यदि \(\sin x-\cos x=\frac{1}{2}\), तो \(\sin x\cos x\) का मान क्या है?
If \(\sin x-\cos x=\frac{1}{2}\), what is the value of \(\sin x\cos x\)?
#trigonometric-identities
#sine-cosine
#product
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A \(\frac{3}{8}\)
B \(\frac{1}{4}\)
C -\(\frac{3}{8}\)
D \(\frac{5}{8}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{3}{8}\)
Step 1
Concept
Use (\(\sin x-\cos x\)2 =1-2\sin x\cos x). This gives \(\sin x\cos x=\frac{3}{8}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{3}{8}\). Use (\(\sin x-\cos x\)2 =1-2\sin x\cos x). This gives \(\sin x\cos x=\frac{3}{8}\).
Step 3
Exam Tip
(\(\sin x-\cos x\)2 =1-2\sin x\cos x) लगाएँ। इससे \(\sin x\cos x=\frac{3}{8}\) मिलता है।
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यदि \(\sin x+\cos x=\frac{6}{5}\), तो \(\sin x-\cos x\) के वर्ग का मान क्या है?
If \(\sin x+\cos x=\frac{6}{5}\), what is the value of the square of \(\sin x-\cos x\)?
#sine-cosine
#square
#identity
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A \(\frac{11}{25}\)
B \(\frac{36}{25}\)
C \(\frac{14}{25}\)
D -\(\frac{11}{25}\)
Explanation opens after your attempt
Correct Answer
C. \(\frac{14}{25}\)
Step 1
Concept
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।
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यदि \(\tan x+\cot x=5\), तो \(\tan^2 x+\cot^2 x\) का मान क्या है?
If \(\tan x+\cot x=5\), what is the value of \(\tan^2 x+\cot^2 x\)?
#tangent
#cotangent
#identity
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A (25)
B (23)
C (24)
D (27)
Explanation opens after your attempt
Step 1
Concept
(\(\tan x+\cot x\)2 =\tan-2 x+\cot-2 x+2). Therefore, the value is (25-2=23).
Step 2
Why this answer is correct
The correct answer is B. (23). (\(\tan x+\cot x\)2 =\tan-2 x+\cot-2 x+2). Therefore, the value is (25-2=23).
Step 3
Exam Tip
(\(\tan x+\cot x\)2 =\tan-2 x+\cot-2 x+2) होता है। इसलिए मान (25-2=23) है।
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यदि \(\sec x-\tan x=\frac{1}{4}\), तो \(\sec x+\tan x\) का मान क्या है?
If \(\sec x-\tan x=\frac{1}{4}\), what is the value of \(\sec x+\tan x\)?
#secant
#tangent
#reciprocal-product
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A \(\frac{1}{4}\)
B (2)
C (4)
D (16)
Explanation opens after your attempt
Step 1
Concept
Since (\(\sec x-\tan x\)\(\sec x+\tan x\)=1). Hence the other factor is (4).
Step 2
Why this answer is correct
The correct answer is C. (4). Since (\(\sec x-\tan x\)\(\sec x+\tan x\)=1). Hence the other factor is (4).
Step 3
Exam Tip
क्योंकि (\(\sec x-\tan x\)\(\sec x+\tan x\)=1) होता है। इसलिए दूसरा गुणनखंड (4) होगा।
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यदि \(\cosec x+\cot x=6\), तो \(\cosec x-\cot x\) का मान क्या है?
If \(\cosec x+\cot x=6\), what is the value of \(\cosec x-\cot x\)?
#cosecant
#cotangent
#identity
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A \(\frac{1}{6}\)
B (6)
C \(\frac{1}{3}\)
D (36)
Explanation opens after your attempt
Correct Answer
A. \(\frac{1}{6}\)
Step 1
Concept
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।
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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 \(\tan \frac{x}{2}\)
B \(\cot \frac{x}{2}\)
C \(\sec x\)
D \(\cosec x\)
Explanation opens after your attempt
Correct Answer
B. \(\cot \frac{x}{2}\)
Step 1
Concept
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।
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\(\frac{1+\cos x}{\sin x}\) किसके बराबर है?
What is \(\frac{1+\cos x}{\sin x}\) equal to?
#half-angle
#cotangent
#identity
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A \(\tan \frac{x}{2}\)
B \(\sec x\)
C \(\cot \frac{x}{2}\)
D \(\cosec x\)
Explanation opens after your attempt
Correct Answer
C. \(\cot \frac{x}{2}\)
Step 1
Concept
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।
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फलन \(4-3\sin x\) का अधिकतम मान क्या है?
What is the maximum value of the function \(4-3\sin x\)?
#maximum-value
#sine
#range
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A (1)
B (4)
C (7)
D (3)
Explanation opens after your attempt
Step 1
Concept
The minimum value of \(\sin x\) is (-1). Hence the maximum value is (4-3(-1)=7).
Step 2
Why this answer is correct
The correct answer is C. (7). The minimum value of \(\sin x\) is (-1). Hence the maximum value is (4-3(-1)=7).
Step 3
Exam Tip
\(\sin x\) का न्यूनतम मान (-1) है। इसलिए अधिकतम मान (4-3(-1)=7) होगा।
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फलन \(2+5\cos x\) का न्यूनतम मान क्या है?
What is the minimum value of the function \(2+5\cos x\)?
#minimum-value
#cosine
#range
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A -(3)
B (2)
C (5)
D (7)
Explanation opens after your attempt
Step 1
Concept
The minimum value of \(\cos x\) is (-1). Therefore, the minimum value is (2+5(-1)=-3).
Step 2
Why this answer is correct
The correct answer is A. -(3). The minimum value of \(\cos x\) is (-1). Therefore, the minimum value is (2+5(-1)=-3).
Step 3
Exam Tip
\(\cos x\) का न्यूनतम मान (-1) होता है। इसलिए न्यूनतम मान (2+5(-1)=-3) है।
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फलन \(3\sin 2x-1\) का परिसर क्या है?
What is the range of the function \(3\sin 2x-1\)?
#range
#sine
#transformation
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A ([-3,3])
B ([-4,2])
C ([-1,3])
D ([0,2])
Explanation opens after your attempt
Correct Answer
B. ([-4,2])
Step 1
Concept
The range of \(\sin 2x\) is ([-1,1]). Multiplying by (3) and adding (-1) gives ([-4,2]).
Step 2
Why this answer is correct
The correct answer is B. ([-4,2]). The range of \(\sin 2x\) is ([-1,1]). Multiplying by (3) and adding (-1) gives ([-4,2]).
Step 3
Exam Tip
\(\sin 2x\) का परिसर ([-1,1]) है। (3) से गुणा और (-1) जोड़ने पर परिसर ([-4,2]) मिलता है।
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फलन \(2\cos 3x+4\) का काल क्या है?
What is the period of the function \(2\cos 3x+4\)?
#period
#cosine
#transformation
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A \(2\pi\)
B \(\frac{2\pi}{3}\)
C \(\frac{\pi}{3}\)
D \(3\pi\)
Explanation opens after your attempt
Correct Answer
B. \(\frac{2\pi}{3}\)
Step 1
Concept
Vertical shift and amplitude do not change the period. The period of \(\cos 3x\) is \(\frac{2\pi}{3}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{2\pi}{3}\). Vertical shift and amplitude do not change the period. The period of \(\cos 3x\) is \(\frac{2\pi}{3}\).
Step 3
Exam Tip
ऊर्ध्व बदलाव और आयाम काल नहीं बदलते। \(\cos 3x\) का काल \(\frac{2\pi}{3}\) है।
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फलन (\tan(2x)) का मूल काल क्या है?
What is the fundamental period of the function (\tan(2x))?
#period
#tangent
#fundamental-period
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A \(\frac{\pi}{2}\)
B \(\pi\)
C \(2\pi\)
D \(\frac{2\pi}{3}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{\pi}{2}\)
Step 1
Concept
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।
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यदि (x) दूसरे चतुर्थांश में है और \(\cos x=-\frac{3}{5}\), तो \(\tan x\) का मान क्या है?
If (x) is in the second quadrant and \(\cos x=-\frac{3}{5}\), what is the value of \(\tan x\)?
#quadrants
#tangent
#identity
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A \(\frac{4}{3}\)
B -\(\frac{4}{3}\)
C \(\frac{3}{4}\)
D -\(\frac{3}{4}\)
Explanation opens after your attempt
Correct Answer
B. -\(\frac{4}{3}\)
Step 1
Concept
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।
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यदि (x) तीसरे चतुर्थांश में है और \(\sin x=-\frac{5}{13}\), तो \(\sec x\) का मान क्या है?
If (x) is in the third quadrant and \(\sin x=-\frac{5}{13}\), what is the value of \(\sec x\)?
#quadrants
#secant
#sine
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A \(\frac{13}{12}\)
B -\(\frac{13}{12}\)
C -\(\frac{12}{13}\)
D \(\frac{12}{13}\)
Explanation opens after your attempt
Correct Answer
B. -\(\frac{13}{12}\)
Step 1
Concept
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।
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यदि (x) चौथे चतुर्थांश में है और \(\tan x=-\frac{24}{7}\), तो \(\cos x\) का मान क्या है?
If (x) is in the fourth quadrant and \(\tan x=-\frac{24}{7}\), what is the value of \(\cos x\)?
#quadrants
#cosine
#tangent
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A \(\frac{7}{25}\)
B -\(\frac{7}{25}\)
C \(\frac{24}{25}\)
D -\(\frac{24}{25}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{7}{25}\)
Step 1
Concept
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।
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यदि (x) दूसरे चतुर्थांश में है और \(\sec x=-\frac{17}{8}\), तो \(\sin x\) का मान क्या है?
If (x) is in the second quadrant and \(\sec x=-\frac{17}{8}\), what is the value of \(\sin x\)?
#quadrants
#sine
#secant
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A \(\frac{8}{17}\)
B -\(\frac{8}{17}\)
C \(\frac{15}{17}\)
D -\(\frac{15}{17}\)
Explanation opens after your attempt
Correct Answer
C. \(\frac{15}{17}\)
Step 1
Concept
\(\cos x=-\frac{8}{17}\). In the second quadrant, \(\sin x\) is positive, so \(\sin x=\frac{15}{17}\).
Step 2
Why this answer is correct
The correct answer is C. \(\frac{15}{17}\). \(\cos x=-\frac{8}{17}\). In the second quadrant, \(\sin x\) is positive, so \(\sin x=\frac{15}{17}\).
Step 3
Exam Tip
\(\cos x=-\frac{8}{17}\) होगा। दूसरे चतुर्थांश में \(\sin x\) धनात्मक है, इसलिए \(\sin x=\frac{15}{17}\) है।
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(\sin\(2\pi-x\)) किसके बराबर है?
What is (\sin\(2\pi-x\)) equal to?
#allied-angles
#sine
#quadrants
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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
\(2\pi-x\) is related to the fourth quadrant. There \(\sin x\) is negative, so (\sin\(2\pi-x\)=-\sin x).
Step 2
Why this answer is correct
The correct answer is B. -\(\sin x\). \(2\pi-x\) is related to the fourth quadrant. There \(\sin x\) is negative, so (\sin\(2\pi-x\)=-\sin x).
Step 3
Exam Tip
\(2\pi-x\) चौथे चतुर्थांश से संबंधित है। वहाँ \(\sin x\) ऋणात्मक होता है, इसलिए (\sin\(2\pi-x\)=-\sin x)।
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(\cos\(2\pi-x\)) किसके बराबर है?
What is (\cos\(2\pi-x\)) equal to?
#allied-angles
#cosine
#quadrants
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A \(-\cos x\)
B \(\sin x\)
C \(\cos x\)
D -\(\sin x\)
Explanation opens after your attempt
Correct Answer
C. \(\cos x\)
Step 1
Concept
\(2\pi-x\) lies in the fourth quadrant and \(\cos x\) remains positive. Hence (\cos\(2\pi-x\)=\cos x).
Step 2
Why this answer is correct
The correct answer is C. \(\cos x\). \(2\pi-x\) lies in the fourth quadrant and \(\cos x\) remains positive. Hence (\cos\(2\pi-x\)=\cos x).
Step 3
Exam Tip
\(2\pi-x\) चौथे चतुर्थांश में आता है और \(\cos x\) धनात्मक रहता है। इसलिए (\cos\(2\pi-x\)=\cos x)।
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(\tan\(2\pi-x\)) किसके बराबर है?
What is (\tan\(2\pi-x\)) equal to?
#allied-angles
#tangent
#quadrants
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A \(\tan x\)
B -\(\tan x\)
C \(\cot x\)
D -\(\cot x\)
Explanation opens after your attempt
Correct Answer
B. -\(\tan x\)
Step 1
Concept
In the fourth quadrant, \(\tan x\) is negative. Therefore, (\tan\(2\pi-x\)=-\tan x).
Step 2
Why this answer is correct
The correct answer is B. -\(\tan x\). In the fourth quadrant, \(\tan x\) is negative. Therefore, (\tan\(2\pi-x\)=-\tan x).
Step 3
Exam Tip
चौथे चतुर्थांश में \(\tan x\) ऋणात्मक होता है। इसलिए (\tan\(2\pi-x\)=-\tan x)।
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(\sin\(\frac{3\pi}{2}+x\)) किसके बराबर है?
What is (\sin\(\frac{3\pi}{2}+x\)) equal to?
#allied-angles
#sine
#cofunction
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A -\(\cos x\)
B \(\cos x\)
C \(\sin x\)
D -\(\sin x\)
Explanation opens after your attempt
Correct Answer
A. -\(\cos x\)
Step 1
Concept
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।
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(\cos\(\frac{3\pi}{2}-x\)) किसके बराबर है?
What is (\cos\(\frac{3\pi}{2}-x\)) equal to?
#allied-angles
#cosine
#cofunction
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A \(\cos x\)
B \(\sin x\)
C -\(\sin x\)
D -\(\cos x\)
Explanation opens after your attempt
Correct Answer
C. -\(\sin x\)
Step 1
Concept
\(\frac{3\pi}{2}-x\) is related to the third quadrant. \(\cos\) changes to \(\sin\) with a negative sign.
Step 2
Why this answer is correct
The correct answer is C. -\(\sin x\). \(\frac{3\pi}{2}-x\) is related to the third quadrant. \(\cos\) changes to \(\sin\) with a negative sign.
Step 3
Exam Tip
\(\frac{3\pi}{2}-x\) तीसरे चतुर्थांश से जुड़ा है। \(\cos\) बदलकर \(\sin\) होता है और चिन्ह ऋणात्मक रहता है।
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(\tan\(\frac{3\pi}{2}+x\)) किसके बराबर है?
What is (\tan\(\frac{3\pi}{2}+x\)) equal to?
#allied-angles
#tangent
#cotangent
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A \(\tan x\)
B -\(\tan x\)
C -\(\cot x\)
D \(\cot x\)
Explanation opens after your attempt
Correct Answer
C. -\(\cot x\)
Step 1
Concept
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)।
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(\frac{\sin\(\pi-x\)}{\cos\(\pi+x\)}) का सरल मान क्या है?
What is the simplified value of (\frac{\sin\(\pi-x\)}{\cos\(\pi+x\)})?
#allied-angles
#simplification
#tangent
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A -\(\tan x\)
B \(\tan x\)
C -\(\cot x\)
D \(\cot x\)
Explanation opens after your attempt
Correct Answer
A. -\(\tan x\)
Step 1
Concept
(\sin\(\pi-x\)=\sin x) and (\cos\(\pi+x\)=-\cos x). Therefore, the fraction is \(-\tan x\).
Step 2
Why this answer is correct
The correct answer is A. -\(\tan x\). (\sin\(\pi-x\)=\sin x) and (\cos\(\pi+x\)=-\cos x). Therefore, the fraction is \(-\tan x\).
Step 3
Exam Tip
(\sin\(\pi-x\)=\sin x) और (\cos\(\pi+x\)=-\cos x) होता है। इसलिए भिन्न \(-\tan x\) है।
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(\frac{\cos\(2\pi-x\)}{\sin\(\pi+x\)}) का सरल मान क्या है?
What is the simplified value of (\frac{\cos\(2\pi-x\)}{\sin\(\pi+x\)})?
#allied-angles
#simplification
#cotangent
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A \(\tan x\)
B -\(\cot x\)
C \(\cot x\)
D -\(\tan x\)
Explanation opens after your attempt
Correct Answer
B. -\(\cot x\)
Step 1
Concept
(\cos\(2\pi-x\)=\cos x) and (\sin\(\pi+x\)=-\sin x). Hence the value is \(-\cot x\).
Step 2
Why this answer is correct
The correct answer is B. -\(\cot x\). (\cos\(2\pi-x\)=\cos x) and (\sin\(\pi+x\)=-\sin x). Hence the value is \(-\cot x\).
Step 3
Exam Tip
(\cos\(2\pi-x\)=\cos x) और (\sin\(\pi+x\)=-\sin x) है। इसलिए मान \(-\cot x\) है।
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\(\sec^2 x+\cosec^2 x\) को \(\tan x\) और \(\cot x\) के रूप में कैसे लिखा जा सकता है?
How can \(\sec^2 x+\cosec^2 x\) be written in terms of \(\tan x\) and \(\cot x\)?
#trigonometric-identities
#secant
#cosecant
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A \(\tan^2 x+\cot^2 x\)
B \(2+\tan^2 x+\cot^2 x\)
C \(1+\tan x+\cot x\)
D \(\tan x\cot x\)
Explanation opens after your attempt
Correct Answer
B. \(2+\tan^2 x+\cot^2 x\)
Step 1
Concept
Use \(\sec^2 x=1+\tan^2 x\) and \(\cosec^2 x=1+\cot^2 x\). The sum becomes \(2+\tan^2 x+\cot^2 x\).
Step 2
Why this answer is correct
The correct answer is B. \(2+\tan^2 x+\cot^2 x\). Use \(\sec^2 x=1+\tan^2 x\) and \(\cosec^2 x=1+\cot^2 x\). The sum becomes \(2+\tan^2 x+\cot^2 x\).
Step 3
Exam Tip
\(\sec^2 x=1+\tan^2 x\) और \(\cosec^2 x=1+\cot^2 x\) लगाएँ। योग \(2+\tan^2 x+\cot^2 x\) होगा।
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\(\frac{\tan x+\cot x}{\sec x\cosec x}\) का सरल मान क्या है?
What is the simplified value of \(\frac{\tan x+\cot x}{\sec x\cosec x}\)?
#simplification
#tangent
#cotangent
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A (1)
B \(\sin x\cos x\)
C \(\tan x\cot x\)
D (0)
Explanation opens after your attempt
Step 1
Concept
\(\tan x+\cot x=\frac{1}{\sin x\cos x}\) and \(\sec x\cosec x=\frac{1}{\sin x\cos x}\). Hence the ratio is (1).
Step 2
Why this answer is correct
The correct answer is A. (1). \(\tan x+\cot x=\frac{1}{\sin x\cos x}\) and \(\sec x\cosec x=\frac{1}{\sin x\cos x}\). Hence the ratio is (1).
Step 3
Exam Tip
\(\tan x+\cot x=\frac{1}{\sin x\cos x}\) और \(\sec x\cosec x=\frac{1}{\sin x\cos x}\) होता है। इसलिए अनुपात (1) है।
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\(\frac{\sec x}{\tan x}\) का सरल मान क्या है?
What is the simplified value of \(\frac{\sec x}{\tan x}\)?
#simplification
#secant
#tangent
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A \(\sin x\)
B \(\cosec x\)
C \(\cos x\)
D \(\sec x\)
Explanation opens after your attempt
Correct Answer
B. \(\cosec x\)
Step 1
Concept
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।
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\(\frac{\cosec x}{\cot x}\) का सरल मान क्या है?
What is the simplified value of \(\frac{\cosec x}{\cot x}\)?
#simplification
#cosecant
#cotangent
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A \(\sin x\)
B \(\sec x\)
C \(\cos x\)
D \(\cosec x\)
Explanation opens after your attempt
Correct Answer
B. \(\sec x\)
Step 1
Concept
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 \(\frac{1}{\cos x}=\sec x\).
Step 3
Exam Tip
\(\cosec x=\frac{1}{\sin x}\) और \(\cot x=\frac{\cos x}{\sin x}\) रखें। अनुपात \(\frac{1}{\cos x}=\sec x\) होगा।
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(\sin-2 x\(1+\cot^2 x\)) का सरल मान क्या है?
What is the simplified value of (\sin-2 x\(1+\cot^2 x\))?
#trigonometric-identities
#cotangent
#simplification
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A \(\sin^2 x\)
B (1)
C \(\cos^2 x\)
D \(\tan^2 x\)
Explanation opens after your attempt
Step 1
Concept
Since \(1+\cot^2 x=\cosec^2 x\). Therefore, \(\sin^2 x\cosec^2 x=1\).
Step 2
Why this answer is correct
The correct answer is B. (1). Since \(1+\cot^2 x=\cosec^2 x\). Therefore, \(\sin^2 x\cosec^2 x=1\).
Step 3
Exam Tip
क्योंकि \(1+\cot^2 x=\cosec^2 x\)। इसलिए \(\sin^2 x\cosec^2 x=1\) होगा।
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(\cos-2 x\(1+\tan^2 x\)) का सरल मान क्या है?
What is the simplified value of (\cos-2 x\(1+\tan^2 x\))?
#trigonometric-identities
#tangent
#simplification
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A \(\tan^2 x\)
B \(\sin^2 x\)
C (1)
D \(\sec^2 x\)
Explanation opens after your attempt
Step 1
Concept
Since \(1+\tan^2 x=\sec^2 x\). Therefore, \(\cos^2 x\sec^2 x=1\).
Step 2
Why this answer is correct
The correct answer is C. (1). Since \(1+\tan^2 x=\sec^2 x\). Therefore, \(\cos^2 x\sec^2 x=1\).
Step 3
Exam Tip
क्योंकि \(1+\tan^2 x=\sec^2 x\)। इसलिए \(\cos^2 x\sec^2 x=1\) है।
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यदि \(\tan x=\frac{1}{2}\), तो \(\frac{1-\tan^2 x}{1+\tan^2 x}\) का मान क्या है?
If \(\tan x=\frac{1}{2}\), what is the value of \(\frac{1-\tan^2 x}{1+\tan^2 x}\)?
#tangent
#algebraic-simplification
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A \(\frac{5}{3}\)
B \(\frac{3}{5}\)
C \(\frac{1}{5}\)
D \(\frac{4}{5}\)
Explanation opens after your attempt
Correct Answer
B. \(\frac{3}{5}\)
Step 1
Concept
Put \(\tan^2 x=\frac{1}{4}\). Then the value is \(\frac{1-\frac{1}{4}}{1+\frac{1}{4}}=\frac{3}{5}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{3}{5}\). Put \(\tan^2 x=\frac{1}{4}\). Then the value is \(\frac{1-\frac{1}{4}}{1+\frac{1}{4}}=\frac{3}{5}\).
Step 3
Exam Tip
\(\tan^2 x=\frac{1}{4}\) रखें। तब मान \(\frac{1-\frac{1}{4}}{1+\frac{1}{4}}=\frac{3}{5}\) है।
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यदि \(\cot x=\frac{3}{2}\), तो \(\frac{\cot^2 x-1}{\cot^2 x+1}\) का मान क्या है?
If \(\cot x=\frac{3}{2}\), what is the value of \(\frac{\cot^2 x-1}{\cot^2 x+1}\)?
#cotangent
#algebraic-simplification
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A \(\frac{5}{13}\)
B \(\frac{13}{5}\)
C \(\frac{3}{13}\)
D \(\frac{4}{13}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{5}{13}\)
Step 1
Concept
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।
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\(\sin^4 x+\cos^4 x\) किसके बराबर है?
What is \(\sin^4 x+\cos^4 x\) equal to?
#powers
#sine
#cosine
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A \(1+2\sin^2 x\cos^2 x\)
B \(1-2\sin^2 x\cos^2 x\)
C \(\sin^2 x-\cos^2 x\)
D \(2\sin^2 x\cos^2 x\)
Explanation opens after your attempt
Correct Answer
B. \(1-2\sin^2 x\cos^2 x\)
Step 1
Concept
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).
Step 2
Why this answer is correct
The correct answer is B. \(1-2\sin^2 x\cos^2 x\). 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).
Step 3
Exam Tip
(\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।
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यदि \(\sin x\cos x=\frac{1}{4}\), तो \(\sin^4 x+\cos^4 x\) का मान क्या है?
If \(\sin x\cos x=\frac{1}{4}\), what is the value of \(\sin^4 x+\cos^4 x\)?
#powers
#identity
#sine-cosine
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A \(\frac{7}{8}\)
B \(\frac{1}{2}\)
C \(\frac{3}{4}\)
D \(\frac{5}{8}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{7}{8}\)
Step 1
Concept
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।
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यदि \(\sin x+\cos x=\sqrt{3}\), तो \(\tan x+\cot x\) का मान क्या है?
If \(\sin x+\cos x=\sqrt{3}\), what is the value of \(\tan x+\cot x\)?
#identity
#error-check
#sine-cosine
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A (2)
B \(\frac{1}{2}\)
C (4)
D (3)
Explanation opens after your attempt
Step 1
Concept
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।
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यदि \(\sin x+\cos x=\frac{7}{5}\), तो \(\tan x+\cot x\) का मान क्या है?
If \(\sin x+\cos x=\frac{7}{5}\), what is the value of \(\tan x+\cot x\)?
#sine-cosine
#tangent
#cotangent
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A \(\frac{50}{24}\)
B \(\frac{25}{12}\)
C \(\frac{24}{25}\)
D \(\frac{12}{25}\)
Explanation opens after your attempt
Correct Answer
B. \(\frac{25}{12}\)
Step 1
Concept
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 \(\tan x+\cot x=\frac{1}{\sin x\cos x}=\frac{25}{12}\)।
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यदि \(\sin x-\cos x=\frac{1}{3}\), तो \(\tan x+\cot x\) का मान क्या है?
If \(\sin x-\cos x=\frac{1}{3}\), what is the value of \(\tan x+\cot x\)?
#sine-cosine
#identity
#cotangent
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A \(\frac{9}{4}\)
B \(\frac{4}{9}\)
C \(\frac{3}{2}\)
D \(\frac{9}{5}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{9}{4}\)
Step 1
Concept
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 2
Why this answer is correct
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 \(\tan x+\cot x=\frac{9}{4}\)।
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यदि \(\sec x+\tan x=3\), तो \(\sec x\) का मान क्या है?
If \(\sec x+\tan x=3\), what is the value of \(\sec x\)?
#secant
#tangent
#identity
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A \(\frac{5}{3}\)
B \(\frac{3}{5}\)
C \(\frac{4}{3}\)
D \(\frac{10}{3}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{5}{3}\)
Step 1
Concept
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 \(\sec x=\frac{5}{3}\)।
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यदि \(\cosec x+\cot x=4\), तो \(\cot x\) का मान क्या है?
If \(\cosec x+\cot x=4\), what is the value of \(\cot x\)?
#cosecant
#cotangent
#identity
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A \(\frac{17}{8}\)
B \(\frac{15}{8}\)
C \(\frac{8}{15}\)
D \(\frac{8}{17}\)
Explanation opens after your attempt
Correct Answer
B. \(\frac{15}{8}\)
Step 1
Concept
\(\cosec x-\cot x=\frac{1}{4}\). Subtracting gives \(2\cot x=4-\frac{1}{4}\), so \(\cot x=\frac{15}{8}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{15}{8}\). \(\cosec x-\cot x=\frac{1}{4}\). Subtracting gives \(2\cot x=4-\frac{1}{4}\), so \(\cot x=\frac{15}{8}\).
Step 3
Exam Tip
\(\cosec x-\cot x=\frac{1}{4}\) होगा। घटाने पर \(2\cot x=4-\frac{1}{4}\), इसलिए \(\cot x=\frac{15}{8}\)।
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\(\frac{1}{1+\sin x}+\frac{1}{1-\sin x}\) का सरल मान क्या है?
What is the simplified value of \(\frac{1}{1+\sin x}+\frac{1}{1-\sin x}\)?
#rational-expression
#sine
#secant
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A \(2\sec^2 x\)
B \(2\cosec^2 x\)
C \(\sec^2 x\)
D (2)
Explanation opens after your attempt
Correct Answer
A. \(2\sec^2 x\)
Step 1
Concept
Combining denominators gives \(\frac{2}{1-\sin^2 x}\). This is \(\frac{2}{\cos^2 x}=2\sec^2 x\).
Step 2
Why this answer is correct
The correct answer is A. \(2\sec^2 x\). Combining denominators gives \(\frac{2}{1-\sin^2 x}\). This is \(\frac{2}{\cos^2 x}=2\sec^2 x\).
Step 3
Exam Tip
हरों को मिलाने पर \(\frac{2}{1-\sin^2 x}\) मिलेगा। यह \(\frac{2}{\cos^2 x}=2\sec^2 x\) है।
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\(\frac{1}{1+\cos x}+\frac{1}{1-\cos x}\) का सरल मान क्या है?
What is the simplified value of \(\frac{1}{1+\cos x}+\frac{1}{1-\cos x}\)?
#rational-expression
#cosine
#cosecant
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A \(2\sec^2 x\)
B \(\cosec^2 x\)
C \(2\cosec^2 x\)
D (1)
Explanation opens after your attempt
Correct Answer
C. \(2\cosec^2 x\)
Step 1
Concept
Combining denominators gives \(\frac{2}{1-\cos^2 x}\). This is \(\frac{2}{\sin^2 x}=2\cosec^2 x\).
Step 2
Why this answer is correct
The correct answer is C. \(2\cosec^2 x\). Combining denominators gives \(\frac{2}{1-\cos^2 x}\). This is \(\frac{2}{\sin^2 x}=2\cosec^2 x\).
Step 3
Exam Tip
हरों को मिलाने पर \(\frac{2}{1-\cos^2 x}\) मिलता है। यह \(\frac{2}{\sin^2 x}=2\cosec^2 x\) है।
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\(\frac{1}{\sec x+\tan x}\) किसके बराबर है?
What is \(\frac{1}{\sec x+\tan x}\) equal to?
#reciprocal-product
#secant
#tangent
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A \(\sec x+\tan x\)
B \(\sec x-\tan x\)
C \(\tan x-\sec x\)
D \(\frac{1}{\sec x-\tan x}\)
Explanation opens after your attempt
Correct Answer
B. \(\sec x-\tan x\)
Step 1
Concept
Since (\(\sec x+\tan x\)\(\sec x-\tan x\)=1). Therefore, the reciprocal is \(\sec x-\tan x\).
Step 2
Why this answer is correct
The correct answer is B. \(\sec x-\tan x\). Since (\(\sec x+\tan x\)\(\sec x-\tan x\)=1). Therefore, the reciprocal is \(\sec x-\tan x\).
Step 3
Exam Tip
क्योंकि (\(\sec x+\tan x\)\(\sec x-\tan x\)=1)। इसलिए व्युत्क्रम \(\sec x-\tan x\) होगा।
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\(\frac{1}{\cosec x-\cot x}\) किसके बराबर है?
What is \(\frac{1}{\cosec x-\cot x}\) equal to?
#reciprocal-product
#cosecant
#cotangent
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A \(\cosec x-\cot x\)
B \(\cot x-\cosec x\)
C \(\cosec x+\cot x\)
D \(\frac{1}{\cosec x+\cot x}\)
Explanation opens after your attempt
Correct Answer
C. \(\cosec x+\cot x\)
Step 1
Concept
Since (\(\cosec x-\cot x\)\(\cosec x+\cot x\)=1). Hence the required reciprocal is \(\cosec x+\cot x\).
Step 2
Why this answer is correct
The correct answer is C. \(\cosec x+\cot x\). Since (\(\cosec x-\cot x\)\(\cosec x+\cot x\)=1). Hence the required reciprocal is \(\cosec x+\cot x\).
Step 3
Exam Tip
क्योंकि (\(\cosec x-\cot x\)\(\cosec x+\cot x\)=1)। इसलिए आवश्यक व्युत्क्रम \(\cosec x+\cot x\) है।
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फलन \(6-2\sin 4x\) का आयाम क्या है?
What is the amplitude of the function \(6-2\sin 4x\)?
#amplitude
#sine
#transformation
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A (6)
B (4)
C (2)
D (8)
Explanation opens after your attempt
Step 1
Concept
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।
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फलन (5\cos\(\frac{x}{2}\)) का काल क्या है?
What is the period of the function (5\cos\(\frac{x}{2}\))?
#period
#cosine
#transformation
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A \(\pi\)
B \(2\pi\)
C \(4\pi\)
D \(\frac{\pi}{2}\)
Explanation opens after your attempt
Correct Answer
C. \(4\pi\)
Step 1
Concept
The period of \(\cos kx\) is \(\frac{2\pi}{k}\). Here \(k=\frac{1}{2}\), so the period is \(4\pi\).
Step 2
Why this answer is correct
The correct answer is C. \(4\pi\). The period of \(\cos kx\) is \(\frac{2\pi}{k}\). Here \(k=\frac{1}{2}\), so the period is \(4\pi\).
Step 3
Exam Tip
\(\cos kx\) का काल \(\frac{2\pi}{k}\) होता है। यहाँ \(k=\frac{1}{2}\), इसलिए काल \(4\pi\) है।
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फलन \(\sin^2 x\) का मूल काल क्या है?
What is the fundamental period of the function \(\sin^2 x\)?
#period
#sine-square
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A \(2\pi\)
B \(\pi\)
C \(\frac{\pi}{2}\)
D \(4\pi\)
Explanation opens after your attempt
Correct Answer
B. \(\pi\)
Step 1
Concept
(\sin-2 \(x+\pi\)=\sin-2 x). Hence the fundamental period of \(\sin^2 x\) is \(\pi\).
Step 2
Why this answer is correct
The correct answer is B. \(\pi\). (\sin-2 \(x+\pi\)=\sin-2 x). Hence the fundamental period of \(\sin^2 x\) is \(\pi\).
Step 3
Exam Tip
(\sin-2 \(x+\pi\)=\sin-2 x) होता है। इसलिए \(\sin^2 x\) का मूल काल \(\pi\) है।
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फलन \(\cos^2 x\) का परिसर क्या है?
What is the range of the function \(\cos^2 x\)?
#range
#cosine-square
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A ([-1,1])
B ((0,1))
C ([0,1])
D \([1,\infty\))
Explanation opens after your attempt
Correct Answer
C. ([0,1])
Step 1
Concept
The value of \(\cos x\) lies in ([-1,1]). Squaring gives the range ([0,1]).
Step 2
Why this answer is correct
The correct answer is C. ([0,1]). The value of \(\cos x\) lies in ([-1,1]). Squaring gives the range ([0,1]).
Step 3
Exam Tip
\(\cos x\) का मान ([-1,1]) में होता है। वर्ग करने पर परिसर ([0,1]) बनता है।
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यदि \(\sec x-\tan x=\frac{2}{5}\), तो \(\tan x\) का मान क्या है?
If \(\sec x-\tan x=\frac{2}{5}\), what is the value of \(\tan x\)?
#secant
#tangent
#identity
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A \(\frac{21}{20}\)
B \(\frac{20}{21}\)
C \(\frac{29}{20}\)
D \(\frac{5}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{21}{20}\)
Step 1
Concept
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।
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\(\frac{\sin x-\sin^3 x}{\cos^2 x}\) का सरल मान क्या है?
What is the simplified value of \(\frac{\sin x-\sin^3 x}{\cos^2 x}\)?
#simplification
#sine
#identity
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A \(\cos x\)
B \(\tan x\)
C \(\sin x\)
D \(\sec x\)
Explanation opens after your attempt
Correct Answer
C. \(\sin x\)
Step 1
Concept
Write the numerator as (\sin x\(1-\sin^2 x\)). Since \(1-\sin^2 x=\cos^2 x\), the value is \(\sin x\).
Step 2
Why this answer is correct
The correct answer is C. \(\sin x\). Write the numerator as (\sin x\(1-\sin^2 x\)). Since \(1-\sin^2 x=\cos^2 x\), the value is \(\sin x\).
Step 3
Exam Tip
अंश को (\sin x\(1-\sin^2 x\)) लिखें। क्योंकि \(1-\sin^2 x=\cos^2 x\), इसलिए मान \(\sin x\) है।
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फलन \(2-4\cos^2 x\) का परिसर क्या है?
What is the range of the function \(2-4\cos^2 x\)?
#range
#cosine-square
#transformation
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A ([-4,2])
B ([-2,2])
C ([0,2])
D ([-2,6])
Explanation opens after your attempt
Correct Answer
B. ([-2,2])
Step 1
Concept
The range of \(\cos^2 x\) is ([0,1]). Therefore, the range of \(2-4\cos^2 x\) is ([-2,2]).
Step 2
Why this answer is correct
The correct answer is B. ([-2,2]). The range of \(\cos^2 x\) is ([0,1]). Therefore, the range of \(2-4\cos^2 x\) is ([-2,2]).
Step 3
Exam Tip
\(\cos^2 x\) का परिसर ([0,1]) है। इसलिए \(2-4\cos^2 x\) का परिसर ([-2,2]) होगा।
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