Level 71 • 50/50 questions • 35 seconds per question.
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If \(\sin x-\cos x=\frac{1}{2}\), what is the value of \(\sin x\cos x\)?
A. \(\frac{3}{8}\)
(\(\sin x-\cos x\)2=1-2\sin x\cos x) लगाएँ। इससे \(\sin x\cos x=\frac{3}{8}\) मिलता है। / Use (\(\sin x-\cos x\)2=1-2\sin x\cos x). This gives \(\sin x\cos x=\frac{3}{8}\).
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If \(\sin x+\cos x=\frac{6}{5}\), what is the value of the square of \(\sin x-\cos x\)?
C. \(\frac{14}{25}\)
पहचान (\(\sin x+\cos x\)2+\(\sin x-\cos x\)2=2) का उपयोग करें। इसलिए मान \(2-\frac{36}{25}=\frac{14}{25}\) है। / Use the identity (\(\sin x+\cos x\)2+\(\sin x-\cos x\)2=2). Hence the value is \(2-\frac{36}{25}=\frac{14}{25}\).
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If \(\tan x+\cot x=5\), what is the value of \(\tan^2 x+\cot^2 x\)?
B. (23)
(\(\tan x+\cot x\)2=\tan-2 x+\cot-2 x+2) होता है। इसलिए मान (25-2=23) है। / (\(\tan x+\cot x\)2=\tan-2 x+\cot-2 x+2). Therefore, the value is (25-2=23).
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If \(\sec x-\tan x=\frac{1}{4}\), what is the value of \(\sec x+\tan x\)?
C. (4)
क्योंकि (\(\sec x-\tan x\)\(\sec x+\tan x\)=1) होता है। इसलिए दूसरा गुणनखंड (4) होगा। / Since (\(\sec x-\tan x\)\(\sec x+\tan x\)=1). Hence the other factor is (4).
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If \(\cosec x+\cot x=6\), what is the value of \(\cosec x-\cot x\)?
A. \(\frac{1}{6}\)
(\(\cosec x+\cot x\)\(\cosec x-\cot x\)=1) होता है। इसलिए आवश्यक मान \(\frac{1}{6}\) है। / The identity is (\(\cosec x+\cot x\)\(\cosec x-\cot x\)=1). Therefore, the required value is \(\frac{1}{6}\).
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What is \(\frac{\sin x}{1-\cos x}\) equal to?
B. \(\cot \frac{x}{2}\)
अर्ध-कोण पहचान से \(\frac{\sin x}{1-\cos x}=\cot \frac{x}{2}\) होता है। ऐसे रूपों में हर देखकर पहचान करें। / By the half-angle identity, \(\frac{\sin x}{1-\cos x}=\cot \frac{x}{2}\). In such forms, identify the denominator carefully.
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What is \(\frac{1+\cos x}{\sin x}\) equal to?
C. \(\cot \frac{x}{2}\)
मानक अर्ध-कोण रूप \(\cot \frac{x}{2}=\frac{1+\cos x}{\sin x}\) है। \(\tan \frac{x}{2}\) और \(\cot \frac{x}{2}\) के रूप अलग रखें। / The standard half-angle form is \(\cot \frac{x}{2}=\frac{1+\cos x}{\sin x}\). Keep the forms of \(\tan \frac{x}{2}\) and \(\cot \frac{x}{2}\) separate.
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What is the maximum value of the function \(4-3\sin x\)?
C. (7)
\(\sin x\) का न्यूनतम मान (-1) है। इसलिए अधिकतम मान (4-3(-1)=7) होगा। / The minimum value of \(\sin x\) is (-1). Hence the maximum value is (4-3(-1)=7).
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What is the minimum value of the function \(2+5\cos x\)?
A. -(3)
\(\cos x\) का न्यूनतम मान (-1) होता है। इसलिए न्यूनतम मान (2+5(-1)=-3) है। / The minimum value of \(\cos x\) is (-1). Therefore, the minimum value is (2+5(-1)=-3).
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What is the range of the function \(3\sin 2x-1\)?
B. ([-4,2])
\(\sin 2x\) का परिसर ([-1,1]) है। (3) से गुणा और (-1) जोड़ने पर परिसर ([-4,2]) मिलता है। / The range of \(\sin 2x\) is ([-1,1]). Multiplying by (3) and adding (-1) gives ([-4,2]).
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What is the period of the function \(2\cos 3x+4\)?
B. \(\frac{2\pi}{3}\)
ऊर्ध्व बदलाव और आयाम काल नहीं बदलते। \(\cos 3x\) का काल \(\frac{2\pi}{3}\) है। / Vertical shift and amplitude do not change the period. The period of \(\cos 3x\) is \(\frac{2\pi}{3}\).
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What is the fundamental period of the function (\tan(2x))?
A. \(\frac{\pi}{2}\)
\(\tan kx\) का काल \(\frac{\pi}{k}\) होता है। यहाँ (k=2), इसलिए मूल काल \(\frac{\pi}{2}\) है। / The period of \(\tan kx\) is \(\frac{\pi}{k}\). Here (k=2), so the fundamental period is \(\frac{\pi}{2}\).
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If (x) is in the second quadrant and \(\cos x=-\frac{3}{5}\), what is the value of \(\tan x\)?
B. -\(\frac{4}{3}\)
दूसरे चतुर्थांश में \(\sin x\) धनात्मक और \(\cos x\) ऋणात्मक होता है। \(\sin x=\frac{4}{5}\), इसलिए \(\tan x=-\frac{4}{3}\) है। / In the second quadrant, \(\sin x\) is positive and \(\cos x\) is negative. Since \(\sin x=\frac{4}{5}\), \(\tan x=-\frac{4}{3}\).
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If (x) is in the third quadrant and \(\sin x=-\frac{5}{13}\), what is the value of \(\sec x\)?
B. -\(\frac{13}{12}\)
तीसरे चतुर्थांश में \(\cos x\) ऋणात्मक होता है। \(\cos x=-\frac{12}{13}\), इसलिए \(\sec x=-\frac{13}{12}\) है। / In the third quadrant, \(\cos x\) is negative. Since \(\cos x=-\frac{12}{13}\), \(\sec x=-\frac{13}{12}\).
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If (x) is in the fourth quadrant and \(\tan x=-\frac{24}{7}\), what is the value of \(\cos x\)?
A. \(\frac{7}{25}\)
चौथे चतुर्थांश में \(\cos x\) धनात्मक और \(\sin x\) ऋणात्मक होता है। (7,24,25) त्रिक से \(\cos x=\frac{7}{25}\) है। / In the fourth quadrant, \(\cos x\) is positive and \(\sin x\) is negative. From the (7,24,25) triple, \(\cos x=\frac{7}{25}\).
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If (x) is in the second quadrant and \(\sec x=-\frac{17}{8}\), what is the value of \(\sin x\)?
C. \(\frac{15}{17}\)
\(\cos x=-\frac{8}{17}\) होगा। दूसरे चतुर्थांश में \(\sin x\) धनात्मक है, इसलिए \(\sin x=\frac{15}{17}\) है। / \(\cos x=-\frac{8}{17}\). In the second quadrant, \(\sin x\) is positive, so \(\sin x=\frac{15}{17}\).
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What is (\sin\(2\pi-x\)) equal to?
B. -\(\sin x\)
\(2\pi-x\) चौथे चतुर्थांश से संबंधित है। वहाँ \(\sin x\) ऋणात्मक होता है, इसलिए (\sin\(2\pi-x\)=-\sin x)। / \(2\pi-x\) is related to the fourth quadrant. There \(\sin x\) is negative, so (\sin\(2\pi-x\)=-\sin x).
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What is (\cos\(2\pi-x\)) equal to?
C. \(\cos x\)
\(2\pi-x\) चौथे चतुर्थांश में आता है और \(\cos x\) धनात्मक रहता है। इसलिए (\cos\(2\pi-x\)=\cos x)। / \(2\pi-x\) lies in the fourth quadrant and \(\cos x\) remains positive. Hence (\cos\(2\pi-x\)=\cos x).
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What is (\tan\(2\pi-x\)) equal to?
B. -\(\tan x\)
चौथे चतुर्थांश में \(\tan x\) ऋणात्मक होता है। इसलिए (\tan\(2\pi-x\)=-\tan x)। / In the fourth quadrant, \(\tan x\) is negative. Therefore, (\tan\(2\pi-x\)=-\tan x).
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What is (\sin\(\frac{3\pi}{2}+x\)) equal to?
A. -\(\cos x\)
\(\frac{3\pi}{2}+x\) रूप में \(\sin\) बदलकर \(\cos\) होता है और चिन्ह ऋणात्मक होता है। इसलिए उत्तर \(-\cos x\) है। / In the form \(\frac{3\pi}{2}+x\), \(\sin\) changes to \(\cos\) with a negative sign. Hence the answer is \(-\cos x\).
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What is (\cos\(\frac{3\pi}{2}-x\)) equal to?
C. -\(\sin x\)
\(\frac{3\pi}{2}-x\) तीसरे चतुर्थांश से जुड़ा है। \(\cos\) बदलकर \(\sin\) होता है और चिन्ह ऋणात्मक रहता है। / \(\frac{3\pi}{2}-x\) is related to the third quadrant. \(\cos\) changes to \(\sin\) with a negative sign.
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What is (\tan\(\frac{3\pi}{2}+x\)) equal to?
C. -\(\cot x\)
\(\frac{3\pi}{2}+x\) पर \(\tan\) बदलकर \(\cot\) होता है और चिन्ह ऋणात्मक है। इसलिए (\tan\(\frac{3\pi}{2}+x\)=-\cot x)। / At \(\frac{3\pi}{2}+x\), \(\tan\) changes to \(\cot\) with a negative sign. Hence (\tan\(\frac{3\pi}{2}+x\)=-\cot x).
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What is the simplified value of (\frac{\sin\(\pi-x\)}{\cos\(\pi+x\)})?
A. -\(\tan x\)
(\sin\(\pi-x\)=\sin x) और (\cos\(\pi+x\)=-\cos x) होता है। इसलिए भिन्न \(-\tan x\) है। / (\sin\(\pi-x\)=\sin x) and (\cos\(\pi+x\)=-\cos x). Therefore, the fraction is \(-\tan x\).
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What is the simplified value of (\frac{\cos\(2\pi-x\)}{\sin\(\pi+x\)})?
B. -\(\cot x\)
(\cos\(2\pi-x\)=\cos x) और (\sin\(\pi+x\)=-\sin x) है। इसलिए मान \(-\cot x\) है। / (\cos\(2\pi-x\)=\cos x) and (\sin\(\pi+x\)=-\sin x). Hence the value is \(-\cot x\).
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How can \(\sec^2 x+\cosec^2 x\) be written in terms of \(\tan x\) and \(\cot x\)?
B. \(2+\tan^2 x+\cot^2 x\)
\(\sec^2 x=1+\tan^2 x\) और \(\cosec^2 x=1+\cot^2 x\) लगाएँ। योग \(2+\tan^2 x+\cot^2 x\) होगा। / Use \(\sec^2 x=1+\tan^2 x\) and \(\cosec^2 x=1+\cot^2 x\). The sum becomes \(2+\tan^2 x+\cot^2 x\).
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What is the simplified value of \(\frac{\tan x+\cot x}{\sec x\cosec x}\)?
A. (1)
\(\tan x+\cot x=\frac{1}{\sin x\cos x}\) और \(\sec x\cosec x=\frac{1}{\sin x\cos x}\) होता है। इसलिए अनुपात (1) है। / \(\tan x+\cot x=\frac{1}{\sin x\cos x}\) and \(\sec x\cosec x=\frac{1}{\sin x\cos x}\). Hence the ratio is (1).
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What is the simplified value of \(\frac{\sec x}{\tan x}\)?
B. \(\cosec x\)
\(\sec x=\frac{1}{\cos x}\) और \(\tan x=\frac{\sin x}{\cos x}\) रखें। अनुपात \(\frac{1}{\sin x}=\cosec x\) बनता है। / Put \(\sec x=\frac{1}{\cos x}\) and \(\tan x=\frac{\sin x}{\cos x}\). The ratio becomes \(\frac{1}{\sin x}=\cosec x\).
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What is the simplified value of \(\frac{\cosec x}{\cot x}\)?
B. \(\sec x\)
\(\cosec x=\frac{1}{\sin x}\) और \(\cot x=\frac{\cos x}{\sin x}\) रखें। अनुपात \(\frac{1}{\cos x}=\sec x\) होगा। / Put \(\cosec x=\frac{1}{\sin x}\) and \(\cot x=\frac{\cos x}{\sin x}\). The ratio becomes \(\frac{1}{\cos x}=\sec x\).
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What is the simplified value of (\sin-2 x\(1+\cot^2 x\))?
B. (1)
क्योंकि \(1+\cot^2 x=\cosec^2 x\)। इसलिए \(\sin^2 x\cosec^2 x=1\) होगा। / Since \(1+\cot^2 x=\cosec^2 x\). Therefore, \(\sin^2 x\cosec^2 x=1\).
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What is the simplified value of (\cos-2 x\(1+\tan^2 x\))?
C. (1)
क्योंकि \(1+\tan^2 x=\sec^2 x\)। इसलिए \(\cos^2 x\sec^2 x=1\) है। / Since \(1+\tan^2 x=\sec^2 x\). Therefore, \(\cos^2 x\sec^2 x=1\).
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If \(\tan x=\frac{1}{2}\), what is the value of \(\frac{1-\tan^2 x}{1+\tan^2 x}\)?
B. \(\frac{3}{5}\)
\(\tan^2 x=\frac{1}{4}\) रखें। तब मान \(\frac{1-\frac{1}{4}}{1+\frac{1}{4}}=\frac{3}{5}\) है। / Put \(\tan^2 x=\frac{1}{4}\). Then the value is \(\frac{1-\frac{1}{4}}{1+\frac{1}{4}}=\frac{3}{5}\).
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If \(\cot x=\frac{3}{2}\), what is the value of \(\frac{\cot^2 x-1}{\cot^2 x+1}\)?
A. \(\frac{5}{13}\)
\(\cot^2 x=\frac{9}{4}\) रखकर सरल करें। मान \(\frac{\frac{9}{4}-1}{\frac{9}{4}+1}=\frac{5}{13}\) है। / Substitute \(\cot^2 x=\frac{9}{4}\) and simplify. The value is \(\frac{\frac{9}{4}-1}{\frac{9}{4}+1}=\frac{5}{13}\).
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What is \(\sin^4 x+\cos^4 x\) equal to?
B. \(1-2\sin^2 x\cos^2 x\)
(\sin-4 x+\cos-4 x=\(\sin^2 x+\cos^2 x\)2-2\sin-2 x\cos-2 x) लिखें। पहला वर्ग (1) है। / Write (\sin-4 x+\cos-4 x=\(\sin^2 x+\cos^2 x\)2-2\sin-2 x\cos-2 x). The first square is (1).
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If \(\sin x\cos x=\frac{1}{4}\), what is the value of \(\sin^4 x+\cos^4 x\)?
A. \(\frac{7}{8}\)
पहचान \(\sin^4 x+\cos^4 x=1-2\sin^2 x\cos^2 x\) लगाएँ। \(\sin^2 x\cos^2 x=\frac{1}{16}\), इसलिए मान \(\frac{7}{8}\) है। / Use \(\sin^4 x+\cos^4 x=1-2\sin^2 x\cos^2 x\). Since \(\sin^2 x\cos^2 x=\frac{1}{16}\), the value is \(\frac{7}{8}\).
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If \(\sin x+\cos x=\sqrt{3}\), what is the value of \(\tan x+\cot x\)?
C. (4)
वर्ग करने पर \(1+2\sin x\cos x=3\), इसलिए \(\sin x\cos x=1\) मिलता है। फिर \(\tan x+\cot x=\frac{1}{\sin x\cos x}\), इसलिए मान (1) नहीं बल्कि विकल्पों में कोई सही नहीं होता। / Squaring gives \(1+2\sin x\cos x=3\), so \(\sin x\cos x=1\). Then \(\tan x+\cot x=\frac{1}{\sin x\cos x}\), so the value is (1), but none of the options is correct.
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If \(\sin x+\cos x=\frac{7}{5}\), what is the value of \(\tan x+\cot x\)?
B. \(\frac{25}{12}\)
वर्ग करने पर \(1+2\sin x\cos x=\frac{49}{25}\), इसलिए \(\sin x\cos x=\frac{12}{25}\)। अब \(\tan x+\cot x=\frac{1}{\sin x\cos x}=\frac{25}{12}\)। / Squaring gives \(1+2\sin x\cos x=\frac{49}{25}\), so \(\sin x\cos x=\frac{12}{25}\). Now \(\tan x+\cot x=\frac{1}{\sin x\cos x}=\frac{25}{12}\).
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If \(\sin x-\cos x=\frac{1}{3}\), what is the value of \(\tan x+\cot x\)?
A. \(\frac{9}{4}\)
वर्ग करने पर \(1-2\sin x\cos x=\frac{1}{9}\) मिलता है। इसलिए \(\sin x\cos x=\frac{4}{9}\) और \(\tan x+\cot x=\frac{9}{4}\)। / Squaring gives \(1-2\sin x\cos x=\frac{1}{9}\). Thus \(\sin x\cos x=\frac{4}{9}\) and \(\tan x+\cot x=\frac{9}{4}\).
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If \(\sec x+\tan x=3\), what is the value of \(\sec x\)?
A. \(\frac{5}{3}\)
क्योंकि \(\sec x-\tan x=\frac{1}{3}\) होगा। दोनों समीकरण जोड़ने पर \(2\sec x=3+\frac{1}{3}\), इसलिए \(\sec x=\frac{5}{3}\)। / Since \(\sec x-\tan x=\frac{1}{3}\). Adding both equations gives \(2\sec x=3+\frac{1}{3}\), so \(\sec x=\frac{5}{3}\).
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If \(\cosec x+\cot x=4\), what is the value of \(\cot x\)?
B. \(\frac{15}{8}\)
\(\cosec x-\cot x=\frac{1}{4}\) होगा। घटाने पर \(2\cot x=4-\frac{1}{4}\), इसलिए \(\cot x=\frac{15}{8}\)। / \(\cosec x-\cot x=\frac{1}{4}\). Subtracting gives \(2\cot x=4-\frac{1}{4}\), so \(\cot x=\frac{15}{8}\).
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What is the simplified value of \(\frac{1}{1+\sin x}+\frac{1}{1-\sin x}\)?
A. \(2\sec^2 x\)
हरों को मिलाने पर \(\frac{2}{1-\sin^2 x}\) मिलेगा। यह \(\frac{2}{\cos^2 x}=2\sec^2 x\) है। / Combining denominators gives \(\frac{2}{1-\sin^2 x}\). This is \(\frac{2}{\cos^2 x}=2\sec^2 x\).
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What is the simplified value of \(\frac{1}{1+\cos x}+\frac{1}{1-\cos x}\)?
C. \(2\cosec^2 x\)
हरों को मिलाने पर \(\frac{2}{1-\cos^2 x}\) मिलता है। यह \(\frac{2}{\sin^2 x}=2\cosec^2 x\) है। / Combining denominators gives \(\frac{2}{1-\cos^2 x}\). This is \(\frac{2}{\sin^2 x}=2\cosec^2 x\).
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What is \(\frac{1}{\sec x+\tan x}\) equal to?
B. \(\sec x-\tan x\)
क्योंकि (\(\sec x+\tan x\)\(\sec x-\tan x\)=1)। इसलिए व्युत्क्रम \(\sec x-\tan x\) होगा। / Since (\(\sec x+\tan x\)\(\sec x-\tan x\)=1). Therefore, the reciprocal is \(\sec x-\tan x\).
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What is \(\frac{1}{\cosec x-\cot x}\) equal to?
C. \(\cosec x+\cot x\)
क्योंकि (\(\cosec x-\cot x\)\(\cosec x+\cot x\)=1)। इसलिए आवश्यक व्युत्क्रम \(\cosec x+\cot x\) है। / Since (\(\cosec x-\cot x\)\(\cosec x+\cot x\)=1). Hence the required reciprocal is \(\cosec x+\cot x\).
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What is the amplitude of the function \(6-2\sin 4x\)?
C. (2)
आयाम गुणांक के परिमाण के बराबर होता है। यहाँ \(\sin 4x\) का गुणांक (-2) है, इसलिए आयाम (2) है। / Amplitude equals the absolute value of the coefficient. Here the coefficient of \(\sin 4x\) is (-2), so the amplitude is (2).
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What is the period of the function (5\cos\(\frac{x}{2}\))?
C. \(4\pi\)
\(\cos kx\) का काल \(\frac{2\pi}{k}\) होता है। यहाँ \(k=\frac{1}{2}\), इसलिए काल \(4\pi\) है। / The period of \(\cos kx\) is \(\frac{2\pi}{k}\). Here \(k=\frac{1}{2}\), so the period is \(4\pi\).
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What is the fundamental period of the function \(\sin^2 x\)?
B. \(\pi\)
(\sin-2\(x+\pi\)=\sin-2 x) होता है। इसलिए \(\sin^2 x\) का मूल काल \(\pi\) है। / (\sin-2\(x+\pi\)=\sin-2 x). Hence the fundamental period of \(\sin^2 x\) is \(\pi\).
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What is the range of the function \(\cos^2 x\)?
C. ([0,1])
\(\cos x\) का मान ([-1,1]) में होता है। वर्ग करने पर परिसर ([0,1]) बनता है। / The value of \(\cos x\) lies in ([-1,1]). Squaring gives the range ([0,1]).
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If \(\sec x-\tan x=\frac{2}{5}\), what is the value of \(\tan x\)?
A. \(\frac{21}{20}\)
क्योंकि (\(\sec x-\tan x\)\(\sec x+\tan x\)=1), इसलिए \(\sec x+\tan x=\frac{5}{2}\)। दोनों समीकरण घटाने पर \(\tan x=\frac{21}{20}\) मिलता है। / Since (\(\sec x-\tan x\)\(\sec x+\tan x\)=1), \(\sec x+\tan x=\frac{5}{2}\). Subtracting the two equations gives \(\tan x=\frac{21}{20}\).
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What is the simplified value of \(\frac{\sin x-\sin^3 x}{\cos^2 x}\)?
C. \(\sin x\)
अंश को (\sin x\(1-\sin^2 x\)) लिखें। क्योंकि \(1-\sin^2 x=\cos^2 x\), इसलिए मान \(\sin x\) है। / Write the numerator as (\sin x\(1-\sin^2 x\)). Since \(1-\sin^2 x=\cos^2 x\), the value is \(\sin x\).
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What is the range of the function \(2-4\cos^2 x\)?
B. ([-2,2])
\(\cos^2 x\) का परिसर ([0,1]) है। इसलिए \(2-4\cos^2 x\) का परिसर ([-2,2]) होगा। / The range of \(\cos^2 x\) is ([0,1]). Therefore, the range of \(2-4\cos^2 x\) is ([-2,2]).
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