यदि \(\sin x+\cos x=\sqrt{2}\), तो \(\sin x\cos x\) का मान क्या होगा?
If \(\sin x+\cos x=\sqrt{2}\), what is the value of \(\sin x\cos x\)?
#trigonometric-identities
#medium
#sine-cosine
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A (0)
B \(\frac{1}{2}\)
C (1)
D \(-\frac{1}{2}\)
Explanation opens after your attempt
Correct Answer
B. \(\frac{1}{2}\)
Step 1
Concept
Squaring both sides gives \(1+2\sin x\cos x=2\). Hence \(\sin x\cos x=\frac{1}{2}\); squaring is useful in such questions.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{1}{2}\). Squaring both sides gives \(1+2\sin x\cos x=2\). Hence \(\sin x\cos x=\frac{1}{2}\); squaring is useful in such questions.
Step 3
Exam Tip
दोनों पक्षों का वर्ग करने पर \(1+2\sin x\cos x=2\) मिलता है। इसलिए \(\sin x\cos x=\frac{1}{2}\); ऐसे प्रश्नों में वर्ग करना उपयोगी है।
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यदि \(\tan x=2\), तो \(\frac{\sin x}{\cos x}\) का मान क्या है?
If \(\tan x=2\), what is the value of \(\frac{\sin x}{\cos x}\)?
#trigonometric-functions
#tangent
#quotient
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A (2)
B \(\frac{1}{2}\)
C (-2)
D (1)
Explanation opens after your attempt
Step 1
Concept
\(\tan x=\frac{\sin x}{\cos x}\). Therefore, the required value is directly (2).
Step 2
Why this answer is correct
The correct answer is A. (2). \(\tan x=\frac{\sin x}{\cos x}\). Therefore, the required value is directly (2).
Step 3
Exam Tip
\(\tan x=\frac{\sin x}{\cos x}\) होता है। इसलिए दिया गया मान सीधे (2) है।
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यदि \(\sec x=5\), तो \(\cos x\) का मान क्या है?
If \(\sec x=5\), what is the value of \(\cos x\)?
#reciprocal-identities
#secant
#cosine
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A (5)
B (-5)
C \(\frac{1}{5}\)
D \(-\frac{1}{5}\)
Explanation opens after your attempt
Correct Answer
C. \(\frac{1}{5}\)
Step 1
Concept
\(\sec x=\frac{1}{\cos x}\). Hence \(\cos x=\frac{1}{5}\); remember reciprocal relations.
Step 2
Why this answer is correct
The correct answer is C. \(\frac{1}{5}\). \(\sec x=\frac{1}{\cos x}\). Hence \(\cos x=\frac{1}{5}\); remember reciprocal relations.
Step 3
Exam Tip
\(\sec x=\frac{1}{\cos x}\) होता है। इसलिए \(\cos x=\frac{1}{5}\); व्युत्क्रम संबंध याद रखें।
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यदि \(\cosec x=4\), तो \(\sin x\) का मान क्या है?
If \(\cosec x=4\), what is the value of \(\sin x\)?
#reciprocal-identities
#cosecant
#sine
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A \(\frac{1}{2}\)
B (4)
C (-4)
D \(\frac{1}{4}\)
Explanation opens after your attempt
Correct Answer
D. \(\frac{1}{4}\)
Step 1
Concept
\(\cosec x=\frac{1}{\sin x}\). Hence \(\sin x=\frac{1}{4}\); learn each reciprocal function as a pair.
Step 2
Why this answer is correct
The correct answer is D. \(\frac{1}{4}\). \(\cosec x=\frac{1}{\sin x}\). Hence \(\sin x=\frac{1}{4}\); learn each reciprocal function as a pair.
Step 3
Exam Tip
\(\cosec x=\frac{1}{\sin x}\) होता है। इसलिए \(\sin x=\frac{1}{4}\); हर व्युत्क्रम फलन को जोड़ी में याद करें।
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\(\frac{1-\cos^2 x}{\sin^2 x}\) का सरल मान क्या है?
What is the simplified value of \(\frac{1-\cos^2 x}{\sin^2 x}\)?
#trigonometric-identities
#simplification
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A (1)
B \(\sin x\)
C \(\cos x\)
D \(\tan x\)
Explanation opens after your attempt
Step 1
Concept
Since \(1-\cos^2 x=\sin^2 x\), the ratio is (1). First convert using identities.
Step 2
Why this answer is correct
The correct answer is A. (1). Since \(1-\cos^2 x=\sin^2 x\), the ratio is (1). First convert using identities.
Step 3
Exam Tip
क्योंकि \(1-\cos^2 x=\sin^2 x\), इसलिए अनुपात (1) है। पहचान को पहले बदलकर देखें।
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\(\frac{1-\sin^2 x}{\cos^2 x}\) का सरल मान क्या है?
What is the simplified value of \(\frac{1-\sin^2 x}{\cos^2 x}\)?
#trigonometric-identities
#simplification
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A \(\sec x\)
B (1)
C \(\tan x\)
D \(\cot x\)
Explanation opens after your attempt
Step 1
Concept
Since \(1-\sin^2 x=\cos^2 x\), the value is (1). \(\sin^2 x+\cos^2 x=1\) is the most important identity.
Step 2
Why this answer is correct
The correct answer is B. (1). Since \(1-\sin^2 x=\cos^2 x\), the value is (1). \(\sin^2 x+\cos^2 x=1\) is the most important identity.
Step 3
Exam Tip
क्योंकि \(1-\sin^2 x=\cos^2 x\), इसलिए मान (1) होगा। \(\sin^2 x+\cos^2 x=1\) सबसे जरूरी पहचान है।
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फलन \(\sin 3x\) का काल क्या है?
What is the period of the function \(\sin 3x\)?
#period
#sine
#transformation
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A \(2\pi\)
B \(\pi\)
C \(\frac{2\pi}{3}\)
D \(\frac{\pi}{3}\)
Explanation opens after your attempt
Correct Answer
C. \(\frac{2\pi}{3}\)
Step 1
Concept
The period of \(\sin kx\) is \(\frac{2\pi}{k}\). Here (k=3), so the period is \(\frac{2\pi}{3}\).
Step 2
Why this answer is correct
The correct answer is C. \(\frac{2\pi}{3}\). The period of \(\sin kx\) is \(\frac{2\pi}{k}\). Here (k=3), so the period is \(\frac{2\pi}{3}\).
Step 3
Exam Tip
\(\sin kx\) का काल \(\frac{2\pi}{k}\) होता है। यहाँ (k=3), इसलिए काल \(\frac{2\pi}{3}\) है।
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फलन \(\tan 4x\) का काल क्या है?
What is the period of the function \(\tan 4x\)?
#period
#tangent
#transformation
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A \(\pi\)
B \(\frac{\pi}{2}\)
C \(\frac{\pi}{3}\)
D \(\frac{\pi}{4}\)
Explanation opens after your attempt
Correct Answer
D. \(\frac{\pi}{4}\)
Step 1
Concept
The period of \(\tan kx\) is \(\frac{\pi}{k}\). Substituting (k=4) gives period \(\frac{\pi}{4}\).
Step 2
Why this answer is correct
The correct answer is D. \(\frac{\pi}{4}\). The period of \(\tan kx\) is \(\frac{\pi}{k}\). Substituting (k=4) gives period \(\frac{\pi}{4}\).
Step 3
Exam Tip
\(\tan kx\) का काल \(\frac{\pi}{k}\) होता है। (k=4) रखने पर काल \(\frac{\pi}{4}\) मिलता है।
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फलन \(2\cos x\) का परिसर क्या है?
What is the range of the function \(2\cos x\)?
#range
#cosine
#transformation
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A ([-2,2])
B ([-1,1])
C ([0,2])
D (\(-\infty,\infty\))
Explanation opens after your attempt
Correct Answer
A. ([-2,2])
Step 1
Concept
The range of \(\cos x\) is ([-1,1]). Multiplying by (2) changes the range to ([-2,2]).
Step 2
Why this answer is correct
The correct answer is A. ([-2,2]). The range of \(\cos x\) is ([-1,1]). Multiplying by (2) changes the range to ([-2,2]).
Step 3
Exam Tip
\(\cos x\) का परिसर ([-1,1]) है। (2) से गुणा करने पर परिसर ([-2,2]) हो जाता है।
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फलन \(3\sin x+1\) का अधिकतम मान क्या है?
What is the maximum value of the function \(3\sin x+1\)?
#maximum-value
#sine
#range
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A (2)
B (3)
C (4)
D (1)
Explanation opens after your attempt
Step 1
Concept
The maximum value of \(\sin x\) is (1). Hence the maximum value of \(3\sin x+1\) is (4).
Step 2
Why this answer is correct
The correct answer is C. (4). The maximum value of \(\sin x\) is (1). Hence the maximum value of \(3\sin x+1\) is (4).
Step 3
Exam Tip
\(\sin x\) का अधिकतम मान (1) है। इसलिए \(3\sin x+1\) का अधिकतम मान (4) होगा।
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फलन \(5-2\cos x\) का न्यूनतम मान क्या है?
What is the minimum value of the function \(5-2\cos x\)?
#minimum-value
#cosine
#range
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A (7)
B (3)
C (5)
D (2)
Explanation opens after your attempt
Step 1
Concept
The maximum value of \(\cos x\) is (1). Therefore, the minimum value is (5-2=3).
Step 2
Why this answer is correct
The correct answer is B. (3). The maximum value of \(\cos x\) is (1). Therefore, the minimum value is (5-2=3).
Step 3
Exam Tip
\(\cos x\) का अधिकतम मान (1) है। इसलिए न्यूनतम मान (5-2=3) होगा।
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यदि \(\sin x=\frac{5}{13}\) और (x) प्रथम चतुर्थांश में है, तो \(\tan x\) का मान क्या है?
If \(\sin x=\frac{5}{13}\) and (x) is in the first quadrant, what is the value of \(\tan x\)?
#first-quadrant
#tangent
#identity
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A \(\frac{12}{13}\)
B \(\frac{13}{12}\)
C \(\frac{5}{12}\)
D \(\frac{12}{5}\)
Explanation opens after your attempt
Correct Answer
C. \(\frac{5}{12}\)
Step 1
Concept
\(\cos x=\frac{12}{13}\) and \(\tan x=\frac{\sin x}{\cos x}\). Hence \(\tan x=\frac{5}{12}\).
Step 2
Why this answer is correct
The correct answer is C. \(\frac{5}{12}\). \(\cos x=\frac{12}{13}\) and \(\tan x=\frac{\sin x}{\cos x}\). Hence \(\tan x=\frac{5}{12}\).
Step 3
Exam Tip
\(\cos x=\frac{12}{13}\) और \(\tan x=\frac{\sin x}{\cos x}\) होगा। इसलिए \(\tan x=\frac{5}{12}\)।
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यदि \(\cos x=\frac{8}{17}\) और (x) प्रथम चतुर्थांश में है, तो \(\cot x\) का मान क्या है?
If \(\cos x=\frac{8}{17}\) and (x) is in the first quadrant, what is the value of \(\cot x\)?
#first-quadrant
#cotangent
#identity
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A \(\frac{8}{15}\)
B \(\frac{15}{8}\)
C \(\frac{17}{8}\)
D \(\frac{8}{17}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{8}{15}\)
Step 1
Concept
\(\sin x=\frac{15}{17}\) and \(\cot x=\frac{\cos x}{\sin x}\). Therefore, \(\cot x=\frac{8}{15}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{8}{15}\). \(\sin x=\frac{15}{17}\) and \(\cot x=\frac{\cos x}{\sin x}\). Therefore, \(\cot x=\frac{8}{15}\).
Step 3
Exam Tip
\(\sin x=\frac{15}{17}\) मिलता है और \(\cot x=\frac{\cos x}{\sin x}\) होता है। इसलिए \(\cot x=\frac{8}{15}\)।
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यदि \(\tan x=\frac{3}{4}\) और (x) प्रथम चतुर्थांश में है, तो \(\sin x\) का मान क्या है?
If \(\tan x=\frac{3}{4}\) and (x) is in the first quadrant, what is the value of \(\sin x\)?
#tangent
#sine
#first-quadrant
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A \(\frac{4}{5}\)
B \(\frac{3}{5}\)
C \(\frac{5}{3}\)
D \(\frac{5}{4}\)
Explanation opens after your attempt
Correct Answer
B. \(\frac{3}{5}\)
Step 1
Concept
From \(\tan x=\frac{3}{4}\), the hypotenuse in a right triangle is (5). Hence \(\sin x=\frac{3}{5}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{3}{5}\). From \(\tan x=\frac{3}{4}\), the hypotenuse in a right triangle is (5). Hence \(\sin x=\frac{3}{5}\).
Step 3
Exam Tip
\(\tan x=\frac{3}{4}\) से समकोण त्रिभुज में कर्ण (5) होगा। इसलिए \(\sin x=\frac{3}{5}\)।
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यदि \(\cot x=\frac{5}{12}\) और (x) प्रथम चतुर्थांश में है, तो \(\cos x\) का मान क्या है?
If \(\cot x=\frac{5}{12}\) and (x) is in the first quadrant, what is the value of \(\cos x\)?
#cotangent
#cosine
#first-quadrant
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A \(\frac{12}{13}\)
B \(\frac{13}{5}\)
C \(\frac{5}{13}\)
D \(\frac{13}{12}\)
Explanation opens after your attempt
Correct Answer
C. \(\frac{5}{13}\)
Step 1
Concept
For \(\cot x=\frac{5}{12}\), take adjacent as (5) and opposite as (12). The hypotenuse is (13), so \(\cos x=\frac{5}{13}\).
Step 2
Why this answer is correct
The correct answer is C. \(\frac{5}{13}\). For \(\cot x=\frac{5}{12}\), take adjacent as (5) and opposite as (12). The hypotenuse is (13), so \(\cos x=\frac{5}{13}\).
Step 3
Exam Tip
\(\cot x=\frac{5}{12}\) में आसन्न (5) और सामने (12) मानें। कर्ण (13) होगा, इसलिए \(\cos x=\frac{5}{13}\)।
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(\sin\(\pi+x\)+\sin\(\pi-x\)) का सरल मान क्या है?
What is the simplified value of (\sin\(\pi+x\)+\sin\(\pi-x\))?
#allied-angles
#sine
#simplification
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A \(2\sin x\)
B (0)
C \(-2\sin x\)
D \(2\cos x\)
Explanation opens after your attempt
Step 1
Concept
(\sin\(\pi+x\)=-\sin x) and (\sin\(\pi-x\)=\sin x). Hence the sum is (0).
Step 2
Why this answer is correct
The correct answer is B. (0). (\sin\(\pi+x\)=-\sin x) and (\sin\(\pi-x\)=\sin x). Hence the sum is (0).
Step 3
Exam Tip
(\sin\(\pi+x\)=-\sin x) और (\sin\(\pi-x\)=\sin x) होते हैं। इसलिए योग (0) है।
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(\cos\(\pi+x\)+\cos\(\pi-x\)) का सरल मान क्या है?
What is the simplified value of (\cos\(\pi+x\)+\cos\(\pi-x\))?
#allied-angles
#cosine
#simplification
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A \(-2\cos x\)
B \(2\cos x\)
C (0)
D \(2\sin x\)
Explanation opens after your attempt
Correct Answer
A. \(-2\cos x\)
Step 1
Concept
(\cos\(\pi+x\)=-\cos x) and (\cos\(\pi-x\)=-\cos x). Therefore, the sum is \(-2\cos x\).
Step 2
Why this answer is correct
The correct answer is A. \(-2\cos x\). (\cos\(\pi+x\)=-\cos x) and (\cos\(\pi-x\)=-\cos x). Therefore, the sum is \(-2\cos x\).
Step 3
Exam Tip
(\cos\(\pi+x\)=-\cos x) और (\cos\(\pi-x\)=-\cos x) होते हैं। इसलिए योग \(-2\cos x\) है।
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(\tan\(\pi+x\)-\tan\(\pi-x\)) का सरल मान क्या है?
What is the simplified value of (\tan\(\pi+x\)-\tan\(\pi-x\))?
#allied-angles
#tangent
#simplification
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A (0)
B \(-2\tan x\)
C \(2\tan x\)
D \(\tan x\)
Explanation opens after your attempt
Correct Answer
C. \(2\tan x\)
Step 1
Concept
(\tan\(\pi+x\)=\tan x) and (\tan\(\pi-x\)=-\tan x). Therefore, the difference is \(2\tan x\).
Step 2
Why this answer is correct
The correct answer is C. \(2\tan x\). (\tan\(\pi+x\)=\tan x) and (\tan\(\pi-x\)=-\tan x). Therefore, the difference is \(2\tan x\).
Step 3
Exam Tip
(\tan\(\pi+x\)=\tan x) और (\tan\(\pi-x\)=-\tan x) होते हैं। इसलिए अंतर \(2\tan x\) है।
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(\sin\(\frac{\pi}{2}+x\)) किसके बराबर है?
What is (\sin\(\frac{\pi}{2}+x\)) equal to?
#cofunction-identities
#sine
#allied-angles
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A \(-\cos x\)
B \(\sin x\)
C \(-\sin x\)
D \(\cos x\)
Explanation opens after your attempt
Correct Answer
D. \(\cos x\)
Step 1
Concept
In the form \(\frac{\pi}{2}+x\), \(\sin\) changes to \(\cos\) and the sign remains positive. Hence the answer is \(\cos x\).
Step 2
Why this answer is correct
The correct answer is D. \(\cos x\). In the form \(\frac{\pi}{2}+x\), \(\sin\) changes to \(\cos\) and the sign remains positive. Hence the answer is \(\cos x\).
Step 3
Exam Tip
\(\frac{\pi}{2}+x\) वाले रूप में \(\sin\) बदलकर \(\cos\) होता है और चिन्ह धनात्मक रहता है। इसलिए उत्तर \(\cos x\) है।
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(\cos\(\frac{\pi}{2}+x\)) किसके बराबर है?
What is (\cos\(\frac{\pi}{2}+x\)) equal to?
#cofunction-identities
#cosine
#allied-angles
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A \(\sin x\)
B \(-\sin x\)
C \(\cos x\)
D \(-\cos x\)
Explanation opens after your attempt
Correct Answer
B. \(-\sin x\)
Step 1
Concept
In the form \(\frac{\pi}{2}+x\), \(\cos\) changes to \(\sin\) with a negative sign. Hence it becomes \(-\sin x\).
Step 2
Why this answer is correct
The correct answer is B. \(-\sin x\). In the form \(\frac{\pi}{2}+x\), \(\cos\) changes to \(\sin\) with a negative sign. Hence it becomes \(-\sin x\).
Step 3
Exam Tip
\(\frac{\pi}{2}+x\) वाले रूप में \(\cos\) बदलकर \(\sin\) होता है और चिन्ह ऋणात्मक होता है। इसलिए \(-\sin x\) मिलता है।
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(\tan\(\frac{\pi}{2}+x\)) किसके बराबर है?
What is (\tan\(\frac{\pi}{2}+x\)) equal to?
#cofunction-identities
#tangent
#cotangent
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A \(-\cot x\)
B \(\cot x\)
C \(\tan x\)
D \(-\tan x\)
Explanation opens after your attempt
Correct Answer
A. \(-\cot x\)
Step 1
Concept
At \(\frac{\pi}{2}+x\), \(\tan\) changes to \(\cot\) with a negative sign. Hence (\tan\(\frac{\pi}{2}+x\)=-\cot x).
Step 2
Why this answer is correct
The correct answer is A. \(-\cot x\). At \(\frac{\pi}{2}+x\), \(\tan\) changes to \(\cot\) with a negative sign. Hence (\tan\(\frac{\pi}{2}+x\)=-\cot x).
Step 3
Exam Tip
\(\frac{\pi}{2}+x\) पर \(\tan\) बदलकर \(\cot\) होता है और चिन्ह ऋणात्मक होता है। इसलिए (\tan\(\frac{\pi}{2}+x\)=-\cot x)।
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(\sec\(\frac{\pi}{2}-x\)) किसके बराबर है?
What is (\sec\(\frac{\pi}{2}-x\)) equal to?
#cofunction-identities
#secant
#cosecant
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A \(\sec x\)
B \(-\cosec x\)
C \(\cosec x\)
D \(-\sec x\)
Explanation opens after your attempt
Correct Answer
C. \(\cosec x\)
Step 1
Concept
(\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।
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(\cosec\(\frac{\pi}{2}-x\)) किसके बराबर है?
What is (\cosec\(\frac{\pi}{2}-x\)) equal to?
#cofunction-identities
#cosecant
#secant
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A \(\sec x\)
B \(\cosec x\)
C \(-\sec x\)
D \(-\cosec x\)
Explanation opens after your attempt
Correct Answer
A. \(\sec x\)
Step 1
Concept
(\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।
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यदि (x) दूसरे चतुर्थांश में है और \(\sin x=\frac{7}{25}\), तो \(\cos x\) का मान क्या है?
If (x) is in the second quadrant and \(\sin x=\frac{7}{25}\), what is the value of \(\cos x\)?
#quadrants
#cosine
#identity
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A \(\frac{24}{25}\)
B \(-\frac{7}{25}\)
C \(-\frac{24}{25}\)
D \(\frac{7}{25}\)
Explanation opens after your attempt
Correct Answer
C. \(-\frac{24}{25}\)
Step 1
Concept
The identity gives \(|\cos x|=\frac{24}{25}\). In the second quadrant, \(\cos x\) is negative.
Step 2
Why this answer is correct
The correct answer is C. \(-\frac{24}{25}\). The identity gives \(|\cos x|=\frac{24}{25}\). In the second quadrant, \(\cos x\) is negative.
Step 3
Exam Tip
पहचान से \(|\cos x|=\frac{24}{25}\) मिलता है। दूसरे चतुर्थांश में \(\cos x\) ऋणात्मक होता है।
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यदि (x) तीसरे चतुर्थांश में है और \(\tan x=\frac{9}{40}\), तो \(\sec x\) का मान क्या है?
If (x) is in the third quadrant and \(\tan x=\frac{9}{40}\), what is the value of \(\sec x\)?
#quadrants
#secant
#tangent
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A \(\frac{41}{40}\)
B \(-\frac{41}{40}\)
C \(\frac{40}{41}\)
D \(-\frac{40}{41}\)
Explanation opens after your attempt
Correct Answer
B. \(-\frac{41}{40}\)
Step 1
Concept
From \(\sec^2 x=1+\tan^2 x\), \(|\sec x|=\frac{41}{40}\). In the third quadrant, \(\sec x\) is negative.
Step 2
Why this answer is correct
The correct answer is B. \(-\frac{41}{40}\). From \(\sec^2 x=1+\tan^2 x\), \(|\sec x|=\frac{41}{40}\). In the third quadrant, \(\sec x\) is negative.
Step 3
Exam Tip
\(\sec^2 x=1+\tan^2 x\) से \(|\sec x|=\frac{41}{40}\) मिलता है। तीसरे चतुर्थांश में \(\sec x\) ऋणात्मक होता है।
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यदि (x) चौथे चतुर्थांश में है और \(\cos x=\frac{15}{17}\), तो \(\sin x\) का मान क्या है?
If (x) is in the fourth quadrant and \(\cos x=\frac{15}{17}\), what is the value of \(\sin x\)?
#quadrants
#sine
#identity
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A \(\frac{8}{17}\)
B \(-\frac{15}{17}\)
C \(-\frac{8}{17}\)
D \(\frac{15}{17}\)
Explanation opens after your attempt
Correct Answer
C. \(-\frac{8}{17}\)
Step 1
Concept
The identity gives \(|\sin x|=\frac{8}{17}\). In the fourth quadrant, \(\sin x\) is negative.
Step 2
Why this answer is correct
The correct answer is C. \(-\frac{8}{17}\). The identity gives \(|\sin x|=\frac{8}{17}\). In the fourth quadrant, \(\sin x\) is negative.
Step 3
Exam Tip
पहचान से \(|\sin x|=\frac{8}{17}\) मिलता है। चौथे चतुर्थांश में \(\sin x\) ऋणात्मक होता है।
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यदि (x) दूसरे चतुर्थांश में है और \(\cot x=-\frac{3}{4}\), तो \(\cosec x\) का मान क्या है?
If (x) is in the second quadrant and \(\cot x=-\frac{3}{4}\), what is the value of \(\cosec x\)?
#quadrants
#cosecant
#cotangent
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A \(\frac{5}{4}\)
B \(-\frac{5}{4}\)
C \(\frac{4}{5}\)
D \(-\frac{4}{5}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{5}{4}\)
Step 1
Concept
From \(\cosec^2 x=1+\cot^2 x\), \(|\cosec x|=\frac{5}{4}\). In the second quadrant, \(\cosec x\) is positive.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{5}{4}\). From \(\cosec^2 x=1+\cot^2 x\), \(|\cosec x|=\frac{5}{4}\). In the second quadrant, \(\cosec x\) is positive.
Step 3
Exam Tip
\(\cosec^2 x=1+\cot^2 x\) से \(|\cosec x|=\frac{5}{4}\) मिलता है। दूसरे चतुर्थांश में \(\cosec x\) धनात्मक होता है।
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\(\frac{\sec^2 x-1}{\tan^2 x}\) का सरल मान क्या है?
What is the simplified value of \(\frac{\sec^2 x-1}{\tan^2 x}\)?
#trigonometric-identities
#secant
#tangent
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A \(\sec x\)
B (1)
C \(\tan x\)
D (0)
Explanation opens after your attempt
Step 1
Concept
\(\sec^2 x-1=\tan^2 x\). Therefore, the whole fraction equals (1).
Step 2
Why this answer is correct
The correct answer is B. (1). \(\sec^2 x-1=\tan^2 x\). Therefore, the whole fraction equals (1).
Step 3
Exam Tip
\(\sec^2 x-1=\tan^2 x\) होता है। इसलिए पूरा भिन्न (1) के बराबर है।
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\(\frac{\cosec^2 x-1}{\cot^2 x}\) का सरल मान क्या है?
What is the simplified value of \(\frac{\cosec^2 x-1}{\cot^2 x}\)?
#trigonometric-identities
#cosecant
#cotangent
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A (0)
B \(\cosec x\)
C (1)
D \(\cot x\)
Explanation opens after your attempt
Step 1
Concept
\(\cosec^2 x-1=\cot^2 x\). Hence the ratio is (1).
Step 2
Why this answer is correct
The correct answer is C. (1). \(\cosec^2 x-1=\cot^2 x\). Hence the ratio is (1).
Step 3
Exam Tip
\(\cosec^2 x-1=\cot^2 x\) होता है। इसलिए अनुपात (1) है।
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(\(\sec x-\tan x\)\(\sec x+\tan x\)) का सरल मान क्या है?
What is the simplified value of (\(\sec x-\tan x\)\(\sec x+\tan x\))?
#trigonometric-identities
#difference-of-squares
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A \(\tan^2 x\)
B \(\sec^2 x\)
C (0)
D (1)
Explanation opens after your attempt
Step 1
Concept
It becomes \(\sec^2 x-\tan^2 x\). By identity, its value is (1).
Step 2
Why this answer is correct
The correct answer is D. (1). It becomes \(\sec^2 x-\tan^2 x\). By identity, its value is (1).
Step 3
Exam Tip
यह \(\sec^2 x-\tan^2 x\) बनता है। पहचान के अनुसार इसका मान (1) है।
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(\(\cosec x-\cot x\)\(\cosec x+\cot x\)) का सरल मान क्या है?
What is the simplified value of (\(\cosec x-\cot x\)\(\cosec x+\cot x\))?
#trigonometric-identities
#difference-of-squares
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A (1)
B \(\cot^2 x\)
C \(\cosec^2 x\)
D (0)
Explanation opens after your attempt
Step 1
Concept
It becomes \(\cosec^2 x-\cot^2 x\). By the standard identity, its value is (1).
Step 2
Why this answer is correct
The correct answer is A. (1). It becomes \(\cosec^2 x-\cot^2 x\). By the standard identity, its value is (1).
Step 3
Exam Tip
यह \(\cosec^2 x-\cot^2 x\) बनता है। मानक पहचान से इसका मान (1) है।
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यदि \(\sin x+\cos x=1\), तो \(\sin x\cos x\) का मान क्या है?
If \(\sin x+\cos x=1\), what is the value of \(\sin x\cos x\)?
#sine-cosine
#identity
#product
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A \(\frac{1}{2}\)
B (0)
C (1)
D -\(\frac{1}{2}\)
Explanation opens after your attempt
Step 1
Concept
Squaring gives \(1+2\sin x\cos x=1\). Therefore, \(\sin x\cos x=0\).
Step 2
Why this answer is correct
The correct answer is B. (0). Squaring gives \(1+2\sin x\cos x=1\). Therefore, \(\sin x\cos x=0\).
Step 3
Exam Tip
वर्ग करने पर \(1+2\sin x\cos x=1\) मिलता है। इसलिए \(\sin x\cos x=0\)।
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यदि \(\sin x-\cos x=0\), तो \(\tan x\) का मान क्या है?
If \(\sin x-\cos x=0\), what is the value of \(\tan x\)?
#tangent
#sine-cosine
#identity
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A (0)
B (-1)
C (1)
D (2)
Explanation opens after your attempt
Step 1
Concept
When \(\sin x=\cos x\), \(\frac{\sin x}{\cos x}=1\). Hence \(\tan x=1\).
Step 2
Why this answer is correct
The correct answer is C. (1). When \(\sin x=\cos x\), \(\frac{\sin x}{\cos x}=1\). Hence \(\tan x=1\).
Step 3
Exam Tip
\(\sin x=\cos x\) होने पर \(\frac{\sin x}{\cos x}=1\) होगा। इसलिए \(\tan x=1\)।
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यदि \(\sin^2 x=\frac{1}{4}\) और (x) प्रथम चतुर्थांश में है, तो \(\cos x\) का मान क्या है?
If \(\sin^2 x=\frac{1}{4}\) and (x) is in the first quadrant, what is the value of \(\cos x\)?
#identity
#first-quadrant
#standard-values
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A \(\frac{1}{2}\)
B \(-\frac{\sqrt{3}}{2}\)
C \(\frac{\sqrt{3}}{2}\)
D \(\frac{3}{4}\)
Explanation opens after your attempt
Correct Answer
C. \(\frac{\sqrt{3}}{2}\)
Step 1
Concept
\(\cos^2 x=1-\frac{1}{4}=\frac{3}{4}\). In the first quadrant, \(\cos x=\frac{\sqrt{3}}{2}\) is taken.
Step 2
Why this answer is correct
The correct answer is C. \(\frac{\sqrt{3}}{2}\). \(\cos^2 x=1-\frac{1}{4}=\frac{3}{4}\). In the first quadrant, \(\cos x=\frac{\sqrt{3}}{2}\) is taken.
Step 3
Exam Tip
\(\cos^2 x=1-\frac{1}{4}=\frac{3}{4}\) होगा। प्रथम चतुर्थांश में \(\cos x=\frac{\sqrt{3}}{2}\) लिया जाता है।
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यदि \(\cos^2 x=\frac{9}{16}\) और (x) चौथे चतुर्थांश में है, तो \(\sin x\) का मान क्या है?
If \(\cos^2 x=\frac{9}{16}\) and (x) is in the fourth quadrant, what is the value of \(\sin x\)?
#identity
#quadrants
#sine
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A \(\frac{\sqrt{7}}{4}\)
B \(-\frac{\sqrt{7}}{4}\)
C \(\frac{3}{4}\)
D \(-\frac{3}{4}\)
Explanation opens after your attempt
Correct Answer
B. \(-\frac{\sqrt{7}}{4}\)
Step 1
Concept
\(\sin^2 x=1-\frac{9}{16}=\frac{7}{16}\). In the fourth quadrant, \(\sin x\) is negative.
Step 2
Why this answer is correct
The correct answer is B. \(-\frac{\sqrt{7}}{4}\). \(\sin^2 x=1-\frac{9}{16}=\frac{7}{16}\). In the fourth quadrant, \(\sin x\) is negative.
Step 3
Exam Tip
\(\sin^2 x=1-\frac{9}{16}=\frac{7}{16}\) होता है। चौथे चतुर्थांश में \(\sin x\) ऋणात्मक होता है।
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फलन \(\cos 5x\) का काल क्या है?
What is the period of the function \(\cos 5x\)?
#period
#cosine
#transformation
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A \(\frac{2\pi}{5}\)
B \(\frac{\pi}{5}\)
C \(5\pi\)
D \(2\pi\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{2\pi}{5}\)
Step 1
Concept
The period of \(\cos kx\) is \(\frac{2\pi}{k}\). Here (k=5), so the period is \(\frac{2\pi}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{2\pi}{5}\). The period of \(\cos kx\) is \(\frac{2\pi}{k}\). Here (k=5), so the period is \(\frac{2\pi}{5}\).
Step 3
Exam Tip
\(\cos kx\) का काल \(\frac{2\pi}{k}\) होता है। यहाँ (k=5), इसलिए काल \(\frac{2\pi}{5}\) है।
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फलन \(4\sin 2x\) का आयाम क्या है?
What is the amplitude of the function \(4\sin 2x\)?
#amplitude
#sine
#transformation
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A (2)
B (4)
C \(\frac{1}{2}\)
D (8)
Explanation opens after your attempt
Step 1
Concept
The amplitude of \(a\sin bx\) is (|a|). Here (|a|=4).
Step 2
Why this answer is correct
The correct answer is B. (4). The amplitude of \(a\sin bx\) is (|a|). Here (|a|=4).
Step 3
Exam Tip
फलन \(a\sin bx\) का आयाम (|a|) होता है। यहाँ (|a|=4) है।
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फलन \(-3\cos x\) का न्यूनतम मान क्या है?
What is the minimum value of the function \(-3\cos x\)?
#minimum-value
#cosine
#range
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A (0)
B (3)
C (-3)
D (-1)
Explanation opens after your attempt
Step 1
Concept
The maximum value of \(\cos x\) is (1). Therefore, the minimum value of \(-3\cos x\) is (-3).
Step 2
Why this answer is correct
The correct answer is C. (-3). The maximum value of \(\cos x\) is (1). Therefore, the minimum value of \(-3\cos x\) is (-3).
Step 3
Exam Tip
\(\cos x\) का अधिकतम मान (1) है। इसलिए \(-3\cos x\) का न्यूनतम मान (-3) होगा।
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फलन \(2+\sin x\) का परिसर क्या है?
What is the range of the function \(2+\sin x\)?
#range
#sine
#vertical-shift
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A ([0,2])
B ([-1,1])
C ([1,3])
D ([2,3])
Explanation opens after your attempt
Correct Answer
C. ([1,3])
Step 1
Concept
The range of \(\sin x\) is ([-1,1]). Adding (2) changes the range to ([1,3]).
Step 2
Why this answer is correct
The correct answer is C. ([1,3]). The range of \(\sin x\) is ([-1,1]). Adding (2) changes the range to ([1,3]).
Step 3
Exam Tip
\(\sin x\) का परिसर ([-1,1]) है। (2) जोड़ने पर परिसर ([1,3]) हो जाता है।
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फलन \(1-\cos x\) का परिसर क्या है?
What is the range of the function \(1-\cos x\)?
#range
#cosine
#vertical-shift
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A ([-1,1])
B ([0,2])
C ([1,2])
D ([-2,0])
Explanation opens after your attempt
Correct Answer
B. ([0,2])
Step 1
Concept
The value of \(\cos x\) lies in ([-1,1]). Therefore, the range of \(1-\cos x\) is ([0,2]).
Step 2
Why this answer is correct
The correct answer is B. ([0,2]). The value of \(\cos x\) lies in ([-1,1]). Therefore, the range of \(1-\cos x\) is ([0,2]).
Step 3
Exam Tip
\(\cos x\) का मान ([-1,1]) में होता है। इसलिए \(1-\cos x\) का परिसर ([0,2]) है।
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\(\frac{\sin x+\cos x}{\sin x-\cos x}\) में यदि \(\tan x=3\), तो मान क्या है?
If \(\tan x=3\), what is the value of \(\frac{\sin x+\cos x}{\sin x-\cos x}\)?
#tangent
#algebraic-simplification
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A (2)
B (-2)
C \(\frac{1}{2}\)
D \(-\frac{1}{2}\)
Explanation opens after your attempt
Step 1
Concept
Dividing numerator and denominator by \(\cos x\) gives \(\frac{\tan x+1}{\tan x-1}\). Substituting \(\tan x=3\) gives (2).
Step 2
Why this answer is correct
The correct answer is A. (2). Dividing numerator and denominator by \(\cos x\) gives \(\frac{\tan x+1}{\tan x-1}\). Substituting \(\tan x=3\) gives (2).
Step 3
Exam Tip
अंश और हर को \(\cos x\) से भाग देने पर \(\frac{\tan x+1}{\tan x-1}\) मिलता है। \(\tan x=3\) रखने पर मान (2) है।
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\(\frac{\sec x+\tan x}{\sec x-\tan x}\) में यदि \(\sec x=2\) और \(\tan x=\sqrt{3}\), तो मान क्या है?
If \(\sec x=2\) and \(\tan x=\sqrt{3}\), what is the value of \(\frac{\sec x+\tan x}{\sec x-\tan x}\)?
#secant
#tangent
#rationalisation
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A \(7+4\sqrt{3}\)
B \(7-4\sqrt{3}\)
C \(4+\sqrt{3}\)
D \(2+\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
A. \(7+4\sqrt{3}\)
Step 1
Concept
The value is \(\frac{2+\sqrt{3}}{2-\sqrt{3}}\). Rationalising the denominator gives \(7+4\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(7+4\sqrt{3}\). The value is \(\frac{2+\sqrt{3}}{2-\sqrt{3}}\). Rationalising the denominator gives \(7+4\sqrt{3}\).
Step 3
Exam Tip
मान \(\frac{2+\sqrt{3}}{2-\sqrt{3}}\) है। हर को परिमेय करने पर \(7+4\sqrt{3}\) मिलता है।
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यदि \(\sec x+\tan x=5\), तो \(\sec x-\tan x\) का मान क्या है?
If \(\sec x+\tan x=5\), what is the value of \(\sec x-\tan x\)?
#secant-tangent
#identity
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A (5)
B \(\frac{1}{5}\)
C (4)
D (-5)
Explanation opens after your attempt
Correct Answer
B. \(\frac{1}{5}\)
Step 1
Concept
Since (\(\sec x+\tan x\)\(\sec x-\tan x\)=1). Therefore, the other factor is \(\frac{1}{5}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{1}{5}\). Since (\(\sec x+\tan x\)\(\sec x-\tan x\)=1). Therefore, the other factor is \(\frac{1}{5}\).
Step 3
Exam Tip
क्योंकि (\(\sec x+\tan x\)\(\sec x-\tan x\)=1)। इसलिए दूसरा गुणनखंड \(\frac{1}{5}\) होगा।
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यदि \(\cosec x-\cot x=\frac{1}{3}\), तो \(\cosec x+\cot x\) का मान क्या है?
If \(\cosec x-\cot x=\frac{1}{3}\), what is the value of \(\cosec x+\cot x\)?
#cosecant-cotangent
#identity
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A (1)
B (3)
C \(\frac{1}{3}\)
D (9)
Explanation opens after your attempt
Step 1
Concept
Since (\(\cosec x-\cot x\)\(\cosec x+\cot x\)=1). Hence \(\cosec x+\cot x=3\).
Step 2
Why this answer is correct
The correct answer is B. (3). Since (\(\cosec x-\cot x\)\(\cosec x+\cot x\)=1). Hence \(\cosec x+\cot x=3\).
Step 3
Exam Tip
क्योंकि (\(\cosec x-\cot x\)\(\cosec x+\cot x\)=1)। इसलिए \(\cosec x+\cot x=3\)।
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\(\sin^4 x-\cos^4 x\) का सरल मान क्या है?
What is the simplified value of \(\sin^4 x-\cos^4 x\)?
#trigonometric-identities
#powers
#simplification
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A \(\sin^2 x+\cos^2 x\)
B \(\cos^2 x-\sin^2 x\)
C \(\sin^2 x-\cos^2 x\)
D (1)
Explanation opens after your attempt
Correct Answer
C. \(\sin^2 x-\cos^2 x\)
Step 1
Concept
Write it as (\(\sin^2 x-\cos^2 x\)\(\sin^2 x+\cos^2 x\)). The second factor is (1).
Step 2
Why this answer is correct
The correct answer is C. \(\sin^2 x-\cos^2 x\). Write it as (\(\sin^2 x-\cos^2 x\)\(\sin^2 x+\cos^2 x\)). The second factor is (1).
Step 3
Exam Tip
इसे (\(\sin^2 x-\cos^2 x\)\(\sin^2 x+\cos^2 x\)) लिखें। दूसरा गुणनखंड (1) है।
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\(\frac{1+\tan^2 x}{1+\cot^2 x}\) का सरल मान क्या है?
What is the simplified value of \(\frac{1+\tan^2 x}{1+\cot^2 x}\)?
#trigonometric-identities
#tangent
#cotangent
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A \(\tan^2 x\)
B \(\cot^2 x\)
C \(\sec^2 x\)
D \(\cosec^2 x\)
Explanation opens after your attempt
Correct Answer
A. \(\tan^2 x\)
Step 1
Concept
The numerator is \(1+\tan^2 x=\sec^2 x\) and the denominator is \(1+\cot^2 x=\cosec^2 x\). Their ratio is \(\frac{\sec^2 x}{\cosec^2 x}=\tan^2 x\).
Step 2
Why this answer is correct
The correct answer is A. \(\tan^2 x\). The numerator is \(1+\tan^2 x=\sec^2 x\) and the denominator is \(1+\cot^2 x=\cosec^2 x\). Their ratio is \(\frac{\sec^2 x}{\cosec^2 x}=\tan^2 x\).
Step 3
Exam Tip
ऊपर \(1+\tan^2 x=\sec^2 x\) और नीचे \(1+\cot^2 x=\cosec^2 x\) है। अनुपात \(\frac{\sec^2 x}{\cosec^2 x}=\tan^2 x\) होता है।
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\(\frac{1+\sin x}{1-\sin x}\) को \(\sec x\) और \(\tan x\) के रूप में किसके बराबर लिखा जा सकता है?
How can \(\frac{1+\sin x}{1-\sin x}\) be written in terms of \(\sec x\) and \(\tan x\)?
#trigonometric-identities
#secant
#tangent
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A (\(\sec x-\tan x\)2 )
B (\(\sec x+\tan x\)2 )
C \(\sec^2 x-\tan^2 x\)
D \(\sec x+\tan^2 x\)
Explanation opens after your attempt
Correct Answer
B. (\(\sec x+\tan x\)2 )
Step 1
Concept
Rationalising the denominator gives (\frac{\(1+\sin x\)2 }{\cos-2 x}). This equals (\(\sec x+\tan x\)2 ).
Step 2
Why this answer is correct
The correct answer is B. (\(\sec x+\tan x\)2 ). Rationalising the denominator gives (\frac{\(1+\sin x\)2 }{\cos-2 x}). This equals (\(\sec x+\tan x\)2 ).
Step 3
Exam Tip
हर को परिमेय करने पर (\frac{\(1+\sin x\)2 }{\cos-2 x}) मिलता है। यह (\(\sec x+\tan x\)2 ) के बराबर है।
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\(\frac{1-\cos x}{1+\cos x}\) किसके बराबर है?
What is \(\frac{1-\cos x}{1+\cos x}\) equal to?
#half-angle
#cosine
#tangent
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A \(\cot^2 \frac{x}{2}\)
B \(\sec^2 x\)
C \(\tan^2 \frac{x}{2}\)
D \(\sin^2 x\)
Explanation opens after your attempt
Correct Answer
C. \(\tan^2 \frac{x}{2}\)
Step 1
Concept
By the half-angle identity, \(\tan^2 \frac{x}{2}=\frac{1-\cos x}{1+\cos x}\). Remember half-angle forms separately.
Step 2
Why this answer is correct
The correct answer is C. \(\tan^2 \frac{x}{2}\). By the half-angle identity, \(\tan^2 \frac{x}{2}=\frac{1-\cos x}{1+\cos x}\). Remember half-angle forms separately.
Step 3
Exam Tip
अर्ध कोण पहचान के अनुसार \(\tan^2 \frac{x}{2}=\frac{1-\cos x}{1+\cos x}\)। अर्ध कोण रूपों को अलग से याद रखें।
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\(\frac{\sin x}{1+\cos x}\) किसके बराबर है?
What is \(\frac{\sin x}{1+\cos x}\) equal to?
#half-angle
#sine
#cosine
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A \(\cot \frac{x}{2}\)
B \(\tan \frac{x}{2}\)
C \(\sec x\)
D \(\cosec x\)
Explanation opens after your attempt
Correct Answer
B. \(\tan \frac{x}{2}\)
Step 1
Concept
The standard half-angle identity is \(\tan \frac{x}{2}=\frac{\sin x}{1+\cos x}\). Identify the half-angle form quickly.
Step 2
Why this answer is correct
The correct answer is B. \(\tan \frac{x}{2}\). The standard half-angle identity is \(\tan \frac{x}{2}=\frac{\sin x}{1+\cos x}\). Identify the half-angle form quickly.
Step 3
Exam Tip
मानक अर्ध कोण पहचान \(\tan \frac{x}{2}=\frac{\sin x}{1+\cos x}\) है। ऐसे रूप में अर्ध कोण तुरंत पहचानें।
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यदि \(\sin x+\cos x=\frac{3}{2}\), तो (\(\sin x-\cos x\)2 ) का मान क्या है?
If \(\sin x+\cos x=\frac{3}{2}\), what is the value of (\(\sin x-\cos x\)2 )?
#trigonometric-identities
#sine-cosine
#square
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A \(\frac{1}{4}\)
B \(\frac{7}{4}\)
C \(\frac{9}{4}\)
D \(\frac{3}{4}\)
Explanation opens after your attempt
Correct Answer
B. \(\frac{7}{4}\)
Step 1
Concept
First, (\(\sin x+\cos x\)2 =1+2\sin x\cos x) gives \(\sin x\cos x=\frac{5}{8}\). Then (\(\sin x-\cos x\)2 =1-2\sin x\cos x=\frac{7}{4}).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{7}{4}\). First, (\(\sin x+\cos x\)2 =1+2\sin x\cos x) gives \(\sin x\cos x=\frac{5}{8}\). Then (\(\sin x-\cos x\)2 =1-2\sin x\cos x=\frac{7}{4}).
Step 3
Exam Tip
पहले (\(\sin x+\cos x\)2 =1+2\sin x\cos x) से \(\sin x\cos x=\frac{5}{8}\) मिलता है। फिर (\(\sin x-\cos x\)2 =1-2\sin x\cos x=\frac{7}{4})।
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