Level 31 • 10/50 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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