The periods of \(\sin 3x\) and \(\cos 5x\) are \( \frac{2\pi}{3} \) and \( \frac{2\pi}{5} \), whose common period is \(2\pi\). In exams, find each period separately first.
Step 2
Why this answer is correct
The correct answer is C. \( 2\pi \). The periods of \(\sin 3x\) and \(\cos 5x\) are \( \frac{2\pi}{3} \) and \( \frac{2\pi}{5} \), whose common period is \(2\pi\). In exams, find each period separately first.
Step 3
Exam Tip
\(\sin 3x\) और \(\cos 5x\) के आवर्त क्रमशः \( \frac{2\pi}{3} \) और \( \frac{2\pi}{5} \) हैं, जिनका सामान्य आवर्त \(2\pi\) है। परीक्षा में गुणांकों के लिए आवर्त अलग-अलग निकालें।
From \(\sin x=\cos x\), we get \(\tan x=1\), so \(x=\frac{\pi}{4},\frac{5\pi}{4}\). In exams, check the given interval carefully.
Step 2
Why this answer is correct
The correct answer is C. (2). From \(\sin x=\cos x\), we get \(\tan x=1\), so \(x=\frac{\pi}{4},\frac{5\pi}{4}\). In exams, check the given interval carefully.
Step 3
Exam Tip
\(\sin x=\cos x\) से \(\tan x=1\) मिलता है, इसलिए \(x=\frac{\pi}{4},\frac{5\pi}{4}\) हैं। परीक्षा में दिए गए अंतराल को ध्यान से देखें।
The minimum value of \(\cos 2x\) is (-1), so (4+3(-1)=1). In exams, remember the range of \(\cos x\) for maximum and minimum.
Step 2
Why this answer is correct
The correct answer is D. (1). The minimum value of \(\cos 2x\) is (-1), so (4+3(-1)=1). In exams, remember the range of \(\cos x\) for maximum and minimum.
Step 3
Exam Tip
\(\cos 2x\) का न्यूनतम मान (-1) है, इसलिए (4+3(-1)=1)। परीक्षा में अधिकतम और न्यूनतम के लिए \(\cos x\) की सीमा याद रखें।
Since \(\sin 70^\circ=\cos 20^\circ\), the sum is \(\sin^2 20^\circ+\cos^2 20^\circ=1\). In exams, identify complementary angles.
Step 2
Why this answer is correct
The correct answer is B. (1). Since \(\sin 70^\circ=\cos 20^\circ\), the sum is \(\sin^2 20^\circ+\cos^2 20^\circ=1\). In exams, identify complementary angles.
Step 3
Exam Tip
क्योंकि \(\sin 70^\circ=\cos 20^\circ\), इसलिए योग \(\sin^2 20^\circ+\cos^2 20^\circ=1\) है। परीक्षा में complementary angles पहचानें।
Both \(\tan\frac{x}{2}\) and \(\cot\frac{x}{2}\) have period \(2\pi\). In exams, use the period of \(\tan ax\) as \( \frac{\pi}{|a|} \).
Step 2
Why this answer is correct
The correct answer is C. \( 2\pi \). Both \(\tan\frac{x}{2}\) and \(\cot\frac{x}{2}\) have period \(2\pi\). In exams, use the period of \(\tan ax\) as \( \frac{\pi}{|a|} \).
Step 3
Exam Tip
\(\tan\frac{x}{2}\) और \(\cot\frac{x}{2}\) दोनों का आवर्त \(2\pi\) है। परीक्षा में \(\tan ax\) का आवर्त \( \frac{\pi}{|a|} \) लगाएं।
We have (\sin x+\cos x=\sqrt{2}\sin\left\(x+\frac{\pi}{4}\right\)). Hence the range is \( [-\sqrt{2},\sqrt{2}] \).
Step 2
Why this answer is correct
The correct answer is D. \( [-\sqrt{2},\sqrt{2}] \). We have (\sin x+\cos x=\sqrt{2}\sin\left\(x+\frac{\pi}{4}\right\)). Hence the range is \( [-\sqrt{2},\sqrt{2}] \).
Step 3
Exam Tip
(\sin x+\cos x=\sqrt{2}\sin\left\(x+\frac{\pi}{4}\right\)) होता है। इसलिए परिसर \( [-\sqrt{2},\sqrt{2}] \) है।
We get \(\cos x=\frac{4}{5}\), so \(\tan x=\frac{\sin x}{\cos x}=\frac{3}{4}\). In exams, decide the sign from the quadrant.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{3}{4} \). We get \(\cos x=\frac{4}{5}\), so \(\tan x=\frac{\sin x}{\cos x}=\frac{3}{4}\). In exams, decide the sign from the quadrant.
Step 3
Exam Tip
\(\cos x=\frac{4}{5}\) होगा, इसलिए \(\tan x=\frac{\sin x}{\cos x}=\frac{3}{4}\)। परीक्षा में चतुर्थांश से चिन्ह तय करें।
For \(\frac{\pi}{2}+x\), sine changes to cosine and remains positive in the second quadrant. Hence the value is \(\cos x\).
Step 2
Why this answer is correct
The correct answer is C. \(\cos x\). For \(\frac{\pi}{2}+x\), sine changes to cosine and remains positive in the second quadrant. Hence the value is \(\cos x\).
Step 3
Exam Tip
\(\frac{\pi}{2}+x\) के लिए sine, cosine में बदलता है और दूसरे चतुर्थांश में धनात्मक रहता है। इसलिए मान \(\cos x\) है।
The formula is \(\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}\). Hence the answer is \(-2\sin 3x\sin x\).
Step 2
Why this answer is correct
The correct answer is B. \(-2\sin 3x\sin x\). The formula is \(\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}\). Hence the answer is \(-2\sin 3x\sin x\).
Step 3
Exam Tip
सूत्र \(\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}\) है। इसलिए उत्तर \(-2\sin 3x\sin x\) है।
In \(|\sin x|\), the negative part becomes positive, so the pattern repeats every \(\pi\). In exams, modulus can reduce the period.
Step 2
Why this answer is correct
The correct answer is D. \( \pi \). In \(|\sin x|\), the negative part becomes positive, so the pattern repeats every \(\pi\). In exams, modulus can reduce the period.
Step 3
Exam Tip
\(|\sin x|\) में ऋणात्मक भाग धनात्मक हो जाता है, इसलिए पैटर्न हर \(\pi\) पर दोहरता है। परीक्षा में modulus से आवर्त घट सकता है।
Since (\sin-2\(x+\pi\)=\sin-2 x), the fundamental period is \(\pi\). In exams, remember the period of squared sine and cosine.
Step 2
Why this answer is correct
The correct answer is A. \( \pi \). Since (\sin-2\(x+\pi\)=\sin-2 x), the fundamental period is \(\pi\). In exams, remember the period of squared sine and cosine.
Step 3
Exam Tip
क्योंकि (\sin-2\(x+\pi\)=\sin-2 x), इसलिए मूल आवर्त \(\pi\) है। परीक्षा में वर्ग वाले sine और cosine का आवर्त याद रखें।
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।
This is the form of (\sin\(75^\circ-15^\circ\)). Hence the value is \(\sin 60^\circ=\frac{\sqrt{3}}{2}\).
Step 2
Why this answer is correct
The correct answer is D. \( \frac{\sqrt{3}}{2} \). This is the form of (\sin\(75^\circ-15^\circ\)). Hence the value is \(\sin 60^\circ=\frac{\sqrt{3}}{2}\).
Step 3
Exam Tip
यह (\sin\(75^\circ-15^\circ\)) का रूप है। इसलिए मान \(\sin 60^\circ=\frac{\sqrt{3}}{2}\) है।
A. \(x=2n\pi+y\) या (x=(2n+1)\pi-y)/\(x=2n\pi+y\) or (x=(2n+1)\pi-y)
Step 1
Concept
For \(\sin x=\sin y\), (x=n\pi+(-1)^n y). This gives the two stated forms.
Step 2
Why this answer is correct
The correct answer is A. \(x=2n\pi+y\) या (x=(2n+1)\pi-y) / \(x=2n\pi+y\) or (x=(2n+1)\pi-y). For \(\sin x=\sin y\), (x=n\pi+(-1)^n y). This gives the two stated forms.
Step 3
Exam Tip
\(\sin x=\sin y\) के लिए (x=n\pi+(-1)^n y) होता है। इसी से दिए गए दोनों रूप मिलते हैं।
Using the formulas for (\tan\left\(\frac{\pi}{4}+x\right\)) and (\tan\left\(\frac{\pi}{4}-x\right\)), the product is (1). This is a useful standard result.
Step 2
Why this answer is correct
The correct answer is D. (1). Using the formulas for (\tan\left\(\frac{\pi}{4}+x\right\)) and (\tan\left\(\frac{\pi}{4}-x\right\)), the product is (1). This is a useful standard result.
Step 3
Exam Tip
(\tan\left\(\frac{\pi}{4}+x\right\)) और (\tan\left\(\frac{\pi}{4}-x\right\)) के सूत्र लगाने पर गुणनफल (1) आता है। यह एक उपयोगी standard result है।
In a right triangle, perpendicular is (7), base is (24), and hypotenuse is (25). Hence \(\sin x=\frac{7}{25}\).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{7}{25} \). In a right triangle, perpendicular is (7), base is (24), and hypotenuse is (25). Hence \(\sin x=\frac{7}{25}\).
Step 3
Exam Tip
समकोण त्रिभुज में लम्ब (7), आधार (24), कर्ण (25) होगा। इसलिए \(\sin x=\frac{7}{25}\)।
The period of \(\cos x\) is \(2\pi\) and that of \(\cos 2x\) is \(\pi\). Their common fundamental period is \(2\pi\).
Step 2
Why this answer is correct
The correct answer is B. \(2\pi\). The period of \(\cos x\) is \(2\pi\) and that of \(\cos 2x\) is \(\pi\). Their common fundamental period is \(2\pi\).
Step 3
Exam Tip
\(\cos x\) का आवर्त \(2\pi\) और \(\cos 2x\) का \(\pi\) है। इनका सामान्य मूल आवर्त \(2\pi\) है।
Since \(\tan x=\frac{\sin x}{\cos x}\), it is undefined when \(\cos x=0\). Thus \(x=\frac{\pi}{2}+n\pi\).
Step 2
Why this answer is correct
The correct answer is D. \(x=\frac{\pi}{2}+n\pi\). Since \(\tan x=\frac{\sin x}{\cos x}\), it is undefined when \(\cos x=0\). Thus \(x=\frac{\pi}{2}+n\pi\).
Step 3
Exam Tip
\(\tan x=\frac{\sin x}{\cos x}\) है, इसलिए \(\cos x=0\) पर यह अपरिभाषित होता है। अतः \(x=\frac{\pi}{2}+n\pi\)।
We use (\sin x+\cos x=\sqrt{2}\sin\left\(x+\frac{\pi}{4}\right\)). The maximum \(\sqrt{2}\) occurs at \(x=\frac{\pi}{4}\).
Step 2
Why this answer is correct
The correct answer is C. \( \frac{\pi}{4} \). We use (\sin x+\cos x=\sqrt{2}\sin\left\(x+\frac{\pi}{4}\right\)). The maximum \(\sqrt{2}\) occurs at \(x=\frac{\pi}{4}\).
Step 3
Exam Tip
(\sin x+\cos x=\sqrt{2}\sin\left\(x+\frac{\pi}{4}\right\)) है। अधिकतम \(\sqrt{2}\) तब मिलता है जब \(x=\frac{\pi}{4}\)।
We have \(\tan x+\cot x=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\). Simplifying gives \(\frac{1}{\sin x\cos x}=\sec x\cosec x\).
Step 2
Why this answer is correct
The correct answer is A. \( \sec x\cosec x \). We have \(\tan x+\cot x=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\). Simplifying gives \(\frac{1}{\sin x\cos x}=\sec x\cosec x\).
Step 3
Exam Tip
\(\tan x+\cot x=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\) है। सरलीकरण से \(\frac{1}{\sin x\cos x}=\sec x\cosec x\) मिलता है।
The period of \(\cos ax\) is \( \frac{2\pi}{|a|} \). Hence the period of \(\cos 4x\) is \( \frac{\pi}{2} \).
Step 2
Why this answer is correct
The correct answer is C. \( \frac{\pi}{2} \). The period of \(\cos ax\) is \( \frac{2\pi}{|a|} \). Hence the period of \(\cos 4x\) is \( \frac{\pi}{2} \).
Step 3
Exam Tip
\(\cos ax\) का आवर्त \( \frac{2\pi}{|a|} \) होता है। इसलिए \(\cos 4x\) का आवर्त \( \frac{\pi}{2} \) है।
In the third quadrant, both \(\sin x\) and \(\cos x\) are negative, while \(\tan x\) is positive. Hence \(\sin x=-\frac{4}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{4}{5} \). In the third quadrant, both \(\sin x\) and \(\cos x\) are negative, while \(\tan x\) is positive. Hence \(\sin x=-\frac{4}{5}\).
Step 3
Exam Tip
तृतीय चतुर्थांश में \(\sin x\) और \(\cos x\) दोनों ऋणात्मक होते हैं, पर \(\tan x\) धनात्मक होता है। इसलिए \(\sin x=-\frac{4}{5}\)।
The triple angle formula is \(\cos 3x=4\cos^3 x-3\cos x\). In exams, do not confuse it with \(\sin 3x\).
Step 2
Why this answer is correct
The correct answer is C. \(4\cos^3 x-3\cos x\). The triple angle formula is \(\cos 3x=4\cos^3 x-3\cos x\). In exams, do not confuse it with \(\sin 3x\).
Step 3
Exam Tip
त्रिगुण कोण सूत्र \(\cos 3x=4\cos^3 x-3\cos x\) है। परीक्षा में इसे \(\sin 3x\) से confuse न करें।
\(\sin x\) is positive in the first and second quadrants. Hence \(x=\frac{\pi}{6},\frac{5\pi}{6}\).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{\pi}{6},\frac{5\pi}{6} \). \(\sin x\) is positive in the first and second quadrants. Hence \(x=\frac{\pi}{6},\frac{5\pi}{6}\).
Step 3
Exam Tip
\(\sin x\) प्रथम और द्वितीय चतुर्थांश में धनात्मक है। इसलिए \(x=\frac{\pi}{6},\frac{5\pi}{6}\)।
\(\cos x\) is negative in the second and third quadrants. Hence \(x=\frac{2\pi}{3},\frac{4\pi}{3}\).
Step 2
Why this answer is correct
The correct answer is B. \( \frac{2\pi}{3},\frac{4\pi}{3} \). \(\cos x\) is negative in the second and third quadrants. Hence \(x=\frac{2\pi}{3},\frac{4\pi}{3}\).
Step 3
Exam Tip
\(\cos x\) द्वितीय और तृतीय चतुर्थांश में ऋणात्मक होता है। इसलिए \(x=\frac{2\pi}{3},\frac{4\pi}{3}\)।
The period of \(\sin 2x\) is \(\pi\) and the period of \(\sin 4x\) is \(\frac{\pi}{2}\). Their common fundamental period is \(\pi\).
Step 2
Why this answer is correct
The correct answer is B. \( \pi \). The period of \(\sin 2x\) is \(\pi\) and the period of \(\sin 4x\) is \(\frac{\pi}{2}\). Their common fundamental period is \(\pi\).
Step 3
Exam Tip
\(\sin 2x\) का आवर्त \(\pi\) और \(\sin 4x\) का आवर्त \(\frac{\pi}{2}\) है। इनका सामान्य मूल आवर्त \(\pi\) है।
In the fourth quadrant, \(\sin \theta\) is negative and \(\cos \theta\) is positive. Hence \(\sin \theta=-\frac{15}{17}\) and \(\tan \theta=-\frac{15}{8}\).
Step 2
Why this answer is correct
The correct answer is C. \(-\frac{15}{8} \). In the fourth quadrant, \(\sin \theta\) is negative and \(\cos \theta\) is positive. Hence \(\sin \theta=-\frac{15}{17}\) and \(\tan \theta=-\frac{15}{8}\).
Step 3
Exam Tip
चतुर्थ चतुर्थांश में \(\sin \theta\) ऋणात्मक और \(\cos \theta\) धनात्मक होता है। इसलिए \(\sin \theta=-\frac{15}{17}\) और \(\tan \theta=-\frac{15}{8}\)।
