^{-1}(-2))?
A. \(-\frac{\pi}{6}\)
क्योंकि (\operatorname{cosec}\left\(-\frac{\pi}{6}\right\)=-2) और \(-\frac{\pi}{6}\) प्रधान सीमा में है। ऋणात्मक मानों में सही प्रधान कोण चुनें। / Because (\operatorname{cosec}\left\(-\frac{\pi}{6}\right\)=-2) and \(-\frac{\pi}{6}\) lies in the principal range. For negative values choose the correct principal angle.
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