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Use the identity (\(\sin x+\cos x\)2+\(\sin x-\cos x\)2=2). Hence the value is \(2-\frac{36}{25}=\frac{14}{25}\).
Step 2
Why this answer is correct
The correct answer is C. \(\frac{14}{25}\). Use the identity (\(\sin x+\cos x\)2+\(\sin x-\cos x\)2=2). Hence the value is \(2-\frac{36}{25}=\frac{14}{25}\).
Step 3
Exam Tip
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।
The identity is (\(\cosec x+\cot x\)\(\cosec x-\cot x\)=1). Therefore, the required value is \(\frac{1}{6}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{6}\). The identity is (\(\cosec x+\cot x\)\(\cosec x-\cot x\)=1). Therefore, the required value is \(\frac{1}{6}\).
Step 3
Exam Tip
(\(\cosec x+\cot x\)\(\cosec x-\cot x\)=1) होता है। इसलिए आवश्यक मान \(\frac{1}{6}\) है।
By the half-angle identity, \(\frac{\sin x}{1-\cos x}=\cot \frac{x}{2}\). In such forms, identify the denominator carefully.
Step 2
Why this answer is correct
The correct answer is B. \(\cot \frac{x}{2}\). By the half-angle identity, \(\frac{\sin x}{1-\cos x}=\cot \frac{x}{2}\). In such forms, identify the denominator carefully.
Step 3
Exam Tip
अर्ध-कोण पहचान से \(\frac{\sin x}{1-\cos x}=\cot \frac{x}{2}\) होता है। ऐसे रूपों में हर देखकर पहचान करें।
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.
Step 2
Why this answer is correct
The correct answer is C. \(\cot \frac{x}{2}\). 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.
Step 3
Exam Tip
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 period of \(\tan kx\) is \(\frac{\pi}{k}\). Here (k=2), so the fundamental period is \(\frac{\pi}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\pi}{2}\). The period of \(\tan kx\) is \(\frac{\pi}{k}\). Here (k=2), so the fundamental period is \(\frac{\pi}{2}\).
Step 3
Exam Tip
\(\tan kx\) का काल \(\frac{\pi}{k}\) होता है। यहाँ (k=2), इसलिए मूल काल \(\frac{\pi}{2}\) है।
In the second quadrant, \(\sin x\) is positive and \(\cos x\) is negative. Since \(\sin x=\frac{4}{5}\), \(\tan x=-\frac{4}{3}\).
Step 2
Why this answer is correct
The correct answer is B. -\(\frac{4}{3}\). In the second quadrant, \(\sin x\) is positive and \(\cos x\) is negative. Since \(\sin x=\frac{4}{5}\), \(\tan x=-\frac{4}{3}\).
Step 3
Exam Tip
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 third quadrant, \(\cos x\) is negative. Since \(\cos x=-\frac{12}{13}\), \(\sec x=-\frac{13}{12}\).
Step 2
Why this answer is correct
The correct answer is B. -\(\frac{13}{12}\). In the third quadrant, \(\cos x\) is negative. Since \(\cos x=-\frac{12}{13}\), \(\sec x=-\frac{13}{12}\).
Step 3
Exam Tip
तीसरे चतुर्थांश में \(\cos x\) ऋणात्मक होता है। \(\cos x=-\frac{12}{13}\), इसलिए \(\sec x=-\frac{13}{12}\) है।
In the fourth quadrant, \(\cos x\) is positive and \(\sin x\) is negative. From the (7,24,25) triple, \(\cos x=\frac{7}{25}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{7}{25}\). In the fourth quadrant, \(\cos x\) is positive and \(\sin x\) is negative. From the (7,24,25) triple, \(\cos x=\frac{7}{25}\).
Step 3
Exam Tip
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 form \(\frac{3\pi}{2}+x\), \(\sin\) changes to \(\cos\) with a negative sign. Hence the answer is \(-\cos x\).
Step 2
Why this answer is correct
The correct answer is A. -\(\cos x\). In the form \(\frac{3\pi}{2}+x\), \(\sin\) changes to \(\cos\) with a negative sign. Hence the answer is \(-\cos x\).
Step 3
Exam Tip
\(\frac{3\pi}{2}+x\) रूप में \(\sin\) बदलकर \(\cos\) होता है और चिन्ह ऋणात्मक होता है। इसलिए उत्तर \(-\cos x\) है।
At \(\frac{3\pi}{2}+x\), \(\tan\) changes to \(\cot\) with a negative sign. Hence (\tan\(\frac{3\pi}{2}+x\)=-\cot x).
Step 2
Why this answer is correct
The correct answer is C. -\(\cot x\). At \(\frac{3\pi}{2}+x\), \(\tan\) changes to \(\cot\) with a negative sign. Hence (\tan\(\frac{3\pi}{2}+x\)=-\cot x).
Step 3
Exam Tip
\(\frac{3\pi}{2}+x\) पर \(\tan\) बदलकर \(\cot\) होता है और चिन्ह ऋणात्मक है। इसलिए (\tan\(\frac{3\pi}{2}+x\)=-\cot x)।
Put \(\sec x=\frac{1}{\cos x}\) and \(\tan x=\frac{\sin x}{\cos x}\). The ratio becomes \(\frac{1}{\sin x}=\cosec x\).
Step 2
Why this answer is correct
The correct answer is B. \(\cosec x\). Put \(\sec x=\frac{1}{\cos x}\) and \(\tan x=\frac{\sin x}{\cos x}\). The ratio becomes \(\frac{1}{\sin x}=\cosec x\).
Step 3
Exam Tip
\(\sec x=\frac{1}{\cos x}\) और \(\tan x=\frac{\sin x}{\cos x}\) रखें। अनुपात \(\frac{1}{\sin x}=\cosec x\) बनता है।
Put \(\cosec x=\frac{1}{\sin x}\) and \(\cot x=\frac{\cos x}{\sin x}\). The ratio becomes \(\frac{1}{\cos x}=\sec x\).
Step 2
Why this answer is correct
The correct answer is B. \(\sec x\). Put \(\cosec x=\frac{1}{\sin x}\) and \(\cot x=\frac{\cos x}{\sin x}\). The ratio becomes \(\frac{1}{\cos x}=\sec x\).
Substitute \(\cot^2 x=\frac{9}{4}\) and simplify. The value is \(\frac{\frac{9}{4}-1}{\frac{9}{4}+1}=\frac{5}{13}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{5}{13}\). Substitute \(\cot^2 x=\frac{9}{4}\) and simplify. The value is \(\frac{\frac{9}{4}-1}{\frac{9}{4}+1}=\frac{5}{13}\).
Step 3
Exam Tip
\(\cot^2 x=\frac{9}{4}\) रखकर सरल करें। मान \(\frac{\frac{9}{4}-1}{\frac{9}{4}+1}=\frac{5}{13}\) 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 \(\frac{7}{8}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{7}{8}\). 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 \(\frac{7}{8}\).
Step 3
Exam Tip
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।
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.
Step 2
Why this answer is correct
The correct answer is C. (4). 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.
Step 3
Exam Tip
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 \(1+2\sin x\cos x=\frac{49}{25}\), so \(\sin x\cos x=\frac{12}{25}\). Now \(\tan x+\cot x=\frac{1}{\sin x\cos x}=\frac{25}{12}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{25}{12}\). Squaring gives \(1+2\sin x\cos x=\frac{49}{25}\), so \(\sin x\cos x=\frac{12}{25}\). Now \(\tan x+\cot x=\frac{1}{\sin x\cos x}=\frac{25}{12}\).
Step 3
Exam Tip
वर्ग करने पर \(1+2\sin x\cos x=\frac{49}{25}\), इसलिए \(\sin x\cos x=\frac{12}{25}\)। अब \(\tan x+\cot x=\frac{1}{\sin x\cos x}=\frac{25}{12}\)।
The correct answer is A. \(\frac{9}{4}\). Squaring gives \(1-2\sin x\cos x=\frac{1}{9}\). Thus \(\sin x\cos x=\frac{4}{9}\) and \(\tan x+\cot x=\frac{9}{4}\).
Step 3
Exam Tip
वर्ग करने पर \(1-2\sin x\cos x=\frac{1}{9}\) मिलता है। इसलिए \(\sin x\cos x=\frac{4}{9}\) और \(\tan x+\cot x=\frac{9}{4}\)।
Since \(\sec x-\tan x=\frac{1}{3}\). Adding both equations gives \(2\sec x=3+\frac{1}{3}\), so \(\sec x=\frac{5}{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{5}{3}\). Since \(\sec x-\tan x=\frac{1}{3}\). Adding both equations gives \(2\sec x=3+\frac{1}{3}\), so \(\sec x=\frac{5}{3}\).
Step 3
Exam Tip
क्योंकि \(\sec x-\tan x=\frac{1}{3}\) होगा। दोनों समीकरण जोड़ने पर \(2\sec x=3+\frac{1}{3}\), इसलिए \(\sec x=\frac{5}{3}\)।
Amplitude equals the absolute value of the coefficient. Here the coefficient of \(\sin 4x\) is (-2), so the amplitude is (2).
Step 2
Why this answer is correct
The correct answer is C. (2). Amplitude equals the absolute value of the coefficient. Here the coefficient of \(\sin 4x\) is (-2), so the amplitude is (2).
Step 3
Exam Tip
आयाम गुणांक के परिमाण के बराबर होता है। यहाँ \(\sin 4x\) का गुणांक (-2) है, इसलिए आयाम (2) है।
Since (\(\sec x-\tan x\)\(\sec x+\tan x\)=1), \(\sec x+\tan x=\frac{5}{2}\). Subtracting the two equations gives \(\tan x=\frac{21}{20}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{21}{20}\). Since (\(\sec x-\tan x\)\(\sec x+\tan x\)=1), \(\sec x+\tan x=\frac{5}{2}\). Subtracting the two equations gives \(\tan x=\frac{21}{20}\).
Step 3
Exam Tip
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।
