Topic Quiz • 24 questions • 45 seconds per question.
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What is the value of \(\cos(\cos^{-1}(-\frac{3}{4}))\)?
A. \(-\frac{3}{4}\)
यदि \(x\in\left[-1,1\right]\), तो \(\cos(\cos^{-1}x)=x\)। यहां \(x=-\frac{3}{4}\) प्रांत में है। / If \(x\in\left[-1,1\right]\), then \(\cos(\cos^{-1}x)=x\). Here \(x=-\frac{3}{4}\) lies in the domain.
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If \(\theta=\cos^{-1}\left(\frac{8}{17}\right)\), what is the value of \(\sin\theta\)?
B. \(\frac{15}{17}\)
\(\theta\in\left[0,\pi\right]\) और \(\cos\theta=\frac{8}{17}\), इसलिए \(\theta\) प्रथम चतुर्थांश में है और \(\sin\theta=\frac{15}{17}\)। त्रिभुज विधि तेज है। / Here \(\theta\in\left[0,\pi\right]\) and \(\cos\theta=\frac{8}{17}\), so \(\theta\) is in the first quadrant and \(\sin\theta=\frac{15}{17}\). The triangle method is fast.
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If \(\theta=\tan^{-1}\left(\frac{7}{24}\right)\), what is the value of \(\sin\theta\)?
C. \(\frac{7}{25}\)
\(\tan\theta=\frac{7}{24}\) से कर्ण \(\sqrt{7^2+24^2}=25\), इसलिए \(\sin\theta=\frac{7}{25}\)। अनुपात से त्रिभुज बनाएं। / From \(\tan\theta=\frac{7}{24}\), the hypotenuse is \(\sqrt{7^2+24^2}=25\), so \(\sin\theta=\frac{7}{25}\). Build a triangle from the ratio.
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What is the value of \(\tan\left(\sin^{-1}\frac{4}{5}\right)\)?
B. \(\frac{4}{3}\)
मान लें \(\theta=\sin^{-1}\frac{4}{5}\), तब \(\sin\theta=\frac{4}{5}\) और \(\cos\theta=\frac{3}{5}\)। इसलिए \(\tan\theta=\frac{4}{3}\)। / Let \(\theta=\sin^{-1}\frac{4}{5}\), then \(\sin\theta=\frac{4}{5}\) and \(\cos\theta=\frac{3}{5}\). Therefore \(\tan\theta=\frac{4}{3}\).
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What is the value of \(\sin\left(\tan^{-1}\frac{12}{5}\right)\)?
B. \(\frac{12}{13}\)
यदि \(\theta=\tan^{-1}\frac{12}{5}\), तो \(\tan\theta=\frac{12}{5}\) और कर्ण (13) है। इसलिए \(\sin\theta=\frac{12}{13}\)। / If \(\theta=\tan^{-1}\frac{12}{5}\), then \(\tan\theta=\frac{12}{5}\) and the hypotenuse is (13). Hence \(\sin\theta=\frac{12}{13}\).
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What is the value of \(\cos\left(\tan^{-1}\frac{15}{8}\right)\)?
B. \(\frac{8}{17}\)
यदि \(\tan\theta=\frac{15}{8}\), तो कर्ण (17) है। मुख्य परिसर में \(\theta\) प्रथम चतुर्थांश में है, इसलिए \(\cos\theta=\frac{8}{17}\)। / If \(\tan\theta=\frac{15}{8}\), then the hypotenuse is (17). In the principal range \(\theta\) is in the first quadrant, so \(\cos\theta=\frac{8}{17}\).
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What is the value of \(\sin^{-1}\left(\frac{2}{3}\right)+\cos^{-1}\left(\frac{2}{3}\right)\)?
A. \(\frac{\pi}{2}\)
हर \(x\in\left[-1,1\right]\) के लिए \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\)। यहां \(x=\frac{2}{3}\) मान्य है। / For every \(x\in\left[-1,1\right]\), \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\). Here \(x=\frac{2}{3}\) is valid.
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If \(\theta=\cot^{-1}\left(\frac{3}{4}\right)\), what is the value of \(\cos\theta\)?
A. \(\frac{3}{5}\)
\(\cot\theta=\frac{3}{4}\) से आधार (3), लम्ब (4) और कर्ण (5) होगा। इसलिए \(\cos\theta=\frac{3}{5}\)। / From \(\cot\theta=\frac{3}{4}\), the base is (3), perpendicular is (4), and hypotenuse is (5). Hence \(\cos\theta=\frac{3}{5}\).
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What is the value of \(\tan^{-1}\left(\frac{1}{2}\right)+\tan^{-1}\left(\frac{1}{3}\right)\)?
B. \(\frac{\pi}{4}\)
सूत्र से योग \(\tan^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\right)=\tan^{-1}1=\frac{\pi}{4}\) है। \(\tan^{-1}\) जोड़ते समय \(ab<1\) जांचें। / By the formula, the sum is \(\tan^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\right)=\tan^{-1}1=\frac{\pi}{4}\). While adding \(\tan^{-1}\) terms, check \(ab<1\).
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What is the value of \(\tan\left(\cos^{-1}\frac{5}{13}\right)\)?
B. \(\frac{12}{5}\)
मान लें \(\theta=\cos^{-1}\frac{5}{13}\), तब \(\cos\theta=\frac{5}{13}\) और \(\sin\theta=\frac{12}{13}\)। इसलिए \(\tan\theta=\frac{12}{5}\)। / Let \(\theta=\cos^{-1}\frac{5}{13}\), then \(\cos\theta=\frac{5}{13}\) and \(\sin\theta=\frac{12}{13}\). Therefore \(\tan\theta=\frac{12}{5}\).
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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 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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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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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 \(\theta=\tan^{-1}\left(\frac{5}{12}\right)\), 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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What type of function is f(x)=\cos^{-1}x on its domain \([-1,1]\)?
B. घटतीdecreasing
f(x)=\cos^{-1}x के लिए \\( \frac{d}{dx}\cos^{-1}x=-\frac{1}{\sqrt{1-x^2}}<0 \\) for x\in(-1,1). अवकलज नकारात्मक होने से फलन घटता है। सीमाओं पर भी \(\cos^{-1}(-1)=\pi\) और \(\cos^{-1}(1)=0\) आता है, अतः पूरे बंद अंतराल \([-1,1]\) पर यह घटता क्रम में है। विकल्प 'बढ़ती' गलत है क्योंकि अवकलज का चिन्ह नकारात्मक है; 'स्थिर' व 'आवर्त' भी उपयुक्त नहीं हैं। परीक्षा सुझाव: तुरंत monotonicity जाँचने के लिए अवकलज ज्ञात करें या याद रखें कि \(\cos\theta\) मुख्य-श्रेणी \([0,\pi]\) पर घटता है, अतः inverse घटेगा। / For f(x)=\cos^{-1}x, \\( \frac{d}{dx}\cos^{-1}x=-\frac{1}{\sqrt{1-x^2}}<0 \\) for x in (-1,1). A negative derivative implies the function is decreasing. At endpoints, \(\cos^{-1}(-1)=\pi\) and \(\cos^{-1}(1)=0\), so the function decreases over the whole closed interval [-1,1]. 'Increasing' is wrong because the derivative is negative; 'constant' and 'periodic' are also not applicable. Exam tip: test monotonicity by checking the sign of the derivative or recall that \(\cos\theta\) is decreasing on [0,\pi], so its inverse is decreasing on [-1,1].
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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 \(\sin\left(\cos^{-1}\left(\frac{7}{25}\right)\right)\)?
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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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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