Level 32 • 49/50 questions • 35 seconds per question.
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If \(y=\sin^{-1}x\), which is the correct domain of (x)?
A. \(x\in[-1,1]\)
\(\sin^{-1}x\) तभी परिभाषित है जब \(x\in[-1,1]\) हो। परीक्षा में पहले डोमेन जरूर जांचें। / \(\sin^{-1}x\) is defined only when \(x\in[-1,1]\). Always check the domain first in exams.
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What is the principal value of \(\sin^{-1}\left(\frac{1}{2}\right)\)?
A. \(\frac{\pi}{6}\)
\(\sin\frac{\pi}{6}=\frac{1}{2}\) और \(\frac{\pi}{6}\) मुख्य परिसर में है। मुख्य मान हमेशा range में चुनें। / \(\sin\frac{\pi}{6}=\frac{1}{2}\) and \(\frac{\pi}{6}\) lies in the principal range. Always choose the value inside the principal range.
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Under which condition is \(\sin(\sin^{-1}x)\) equal to \(x\)?
B. केवल \(x\in[-1,1]\) के लिएOnly for \(x\in[-1,1]\)
\(\sin^{-1}x\) का डोमेन \([-1,1]\) है, इसलिए इसी पर \(\sin(\sin^{-1}x)=x\) होगा। composition में inner function का domain देखें। / The domain of \(\sin^{-1}x\) is \([-1,1]\), so \(\sin(\sin^{-1}x)=x\) there. In composition, check the inner function domain.
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When is \(\sin^{-1}(\sin\theta)=\theta\) definitely true?
B. \(\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]\)
\(\sin^{-1}x\) का आउटपुट \([-\frac{\pi}{2},\frac{\pi}{2}]\) में होता है। इसलिए \(\theta\) इसी range में होना चाहिए। / The output of \(\sin^{-1}x\) lies in \([-\frac{\pi}{2},\frac{\pi}{2}]\). Therefore \(\theta\) must lie in this range.
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When is \(\cos^{-1}(\cos\theta)=\theta\) definitely true?
A. \(\theta\in[0,\pi]\)
\(\cos^{-1}x\) का principal output \([0,\pi]\) में होता है। equality के लिए \(\theta\) को उसी interval में लें। / The principal output of \(\cos^{-1}x\) lies in \([0,\pi]\). For equality, take \(\theta\) in that interval.
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When is \(\tan^{-1}(\tan\theta)=\theta\) definitely true?
B. \(\theta\in(-\frac{\pi}{2},\frac{\pi}{2})\)
\(\tan^{-1}x\) का principal range \((-\frac{\pi}{2},\frac{\pi}{2})\) है। \(\tan\) period वाला है, इसलिए principal interval जरूरी है। / The principal range of \(\tan^{-1}x\) is (\(-\frac{\pi}{2},\frac{\pi}{2}\)). Since \(\tan\) is periodic, the principal interval is necessary.
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Which is the correct form of \(\tan^{-1}(-x)\)?
A. \(-\tan^{-1}x\)
\(\tan^{-1}x\) भी विषम function है। इसलिए negative input पर output का sign बदलता है। / \(\tan^{-1}x\) is also an odd function. Therefore a negative input changes the sign of the output.
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Which identity is correct for \(\cos^{-1}(-x)\)?
B. \(\pi-\cos^{-1}x\)
\(\cos^{-1}(-x)=\pi-\cos^{-1}x\) क्योंकि output \([0,\pi]\) में रहता है। इसे odd function न समझें। / \(\cos^{-1}(-x)=\pi-\cos^{-1}x\) because the output remains in \([0,\pi]\). Do not treat it as an odd function.
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What is the principal value of \(\tan^{-1}x+\cot^{-1}x\) when (x>0)?
A. \(\frac{\pi}{2}\)
(x>0) के लिए \(\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}\) होता है। \(\cot^{-1}x\) की range convention जरूर देखें। / For (x>0), \(\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}\). Always check the range convention of \(\cot^{-1}x\).
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What is the principal value of \(cosec^{-1}2\)?
A. \(\frac{\pi}{6}\)
\(cosec\frac{\pi}{6}=2\), इसलिए मुख्य मान \(\frac{\pi}{6}\) है। \(cosec^{-1}x\) को \(\sin^{-1}\frac{1}{x}\) से जोड़कर सोचें। / \(cosec\frac{\pi}{6}=2\), so the principal value is \(\frac{\pi}{6}\). Think of \(cosec^{-1}x\) through \(\sin^{-1}\frac{1}{x}\).
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What is the principal value of \(\cos^{-1}\left(\cos\frac{4\pi}{3}\right)\)?
B. \(\frac{2\pi}{3}\)
\(\cos\frac{4\pi}{3}=-\frac{1}{2}\), इसलिए \(\cos^{-1}\left(-\frac{1}{2}\right)=\frac{2\pi}{3})। (\cos^{-1}x\) का उत्तर \([0,\pi]\) में दें। / \(\cos\frac{4\pi}{3}=-\frac{1}{2}\), so \(\cos^{-1}\left(-\frac{1}{2}\right)=\frac{2\pi}{3}\). Give the answer of \(\cos^{-1}x) in ([0,\pi]\).
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What is the correct value of \(\sin(\cos^{-1}x)\) when \(x\in[-1,1]\)?
A. \(\sqrt{1-x^2}\)
यदि \(\theta=\cos^{-1}x\), तो \(\theta\in[0,\pi]\) और \(\sin\theta\ge0\)। इसलिए मान \(\sqrt{1-x^2}\) है। / If \(\theta=\cos^{-1}x\), then \(\theta\in[0,\pi]\) and \(\sin\theta\ge0\). Hence the value is \(\sqrt{1-x^2}\).
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What is the correct value of \(\cos(\sin^{-1}x)\) when \(x\in[-1,1]\)?
A. \(\sqrt{1-x^2}\)
यदि \(\theta=\sin^{-1}x\), तो \(\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]\) और \(\cos\theta\ge0\)। इसलिए positive root लेना चाहिए। / If \(\theta=\sin^{-1}x\), then \(\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]\) and \(\cos\theta\ge0\). So the positive root should be taken.
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What is the correct value of \(\tan(\sin^{-1}x)\)?
A. \(\frac{x}{\sqrt{1-x^2}}\)
मान लें \(\theta=\sin^{-1}x\), तब \(\sin\theta=x\) और \(\cos\theta=\sqrt{1-x^2}\)। इसलिए \(\tan\theta=\frac{x}{\sqrt{1-x^2}}\)। / Let \(\theta=\sin^{-1}x\), then \(\sin\theta=x\) and \(\cos\theta=\sqrt{1-x^2}\). Hence \(\tan\theta=\frac{x}{\sqrt{1-x^2}}\).
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What is the correct value of \(\sin(\tan^{-1}x)\)?
A. \(\frac{x}{\sqrt{1+x^2}}\)
यदि \(\theta=\tan^{-1}x\), तो \(\tan\theta=x\) और hypotenuse \(\sqrt{1+x^2}\) है। इसलिए \(\sin\theta=\frac{x}{\sqrt{1+x^2}}\)। / If \(\theta=\tan^{-1}x\), then \(\tan\theta=x\) and the hypotenuse is \(\sqrt{1+x^2}\). So \(\sin\theta=\frac{x}{\sqrt{1+x^2}}\).
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What is the correct value of \(\cos(\tan^{-1}x)\)?
A. \(\frac{1}{\sqrt{1+x^2}}\)
\(\theta=\tan^{-1}x\) लेने पर adjacent side (1) और hypotenuse \(\sqrt{1+x^2}\) मिलती है। इसलिए \(\cos\theta=\frac{1}{\sqrt{1+x^2}}\)। / Taking \(\theta=\tan^{-1}x\), the adjacent side is (1) and the hypotenuse is \(\sqrt{1+x^2}\). Thus \(\cos\theta=\frac{1}{\sqrt{1+x^2}}\).
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If \(\tan^{-1}x=-\frac{\pi}{3}\), what is the value of \(x\)?
A. -\(\sqrt{3}\)
\(\tan^{-1}x=-\frac{\pi}{3}\) का अर्थ है \(x=\tan\left(-\frac{\pi}{3}\right)\)। इसलिए \(x=-\sqrt{3}\)। / \(\tan^{-1}x=-\frac{\pi}{3}\) means \(x=\tan\left(-\frac{\pi}{3}\right)\). Hence \(x=-\sqrt{3}\).
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What is the correct conclusion about the equation \(\sin^{-1}x+\cos^{-1}x=\pi\)?
B. कोई हल नहींNo solution
हर \(x\in[-1,1]\) के लिए \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\) होता है। इसलिए \(\pi\) वाला समीकरण असंभव है। / For every \(x\in[-1,1]\), \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\). Therefore the equation with \(\pi\) is impossible.
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For \(\sin^{-1}(2x)\) to be defined, in which interval must \(x\) lie?
A. \([-\frac{1}{2},\frac{1}{2}]\)
\(\sin^{-1}(2x)\) के लिए \(-1\le 2x\le1\) होना चाहिए। इससे \(x\in[-\frac{1}{2},\frac{1}{2}]\) मिलता है। / For \(\sin^{-1}(2x)\), we need \(-1\le 2x\le1\). This gives \(x\in[-\frac{1}{2},\frac{1}{2}]\).
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Which is the correct domain interval of \(\cos^{-1}(3x-1)\)?
A. \([0,\frac{2}{3}]\)
\(\cos^{-1}(3x-1)\) के लिए \(-1\le3x-1\le1\) चाहिए। हल करने पर \(0\le x\le\frac{2}{3}\) मिलता है। / For \(\cos^{-1}(3x-1)\), we need \(-1\le3x-1\le1\). Solving gives \(0\le x\le\frac{2}{3}\).
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What is the domain of \(x\) for \(sin^{-1}(x^2-1)\)?
D. ([-2,2])
\(-1\le x^2-1\le1\) से \(0\le x^2\le2\) मिलता है। इसलिए \(x\in[-\sqrt{2},\sqrt{2}]\), जो दिए विकल्पों में नहीं है। / From \(-1\le x^2-1\le1\), we get \(0\le x^2\le2\). Thus \(x\in[-\sqrt{2},\sqrt{2}]\), so none of the first three is exact.
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What is the domain of \(\tan^{-1}(x^2+1)\)?
A. \(\mathbb{R}\)
\(\tan^{-1}u\) हर real \(u\) के लिए परिभाषित होता है। इसलिए \(x^2+1\) के कारण (x) पर कोई प्रतिबंध नहीं है। / \(\tan^{-1}u\) is defined for every real (u). Hence \(x^2+1\) puts no restriction on (x).
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On which set is \(cosec^{-1}x\) defined?
B. (\(-\infty,-1]\cup[1,\infty\))
\(cosec y=\frac{1}{sin y}\), इसलिए \(cosec y\) के values \(|x|\ge1\) होते हैं। domain याद करते समय reciprocal nature देखें। / \(cosec y=\frac{1}{sin y}\), so values of \(cosec y\) satisfy \(|x|\ge1\). Use the reciprocal nature to remember the domain.
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A. \(\frac{\pi}{2}\)
यह सीधे identity \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\) से मिलता है। यहाँ \(x=\frac{\sqrt{3}}{2}\) valid domain में है। / This follows directly from \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\). Here \(x=\frac{\sqrt{3}}{2}\) is in the valid domain.
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What is the principal value of \(\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)\)?
A. -\(\frac{\pi}{3}\)
\(\sin\left(-\frac{\pi}{3}\right)=-\frac{\sqrt{3}}{2}\) और यह principal range में है। \(\sin^{-1}x\) के लिए answer \([-\frac{\pi}{2},\frac{\pi}{2}]\) में रखें। / \(\sin\left(-\frac{\pi}{3}\right)=-\frac{\sqrt{3}}{2}\) and it lies in the principal range. Keep the answer of \(\sin^{-1}x) in ([-\frac{\pi}{2},\frac{\pi}{2}]\).
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What is the principal value of \(\cos^{-1}(0)\)?
A. \(\frac{\pi}{2}\)
\(\cos\frac{\pi}{2}=0\) और \(\frac{\pi}{2}\in[0,\pi]\)। special values को principal range के साथ याद करें। / \(\cos\frac{\pi}{2}=0\) and \(\frac{\pi}{2}\in[0,\pi]\). Remember special values along with the principal range.
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What is the principal value of \(\tan^{-1}0\)?
A. (0)
\(\tan0=0\), इसलिए \(\tan^{-1}0=0\)। zero value वाले प्रश्नों में भी principal range जांचें। / \(\tan0=0\), so \(\tan^{-1}0=0\). Even in zero-value questions, check the principal range.
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Which statement is correct: what does \(\sin^{-1}x\) mean?
B. यह \(\sin x\) का inverse function हैIt is the inverse function of \(\sin x\)
\(\sin^{-1}x\) inverse sine है, reciprocal नहीं। \(\sin^{-1}x\) और (\(\sin x\)^{-1}) में अंतर रखें। / \(\sin^{-1}x\) is inverse sine, not reciprocal. Keep the difference between \(\sin^{-1}x\) and (\(\sin x\)^{-1}).
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Which statement is correct about \(\sin^{-1}2\)?
C. यह real numbers में परिभाषित नहीं हैIt is not defined in real numbers
\(\sin^{-1}x\) real में तभी defined है जब \(x\in[-1,1]\)। (2) इस interval में नहीं है। / \(\sin^{-1}x\) is real-defined only when \(x\in[-1,1]\). The number (2) is not in this interval.
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What is the correct conclusion about \(\cos^{-1}\left(\tfrac{3}{2}\right)\)?
A. वास्तविक संख्याओं में परिभाषित नहींNot defined in real numbers
\(\cos^{-1}x\) (arccos) केवल तब वास्तविक मान देता है जब \(x\in[-1,1]\)। इसकी वास्तविक छवि मूल श्रेणी \([0,\pi]\) में होती है। यहाँ \(\tfrac{3}{2}=1.5>1\), इसलिए \(\cos^{-1}\left(\tfrac{3}{2}\right)\) का वास्तविक मान नहीं है — यह वास्तविक संख्या नहीं दे सकता। विकल्प \(\dfrac{\pi}{3}\) भ्रम पैदा कर सकता है क्योंकि \(\cos\dfrac{\pi}{3}=\tfrac{1}{2}\), न कि \(\tfrac{3}{2}\)। परीक्षा-सुझाव: किसी भी inverse-trig प्रश्न में पहले फंक्शन का domain चेक करें (उदा. arccos के लिए \([-1,1]\))। / The function \(\cos^{-1}x\) (arccos) has real values only for \(x\in[-1,1]\); its principal range for real inputs is \([0,\pi]\). Since \(\tfrac{3}{2}=1.5>1\), \(\cos^{-1}\left(\tfrac{3}{2}\right)\) has no real value. Option \(\dfrac{\pi}{3}\) is a tempting distractor but incorrect because \(\cos\dfrac{\pi}{3}=\tfrac{1}{2}\), not \(\tfrac{3}{2}\). Exam tip: always check the domain of inverse trigonometric functions before attempting to find a value.
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If \(\theta=\sin^{-1}\left(\frac{3}{5}\right)\), what is the value of \(\cos\theta\)?
A. \(\frac{4}{5}\)
\(\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]\) है, इसलिए \(\cos\theta\) positive होगा। (3,4,5) triangle से \(\cos\theta=\frac{4}{5}\)। / \(\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]\), so \(\cos\theta\) is positive. From the (3,4,5) triangle, \(\cos\theta=\frac{4}{5}\).
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If \(\theta=\cos^{-1}\left(-\dfrac{3}{5}\right)\), what is the value of \(\sin\theta\)?
A. \(\dfrac{4}{5}\)
\(\cos\theta=-\dfrac{3}{5}\) होने पर पहचान \(\sin^2\theta+\cos^2\theta=1\) से \(\sin^2\theta=1-\dfrac{9}{25}=\dfrac{16}{25}\), अतः \(\sin\theta=\pm\dfrac{4}{5}\). परन्तु \(\theta=\cos^{-1}(\cdot)\) की principal range \([0,\pi]\) है और \(\cos\theta<0\) होने पर \(\theta\) द्वितीय क्वाड्रेंट में होगा जहाँ \(\sin\theta\) धनात्मक होता है। इसलिए \(\sin\theta=\dfrac{4}{5}\)। सामान्य भूल: केवल मूल फलन से संकेत का ध्यान रखें—\(-\dfrac{4}{5}\) वही distractor है जो संकेत उलट देने पर चुन लिया जाता है। परीक्षा सुझाव: arccos की range याद रखें और Pythagoras पहचान से मान निकालें। / From \(\cos\theta=-\dfrac{3}{5}\) and \(\sin^2\theta+\cos^2\theta=1\) we get \(\sin^2\theta=1-\dfrac{9}{25}=\dfrac{16}{25}\), so \(\sin\theta=\pm\dfrac{4}{5}\). Since \(\theta=\cos^{-1}(\ldots)\) lies in \([0,\pi]\) and \(\cos\theta<0\) places \(\theta\) in the second quadrant where sine is positive, \(\sin\theta=\dfrac{4}{5}\). Closest trap is \(-\dfrac{4}{5}\) from ignoring the quadrant sign. Exam tip: always use the principal range of inverse cosine and the identity \(\sin^2+\cos^2=1\).
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If \(\theta=\tan^{-1}\left(\frac{5}{12}\right)\), what is \(\sin\theta\)?
A. \(\frac{5}{13}\)
\(\tan\theta=\frac{5}{12}\) से opposite (5), adjacent (12), hypotenuse (13) है। इसलिए \(\sin\theta=\frac{5}{13}\)। / From \(\tan\theta=\frac{5}{12}\), opposite is (5), adjacent is (12), and hypotenuse is (13). Hence \(\sin\theta=\frac{5}{13}\).
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When is \(\cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x\) true?
A. जब \(x\in[-1,1]\)When \(x\in[-1,1]\)
यह identity \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\) से मिलती है। इसका domain \(x\in[-1,1]\) है। / This identity follows from \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\). Its domain is \(x\in[-1,1]\).
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What type of function is \(\cos^{-1}x\) on its domain ([-1,1])?
B. decreasing
\(\cos^{-1}x\) ([-1,1]) पर decreasing है। (x) बढ़ने पर principal angle घटता है। / \(\cos^{-1}x\) is decreasing on ([-1,1]). As (x) increases, the principal angle decreases.
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If (0<a<1), which statement about \(\sin^{-1}a\) and \(\cos^{-1}a\) is always true?
A. उनका योग \(\frac{\pi}{2}\) हैTheir sum is \(\frac{\pi}{2}\)
हर \(a\in[-1,1]\) के लिए \(\sin^{-1}a+\cos^{-1}a=\frac{\pi}{2}\)। बराबर केवल \(a=\frac{1}{\sqrt{2}}\) पर होते हैं। / For every \(a\in[-1,1]\), \(\sin^{-1}a+\cos^{-1}a=\frac{\pi}{2}\). They are equal only at \(a=\frac{1}{\sqrt{2}}\).
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What is the value of \(\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)+\sin^{-1}\left(\frac{1}{2}\right)\)?
A. \(\frac{\pi}{3}\)
\(\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}\) और \(\sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}\)। इसलिए योग \(\frac{\pi}{3}\) है। / Here \(\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}\) and \(\sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}\). Hence the sum is \(\frac{\pi}{3}\).
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A. 0
\(\sin^{-1}x\) का प्रमुख मान क्षेत्र \([-\tfrac{\pi}{2},\tfrac{\pi}{2}])\) है और \(\cos^{-1}x\) का प्रमुख मान क्षेत्र \([0,\pi]\) है। \(\sin\tfrac{\pi}{4}=\cos\tfrac{\pi}{4}=\tfrac{\sqrt{2}}{2}\) होने पर दोनों इनवर्स फ़ंक्शन अपने प्रमुख मान के कारण \(\tfrac{\pi}{4}\) लेते हैं। अतः अंतर \(\tfrac{\pi}{4}-\tfrac{\pi}{4}=0\)। नज़दीकी विचलन \(\tfrac{\pi}{4}\) गलत है क्योंकि वह अंतर नहीं बल्कि घटक का मान है। परीक्षा टिप: इनवर्स त्रिकोणमितीय फलनों के प्रमुख मान (principal values) याद रखें—यही फर्क नतीजा तय करता है। / The principal value range of \(\sin^{-1}x\) is \([-\tfrac{\pi}{2},\tfrac{\pi}{2}])\) and for \(\cos^{-1}x\) it is \([0,\pi]\). Since \(\sin\tfrac{\pi}{4}=\cos\tfrac{\pi}{4}=\tfrac{\sqrt{2}}{2}\), both inverse values equal \(\tfrac{\pi}{4}\). Therefore the difference is \(\tfrac{\pi}{4}-\tfrac{\pi}{4}=0\). The nearest distractor \(\tfrac{\pi}{4}\) is the value of each inverse separately, not their difference. Exam tip: always check the principal value intervals for inverse trig functions before subtracting.
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What is the principal value of \(\sin^{-1}\left(\sin\left(-\frac{2\pi}{3}\right)\right)\)?
B. -\(\frac{\pi}{3}\)
\(\sin\left(-\frac{2\pi}{3}\right)=-\frac{\sqrt{3}}{2}\), इसलिए मुख्य मान \(-\frac{\pi}{3}\) है। \(\sin^{-1}x\) का उत्तर हमेशा \([-\frac{\pi}{2},\frac{\pi}{2}]\) में रखें। / \(\sin\left(-\frac{2\pi}{3}\right)=-\frac{\sqrt{3}}{2}\), so the principal value is -\(\frac{\pi}{3}\). Keep the answer of \(\sin^{-1}x\) in \(-[\frac{\pi}{2},\frac{\pi}{2}]\).
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What is the principal value of \(\cos^{-1}\left(\cos\left(-\frac{\pi}{3}\right)\right)\)?
B. \(\frac{\pi}{3}\)
\(\cos\left(-\frac{\pi}{3}\right)=\frac{1}{2}\) और \(\cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3})। (\cos^{-1}x\) में उत्तर ऋणात्मक नहीं होता। / \(\cos\left(-\frac{\pi}{3}\right)=\frac{1}{2}\) and \(\cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}\). The answer of \(\cos^{-1}x\\) is not negative.
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What is the principal value of \(\tan^{-1}\left(\tan\left(-\frac{5\pi}{6}\right)\right)\)?
B. \(\frac{\pi}{6}\)
\(\tan\left(-\frac{5\pi}{6}\right)=\frac{1}{\sqrt{3}}\), इसलिए principal value \(\frac{\pi}{6}\) है। \(\tan^{-1}x\) का उत्तर (\(-\frac{\pi}{2},\frac{\pi}{2}\)) में दें। / \(\tan\left(-\frac{5\pi}{6}\right)=\frac{1}{\sqrt{3}}\), so the principal value is \(\frac{\pi}{6}\). Give the answer of \(\tan^{-1}x\) in (\(-\frac{\pi}{2},\frac{\pi}{2}\).
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Under which condition is \(\sin^{-1}x=\tan^{-1}\left(\dfrac{x}{\sqrt{1-x^2}}\right)\) correct?
A. \(-1<x<1\)
यदि \(\theta=\sin^{-1}x\) लिया जाए तो \(\theta\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]\) और \(\sin\theta=x\). उसी मुख्य शर्त में \(\cos\theta=\sqrt{1-x^2}\) (धनात्मक शून्य के साथ) और इसलिए \(\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{x}{\sqrt{1-x^2}}\). इसलिए \(\tan^{-1}\big(\dfrac{x}{\sqrt{1-x^2}}\big)=\theta=\sin^{-1}x\) परन्तु केवल तब जब \(\sqrt{1-x^2}\neq0\) और \(\sin^{-1}x\) परिभाषित हो, अर्थात् \(-1<x<1\). यदि \(x=\pm1\) तो हरास्पद भाग में हर (denominator) शून्य हो जाता है और अभिव्यक्ति अव्याख्येय है (सीमाएँ ±\(\tfrac{\pi}{2}\) देती हैं किंतु समानता वास्तविक रूप से परिभाषित नहीं)। \(|x|>1\) पर \(\sin^{-1}x\) ही परिभाषित नहीं है। परीक्षा टिप: inverse त्रिकोणमितीय पहचानें लिखते समय principal-value रेंज और sqrt के संकेत पर ध्यान दें — cos का मुख्य मान धनात्मक लिया जाता है। / Let \(\theta=\sin^{-1}x\) so that \(\theta\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]\) and \(\sin\theta=x\). In this principal range \(\cos\theta=\sqrt{1-x^2}\) (nonnegative), hence \(\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{x}{\sqrt{1-x^2}}\). Thus \(\tan^{-1}\big(\dfrac{x}{\sqrt{1-x^2}}\big)=\theta=\sin^{-1}x\) provided \(\sqrt{1-x^2}\neq0\) and \(\sin^{-1}x\) is defined — i.e. \(-1<x<1\). At \(x=\pm1\) the denominator is zero (the expression is not defined even though the limit gives ±\(\tfrac{\pi}{2}\)), and for \(|x|>1\) the left side \(\sin^{-1}x\) is not defined. Exam tip: always check domains and principal-value ranges when equating inverse trig functions; note the sign of the square root in the principal interval.
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Which statement is correct for \(\tan^{-1}x=\sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)\)?
A. यह सभी \(x\in\mathbb{R}\) के लिए सत्य हैIt is true for all \(x\in\mathbb{R}\)
\(\frac{x}{\sqrt{1+x^2}}\) हमेशा ([-1,1]) में रहता है। इसलिए identity सभी real (x) के लिए मान्य है। / \(\frac{x}{\sqrt{1+x^2}}\) always lies in ([-1,1]). Therefore the identity is valid for all real (x).
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What is the value of \(\cos\left(\sin^{-1}\left(-\frac{5}{13}\right)\right)\)?
A. \(\frac{12}{13}\)
\(\sin^{-1}\) का principal angle चौथे quadrant में हो सकता है, जहां \(\cos\theta\) positive होता है। (5,12,13) triangle से उत्तर \(\frac{12}{13}\) है। / The principal angle of \(\sin^{-1}\) may lie in the fourth quadrant, where \(\cos\theta\) is positive. From the (5,12,13) triangle, the answer is \(\frac{12}{13}\).
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What is the value of \(\sin\left(\cos^{-1}\left(\frac{7}{25}\right)\right)\)?
A. \(\frac{24}{25}\)
\(\theta=\cos^{-1}\left(\frac{7}{25}\right)\) लेने पर \(\theta\in[0,\pi]\) और \(sin\theta\ge0\)। इसलिए (7,24,25) triangle से \(\sin\theta=\frac{24}{25}\)। / Taking \(\theta=\cos^{-1}\left(\frac{7}{25}\right)\), we have \(\theta\in[0,\pi]\) and \(sin\theta\ge0\). Hence from the (7,24,25) triangle, \(\sin\theta=\frac{24}{25}\).
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Which condition is correct for \(sec^{-1}(2x+1)\) to be defined?
A. \(|2x+1|\ge1\)
\(\sec^{-1}u\) के लिए \(|u|\ge1\) होना चाहिए। यहाँ (u=2x+1), इसलिए condition \(|2x+1|\ge1\) है। / For \(\sec^{-1}u\), we need \(|u|\ge1\). Here (u=2x+1), so the condition is \(|2x+1|\ge1\).
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What is the domain of \(cosec^{-1}(x-2)\)?
A. (\(-\infty,1]\cup[3,\infty\))
\(cosec^{-1}(x-2)\) के लिए \(|x-2|\ge1\) चाहिए। इससे \(x\le1\) या \(x\ge3\) मिलता है। / For \(cosec^{-1}(x-2)\), we need \(|x-2|\ge1\). This gives \(x\le1\) or \(x\ge3\).
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What is the value of \(\cos^{-1}\left(\frac{1}{3}\right)+\cos^{-1}\left(-\frac{1}{3}\right)\)?
A. \(\pi\)
(\cos^{-1}(-x)=\pi-\cos^{-1}x) होता है। इसलिए दोनों terms का योग \(\pi\) है। / We have (\cos^{-1}(-x)=\pi-\cos^{-1}x). Therefore the sum of the two terms is \(\pi\).
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What is the correct value of \(\tan\left(\cos^{-1}x\right)\), when \(0<x<1\)?
A. \(\frac{\sqrt{1-x^2}}{x}\)
यदि \(\theta=\cos^{-1}x\), तो \(\cos\theta=x\) और \(\sin\theta=\sqrt{1-x^2}\)। इसलिए \(\tan\theta=\frac{\sqrt{1-x^2}}{x}\)। / If \(\theta=\cos^{-1}x\), then \(\cos\theta=x\) and \(\sin\theta=\sqrt{1-x^2}\). Hence \(\tan\theta=\frac{\sqrt{1-x^2}}{x}\).
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