Since (\(\sec x+\tan x\)\(\sec x-\tan x\)=1), the other factor is \(\frac{1}{3}\). In exams, identify reciprocal pairs.
Step 2
Why this answer is correct
The correct answer is B. \( \frac{1}{3} \). Since (\(\sec x+\tan x\)\(\sec x-\tan x\)=1), the other factor is \(\frac{1}{3}\). In exams, identify reciprocal pairs.
Step 3
Exam Tip
क्योंकि (\(\sec x+\tan x\)\(\sec x-\tan x\)=1), इसलिए दूसरा गुणक \(\frac{1}{3}\) है। परीक्षा में reciprocal pair पहचानें।
\( \sec \theta=\frac{1}{\cos \theta}\), so \( \cos \theta=\frac{1}{2}\). In exams remember reciprocal relations.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{1}{2}\). \( \sec \theta=\frac{1}{\cos \theta}\), so \( \cos \theta=\frac{1}{2}\). In exams remember reciprocal relations.
Step 3
Exam Tip
\(\sec \theta=\frac{1}{\cos \theta}\), इसलिए \(\cos \theta=\frac{1}{2}\) है। परीक्षा में reciprocal relations याद रखें।
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।
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 \(\sec x=\frac{5}{3}\)।
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।
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।
(\sin\(\frac{\pi}{2}-x\)=\cos x), so (\cosec\(\frac{\pi}{2}-x\)=\frac{1}{\cos x}=\sec x). Use reciprocal and cofunction identities together.
Step 2
Why this answer is correct
The correct answer is A. \(\sec x\). (\sin\(\frac{\pi}{2}-x\)=\cos x), so (\cosec\(\frac{\pi}{2}-x\)=\frac{1}{\cos x}=\sec x). Use reciprocal and cofunction identities together.
Step 3
Exam Tip
(\sin\(\frac{\pi}{2}-x\)=\cos x), इसलिए (\cosec\(\frac{\pi}{2}-x\)=\frac{1}{\cos x}=\sec x)। व्युत्क्रम और पूरक पहचान साथ लगाएँ।
(\cos\(\frac{\pi}{2}-x\)=\sin x), so (\sec\(\frac{\pi}{2}-x\)=\frac{1}{\sin x}=\cosec x). Complementary angles give cofunctions.
Step 2
Why this answer is correct
The correct answer is C. \(\cosec x\). (\cos\(\frac{\pi}{2}-x\)=\sin x), so (\sec\(\frac{\pi}{2}-x\)=\frac{1}{\sin x}=\cosec x). Complementary angles give cofunctions.
Step 3
Exam Tip
(\cos\(\frac{\pi}{2}-x\)=\sin x), इसलिए (\sec\(\frac{\pi}{2}-x\)=\frac{1}{\sin x}=\cosec x)। पूरक कोण में सहफलन बनता है।
The correct identity is \(1+\tan^2 x=\sec^2 x\). In identity-based questions, check signs and fractions carefully.
Step 2
Why this answer is correct
The correct answer is D. \(1+\tan^2 x=\sec^2 x\). The correct identity is \(1+\tan^2 x=\sec^2 x\). In identity-based questions, check signs and fractions carefully.
Step 3
Exam Tip
सही पहचान \(1+\tan^2 x=\sec^2 x\) है। पहचान आधारित प्रश्नों में चिन्ह और भिन्न को सावधानी से देखें।
\(\sec x=\frac{1}{\cos x}\), so it is undefined when \(\cos x=0\). In reciprocal functions, the denominator cannot be zero.
Step 2
Why this answer is correct
The correct answer is C. \(\cos x=0\). \(\sec x=\frac{1}{\cos x}\), so it is undefined when \(\cos x=0\). In reciprocal functions, the denominator cannot be zero.
Step 3
Exam Tip
\(\sec x=\frac{1}{\cos x}\) है, इसलिए \(\cos x=0\) पर यह अपरिभाषित होता है। व्युत्क्रम फलनों में हर शून्य नहीं हो सकता।
\(\sec x=\frac{1}{\cos x}\) and \(\cos x\) lies in ([-1,1]). Hence \(\sec x\) is greater than or equal to (1) or less than or equal to (-1).
Step 2
Why this answer is correct
The correct answer is D. (\(-\infty,-1]\cup[1,\infty\)). \(\sec x=\frac{1}{\cos x}\) and \(\cos x\) lies in ([-1,1]). Hence \(\sec x\) is greater than or equal to (1) or less than or equal to (-1).
Step 3
Exam Tip
\(\sec x=\frac{1}{\cos x}\) है और \(\cos x\) का मान ([-1,1]) में होता है। इसलिए \(\sec x\) का मान (1) से बड़ा या (-1) से छोटा होता है।
\(\sec \theta\) is the reciprocal of \(\cos \theta\) and remains positive in the fourth quadrant. So (\sec\(2\pi-\theta\)=\sec \theta).
Step 2
Why this answer is correct
The correct answer is D. \(\sec \theta\). \(\sec \theta\) is the reciprocal of \(\cos \theta\) and remains positive in the fourth quadrant. So (\sec\(2\pi-\theta\)=\sec \theta).
Step 3
Exam Tip
\(\sec \theta\) \(\cos \theta\) का व्युत्क्रम है और चौथे चतुर्थांश में धनात्मक रहता है। इसलिए (\sec\(2\pi-\theta\)=\sec \theta) है।
The period of \(\sec \theta\) is \(2\pi\) because it is the reciprocal of \(\cos \theta\). Hence adding \(2\pi\) keeps the value same.
Step 2
Why this answer is correct
The correct answer is B. \(\sec \theta\). The period of \(\sec \theta\) is \(2\pi\) because it is the reciprocal of \(\cos \theta\). Hence adding \(2\pi\) keeps the value same.
Step 3
Exam Tip
\(\sec \theta\) का काल \(2\pi\) है क्योंकि यह \(\cos \theta\) का व्युत्क्रम है। इसलिए \(2\pi\) जोड़ने पर मान वही रहता है।
The cofunction of \(\sec \theta\) is \(\cosec \theta\), and \(\sec \theta\) is negative in the second quadrant. Hence the value is \(-\cosec \theta\).
Step 2
Why this answer is correct
The correct answer is A. -\(\cosec \theta\). The cofunction of \(\sec \theta\) is \(\cosec \theta\), and \(\sec \theta\) is negative in the second quadrant. Hence the value is \(-\cosec \theta\).
Step 3
Exam Tip
\(\sec \theta\) का सह-फलन \(\cosec \theta\) है और दूसरे चतुर्थांश में \(\sec \theta\) ऋणात्मक होता है। इसलिए मान \(-\cosec \theta\) है।
