The identity \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\) is true for all \(x\in\left[-1,1\right]\). Remember this basic identity for exams.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\pi}{2}\). The identity \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\) is true for all \(x\in\left[-1,1\right]\). Remember this basic identity for exams.
Step 3
Exam Tip
\(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\) सभी \(x\in\left[-1,1\right]\) के लिए सही है। परीक्षा में यह मूल पहचान याद रखें।
The identity \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\) applies because \(\frac{3}{5}\in\left[-1,1\right]\). Use the identity directly in such questions.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\pi}{2}\). The identity \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\) applies because \(\frac{3}{5}\in\left[-1,1\right]\). Use the identity directly in such questions.
Step 3
Exam Tip
पहचान \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\) लागू होती है क्योंकि \(\frac{3}{5}\in\left[-1,1\right]\)। ऐसे प्रश्नों में सीधे पहचान लगाएं।
When (x>0), \(tan^{-1}x+\tan^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{2}\). Always check the sign condition.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\pi}{2}\). When (x>0), \(tan^{-1}x+\tan^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{2}\). Always check the sign condition.
Step 3
Exam Tip
जब (x>0), तब \(tan^{-1}x+\tan^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{2}\)। चिन्ह की शर्त हमेशा देखें।
Here (2>0), so \(tan^{-1}2+\tan^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{2}\). Identify (x) and \(\frac{1}{x}\) in the identity.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{\pi}{2}\). Here (2>0), so \(tan^{-1}2+\tan^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{2}\). Identify (x) and \(\frac{1}{x}\) in the identity.
Step 3
Exam Tip
यहाँ (2>0), इसलिए \(tan^{-1}2+\tan^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{2}\)। पहचान में (x) और \(\frac{1}{x}\) पहचानें।
The identity \(\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}\) is used for all real (x). This is a very useful introductory identity.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\pi}{2}\). The identity \(\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}\) is used for all real (x). This is a very useful introductory identity.
Step 3
Exam Tip
\(\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}\) सभी वास्तविक (x) के लिए लिया जाता है। यह बहुत उपयोगी प्रारंभिक पहचान है।
The identity \(\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}\) holds. For negative (x), the principal value of \(\cot^{-1}x\) is still taken in (\left\(0,\pi\right\)).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\pi}{2}\). The identity \(\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}\) holds. For negative (x), the principal value of \(\cot^{-1}x\) is still taken in (\left\(0,\pi\right\)).
Step 3
Exam Tip
\(\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}\) होता है। ऋणात्मक (x) में भी \(\cot^{-1}x\) का मुख्य मान (\left\(0,\pi\right\)) में लिया जाता है।
The relation \(cos^{-1}\left(-x\right)=\pi-\cos^{-1}x\) is correct because of principal values. Do not treat it as an odd function.
Step 2
Why this answer is correct
The correct answer is B. \(\pi-\cos^{-1}x\). The relation \(cos^{-1}\left(-x\right)=\pi-\cos^{-1}x\) is correct because of principal values. Do not treat it as an odd function.
Step 3
Exam Tip
\(cos^{-1}\left(-x\right)=\pi-\cos^{-1}x\) मुख्य मानों के कारण सही है। इसे विषम फलन की तरह न मानें।
Since \(cot\frac{3\pi}{4}=-1\) and \(frac{3\pi}{4}\in\left(0,\pi\right)\), the principal value is \(frac{3\pi}{4}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{3\pi}{4}\). Since \(cot\frac{3\pi}{4}=-1\) and \(frac{3\pi}{4}\in\left(0,\pi\right)\), the principal value is \(frac{3\pi}{4}\).
Step 3
Exam Tip
\(cot\frac{3\pi}{4}=-1) और \frac{3\pi}{4}\in\left(0,\pi\right)\)। इसलिए मुख्य मान \(\frac{3\pi}{4}\) है।
We have \(cot\frac{5\pi}{6}=-\sqrt{3}\). The principal value of \(cot^{-1}x\) must be taken in \(\left(0,\pi\right)\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{5\pi}{6}\). We have \(cot\frac{5\pi}{6}=-\sqrt{3}\). The principal value of \(cot^{-1}x\) must be taken in \(\left(0,\pi\right)\).
Step 3
Exam Tip
\(cot\frac{5\pi}{6}=-\sqrt{3}\) होता है। \(cot^{-1}x\) का मुख्य मान \(\left(0,\pi\right)\) में लेना चाहिए।
Since \(sec\frac{2\pi}{3}=-2\) and it lies in the principal range of \(sec^{-1}x), the answer is \(\frac{2\pi}{3}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{2\pi}{3}\). Since \(sec\frac{2\pi}{3}=-2\) and it lies in the principal range of \(sec^{-1}x), the answer is \(\frac{2\pi}{3}\).
Step 3
Exam Tip
\(sec\frac{2\pi}{3}=-2\) और यह \(sec^{-1}x\) के मुख्य परिसर में है। इसलिए उत्तर \(\frac{2\pi}{3}\) है।
Since \(sin\frac{5\pi}{6}=\frac{1}{2}\) and \(sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}\). Always choose the angle in the principal range.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{\pi}{6}\). Since \(sin\frac{5\pi}{6}=\frac{1}{2}\) and \(sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}\). Always choose the angle in the principal range.
Step 3
Exam Tip
\(sin\frac{5\pi}{6}=\frac{1}{2}\) और \(sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}\)। हमेशा मुख्य परिसर का कोण चुनें।
The angle \(\frac{5\pi}{6}\) lies in the principal range \(\left[0,\pi\right]\). Hence \(cos^{-1}\left(cos\frac{5\pi}{6}\right)=\frac{5\pi}{6}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{5\pi}{6}\). The angle \(\frac{5\pi}{6}\) lies in the principal range \(\left[0,\pi\right]\). Hence \(cos^{-1}\left(cos\frac{5\pi}{6}\right)=\frac{5\pi}{6}\).
Step 3
Exam Tip
\(\frac{5\pi}{6}\) मुख्य परिसर \(\left[0,\pi\right]\) में है। इसलिए \(cos^{-1}\left(\cos\frac{5\pi}{6}\right)=\frac{5\pi}{6}\)।
The angle \(-\frac{\pi}{3}\) lies in the principal range of \(\tan^{-1}x\). So the value remains the same.
Step 2
Why this answer is correct
The correct answer is A. -\(\frac{\pi}{3}\). The angle \(-\frac{\pi}{3}\) lies in the principal range of \(\tan^{-1}x\). So the value remains the same.
Step 3
Exam Tip
कोण \(-\frac{\pi}{3}\) \(\tan^{-1}x\) के मुख्य परिसर में है। इसलिए मान वही रहेगा।
We have \(tan\frac{5\pi}{6}=-\frac{1}{\sqrt{3}}\). Thus \(tan^{-1}\left(-\frac{1}{\sqrt{3}}\right)=-\frac{\pi}{6}\).
Step 2
Why this answer is correct
The correct answer is B. -\(\frac{\pi}{6}\). We have \(tan\frac{5\pi}{6}=-\frac{1}{\sqrt{3}}\). Thus \(tan^{-1}\left(-\frac{1}{\sqrt{3}}\right)=-\frac{\pi}{6}\).
Step 3
Exam Tip
\(tan\frac{5\pi}{6}=-\frac{1}{\sqrt{3}}\) है। \(tan^{-1}\left(-\frac{1}{\sqrt{3}}\right)=-\frac{\pi}{6}\)।
The statement \(\theta=sin^{-1}\left(\frac{3}{5}\right)\) means \(sin\theta=\frac{3}{5}). Apply the definition directly.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{3}{5}\). The statement \(\theta=sin^{-1}\left(\frac{3}{5}\right)\) means \(sin\theta=\frac{3}{5}). Apply the definition directly.
Step 3
Exam Tip
\(\theta=\sin^{-1}\left(\frac{3}{5}\right)\) का अर्थ \(sin\theta=\frac{3}{5}\) है। परिभाषा को सीधे लगाएं।
The expression \(\cos^{-1}\left(\frac{4}{5}\right)\) is the angle whose cosine is \(\frac{4}{5}\). So \(\cos\theta=\frac{4}{5}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{4}{5}\). The expression \(\cos^{-1}\left(\frac{4}{5}\right)\) is the angle whose cosine is \(\frac{4}{5}\). So \(\cos\theta=\frac{4}{5}\).
Step 3
Exam Tip
\(\cos^{-1}\left(\frac{4}{5}\right)\) वह कोण है जिसका \(\cos\) मान \(\frac{4}{5}\) हो। इसलिए \(\cos\theta=\frac{4}{5}\)।
The statement \(\theta=\tan^{-1}\left(\frac{5}{12}\right)\) means \(\tan\theta=\frac{5}{12}\). Remember the definition of inverse function.
Step 2
Why this answer is correct
The correct answer is C. \(\frac{5}{12}\). The statement \(\theta=\tan^{-1}\left(\frac{5}{12}\right)\) means \(\tan\theta=\frac{5}{12}\). Remember the definition of inverse function.
Step 3
Exam Tip
\(\theta=\tan^{-1}\left(\frac{5}{12}\right)\) का अर्थ \(\tan\theta=\frac{5}{12}\) है। व्युत्क्रम फलन की परिभाषा याद रखें।
Here \(\theta\) is in the principal range and \(\sin\theta=\frac{5}{13}\). From a right triangle, \(\cos\theta=\frac{12}{13}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{12}{13}\). Here \(\theta\) is in the principal range and \(\sin\theta=\frac{5}{13}\). From a right triangle, \(\cos\theta=\frac{12}{13}\).
Step 3
Exam Tip
\(\theta) मुख्य परिसर में है और (\sin\theta=\frac{5}{13}\) है। इसलिए समकोण त्रिभुज से \(\cos\theta=\frac{12}{13}\)।
Here \(\cos\theta=\frac{12}{13}\) and \(\theta\in\left[0,\pi\right]\). In this case, \(\sin\theta=\frac{5}{13}\) is positive.
Step 2
Why this answer is correct
The correct answer is C. \(\frac{5}{13}\). Here \(\cos\theta=\frac{12}{13}\) and \(\theta\in\left[0,\pi\right]\). In this case, \(\sin\theta=\frac{5}{13}\) is positive.
Step 3
Exam Tip
\(\cos\theta=\frac{12}{13}\) और \(\theta\in\left[0,\pi\right]\) है। इस स्थिति में \(\sin\theta=\frac{5}{13}\) धनात्मक होगा।
From \(\tan\theta=\frac{3}{4}\), take opposite side (3) and adjacent side (4). The hypotenuse is (5), so \(\sin\theta=\frac{3}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{3}{5}\). From \(\tan\theta=\frac{3}{4}\), take opposite side (3) and adjacent side (4). The hypotenuse is (5), so \(\sin\theta=\frac{3}{5}\).
Step 3
Exam Tip
\(\tan\theta=\frac{3}{4}\) से लंब (3) और आधार (4) लें। कर्ण (5) होगा, इसलिए \(\sin\theta=\frac{3}{5}\)।
If \(\theta=\cos^{-1}x\), then \(\cos\theta=x\) and \(\theta\in\left[0,\pi\right]\). In this range \(\sin\theta\ge0\), so the value is \(\sqrt{1-x^2}\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{1-x^2}\). If \(\theta=\cos^{-1}x\), then \(\cos\theta=x\) and \(\theta\in\left[0,\pi\right]\). In this range \(\sin\theta\ge0\), so the value is \(\sqrt{1-x^2}\).
Step 3
Exam Tip
यदि \(\theta=\cos^{-1}x\), तो \(\cos\theta=x\) और \(\theta\in\left[0,\pi\right]\)। इस परिसर में \(\sin\theta\ge0\), इसलिए मान \(\sqrt{1-x^2}\) है।
If \(\theta=\sin^{-1}x\), then \(\sin\theta=x\) and \(\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). Here \(cos\theta\ge0\), so \(\sqrt{1-x^2}\) is correct.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{1-x^2}\). If \(\theta=\sin^{-1}x\), then \(\sin\theta=x\) and \(\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). Here \(cos\theta\ge0\), so \(\sqrt{1-x^2}\) is correct.
Step 3
Exam Tip
यदि \(\theta=\sin^{-1}x\), तो \(sin\theta=x\) और \(\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\)। यहाँ \(cos\theta\ge0\), इसलिए \(\sqrt{1-x^2}\) सही है।
If \(\theta=\sin^{-1}\left(\frac{3}{5}\right)\), then \(\sin\theta=\frac{3}{5}\) and \(\cos\theta=\frac{4}{5}\). Hence \(\tan\theta=\frac{3}{4}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{3}{4}\). If \(\theta=\sin^{-1}\left(\frac{3}{5}\right)\), then \(\sin\theta=\frac{3}{5}\) and \(\cos\theta=\frac{4}{5}\). Hence \(\tan\theta=\frac{3}{4}\).
Step 3
Exam Tip
यदि \(\theta=\sin^{-1}\left(\frac{3}{5}\right)\), तो \(\sin\theta=\frac{3}{5}\) और \(\cos\theta=\frac{4}{5}\)। इसलिए \(\tan\theta=\frac{3}{4}\)।
If \(\theta=\cos^{-1}\left(\frac{5}{13}\right)\), then \(\cos\theta=\frac{5}{13}\) and \(\sin\theta=\frac{12}{13}\). Hence \(\cot\theta=\frac{5}{12}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{5}{12}\). If \(\theta=\cos^{-1}\left(\frac{5}{13}\right)\), then \(\cos\theta=\frac{5}{13}\) and \(\sin\theta=\frac{12}{13}\). Hence \(\cot\theta=\frac{5}{12}\).
Step 3
Exam Tip
यदि \(\theta=\cos^{-1}\left(\frac{5}{13}\right)\), तो \(\cos\theta=\frac{5}{13}\) और \(\sin\theta=\frac{12}{13}\)। इसलिए \(\cot\theta=\frac{5}{12}\)।
The function \(\sin^{-1}x\) is increasing on its domain \(\left[-1,1\right]\). In graph questions, check both domain and trend.
Step 2
Why this answer is correct
The correct answer is A. \(\left[-1,1\right]\). The function \(\sin^{-1}x\) is increasing on its domain \(\left[-1,1\right]\). In graph questions, check both domain and trend.
Step 3
Exam Tip
\(\sin^{-1}x\) अपने प्रांत \(\left[-1,1\right]\) में बढ़ता है। ग्राफ संबंधी प्रश्नों में प्रांत और प्रवृत्ति दोनों देखें।
The function \(\cos^{-1}x\) is decreasing on \(\left[-1,1\right]\). As (x) increases, the principal angle decreases.
Step 2
Why this answer is correct
The correct answer is B. घटता हुआ / Decreasing. The function \(\cos^{-1}x\) is decreasing on \(\left[-1,1\right]\). As (x) increases, the principal angle decreases.
Step 3
Exam Tip
\(\cos^{-1}x\) \(\left[-1,1\right]\) पर घटता हुआ फलन है। (x) बढ़ने पर मुख्य कोण घटता है।
A. यह \(\mathbb{R}\) पर बढ़ता है/It is increasing on \(\mathbb{R}\)
Step 1
Concept
The function \(\tan^{-1}x\) is defined and increasing for all real numbers. Its range is (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)).
Step 2
Why this answer is correct
The correct answer is A. यह \(\mathbb{R}\) पर बढ़ता है / It is increasing on \(\mathbb{R}\). The function \(\tan^{-1}x\) is defined and increasing for all real numbers. Its range is (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)).
Step 3
Exam Tip
\(\tan^{-1}x\) सभी वास्तविक संख्याओं पर परिभाषित और बढ़ता हुआ है। इसका परिसर (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)) है।
The function \(\cot^{-1}x\) is decreasing on \(\mathbb{R}\). Its principal value range is (\left\(0,\pi\right\)).
Step 2
Why this answer is correct
The correct answer is B. घटता हुआ / Decreasing. The function \(\cot^{-1}x\) is decreasing on \(\mathbb{R}\). Its principal value range is (\left\(0,\pi\right\)).
Step 3
Exam Tip
\(\cot^{-1}x\) \(\mathbb{R}\) पर घटता हुआ फलन है। इसका मुख्य मान परिसर (\left\(0,\pi\right\)) है।
Since \(\frac{7}{6}\notin\left[-1,1\right]\), \(\sin^{-1}\left(\frac{7}{6}\right)\) is not defined. Check the domain first.
Step 2
Why this answer is correct
The correct answer is B. यह परिभाषित नहीं है / It is not defined. Since \(\frac{7}{6}\notin\left[-1,1\right]\), \(\sin^{-1}\left(\frac{7}{6}\right)\) is not defined. Check the domain first.
Step 3
Exam Tip
\(\frac{7}{6}\notin\left[-1,1\right]\), इसलिए \(\sin^{-1}\left(\frac{7}{6}\right)\) परिभाषित नहीं है। प्रांत पहले जांचें।
The domain of \(\cos^{-1}x\) is \(\left[-1,1\right]\). Since \(\frac{11}{10}>1\), it is not defined.
Step 2
Why this answer is correct
The correct answer is A. परिभाषित नहीं / Not defined. The domain of \(\cos^{-1}x\) is \(\left[-1,1\right]\). Since \(\frac{11}{10}>1\), it is not defined.
Step 3
Exam Tip
\(\cos^{-1}x\) का प्रांत \(\left[-1,1\right]\) है। क्योंकि \(\frac{11}{10}>1\), यह परिभाषित नहीं है।
For \(\sec^{-1}x\), we need \(\left|x\right|\ge1\). Here \(\left|\frac{3}{2}\right|\ge1\), so it is defined.
Step 2
Why this answer is correct
The correct answer is A. यह परिभाषित है / It is defined. For \(\sec^{-1}x\), we need \(\left|x\right|\ge1\). Here \(\left|\frac{3}{2}\right|\ge1\), so it is defined.
Step 3
Exam Tip
\(\sec^{-1}x\) के लिए \(\left|x\right|\ge1\) होना चाहिए। यहाँ \(\left|\frac{3}{2}\right|\ge1\), इसलिए यह परिभाषित है।
The function \(cosec^{-1}x\) is defined when \(\left|x\right|\ge1\). Here \(\left|-\frac{5}{4}\right|\ge1\), so it is defined.
Step 2
Why this answer is correct
The correct answer is A. यह परिभाषित है / It is defined. The function \(cosec^{-1}x\) is defined when \(\left|x\right|\ge1\). Here \(\left|-\frac{5}{4}\right|\ge1\), so it is defined.
Step 3
Exam Tip
\(cosec^{-1}x\) तभी परिभाषित है जब \(\left|x\right|\ge1\)। यहाँ \(\left|-\frac{5}{4}\right|\ge1\), इसलिए यह परिभाषित है।
The domain of \(\sec^{-1}x\) is (\left\(-\infty,-1\right]\cup\left[1,\infty\right\)). The number (0) is not in this domain.
Step 2
Why this answer is correct
The correct answer is B. यह परिभाषित नहीं है / It is not defined. The domain of \(\sec^{-1}x\) is (\left\(-\infty,-1\right]\cup\left[1,\infty\right\)). The number (0) is not in this domain.
Step 3
Exam Tip
\(\sec^{-1}x\) का प्रांत (\left\(-\infty,-1\right]\cup\left[1,\infty\right\)) है। (0) इस प्रांत में नहीं है।
The angle \(-\frac{\pi}{4}\) lies in the principal range of \(\sin^{-1}x\). Hence the value is \(-\frac{\pi}{4}\).
Step 2
Why this answer is correct
The correct answer is A. -\(\frac{\pi}{4}\). The angle \(-\frac{\pi}{4}\) lies in the principal range of \(\sin^{-1}x\). Hence the value is \(-\frac{\pi}{4}\).
Step 3
Exam Tip
कोण \((-\frac{\pi}{4}) (\sin^{-1}x)\) के मुख्य परिसर में है। इसलिए मान \(-\frac{\pi}{4}\) है।
Since \(\cos\frac{\pi}{3}=\frac{1}{2}\) and \(\sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}\). First find the inner value.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\pi}{6}\). Since \(\cos\frac{\pi}{3}=\frac{1}{2}\) and \(\sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}\). First find the inner value.
Step 3
Exam Tip
\(\cos\frac{\pi}{3}=\frac{1}{2}\) और \(\sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}\)। पहले अंदर का मान निकालें।
Since \(\sin\frac{\pi}{6}=\frac{1}{2}\) and \(\cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}\). Remember standard angles.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{\pi}{3}\). Since \(\sin\frac{\pi}{6}=\frac{1}{2}\) and \(\cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}\). Remember standard angles.
Step 3
Exam Tip
\(\sin\frac{\pi}{6}=\frac{1}{2}\) और \(\cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}\)। मानक कोण याद रखें।
Since \(\cot\frac{\pi}{3}=\frac{1}{\sqrt{3}}\) and \(\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}\). In such questions, find the trigonometric value first.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\pi}{6}\). Since \(\cot\frac{\pi}{3}=\frac{1}{\sqrt{3}}\) and \(\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}\). In such questions, find the trigonometric value first.
Step 3
Exam Tip
\(\cot\frac{\pi}{3}=\frac{1}{\sqrt{3}}\) और \(\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}\)। ऐसे प्रश्न में त्रिकोणमितीय मान पहले निकालें।
Since \(sin^{-1}0=0\), the graph passes through \(\left(0,0\right)\). Basic values are useful in graph questions.
Step 2
Why this answer is correct
The correct answer is A. \(\left(0,0\right)\). Since \(sin^{-1}0=0\), the graph passes through \(\left(0,0\right)\). Basic values are useful in graph questions.
Step 3
Exam Tip
\(\sin^{-1}0=0\), इसलिए ग्राफ \(\left(0,0\right)\) से गुजरता है। ग्राफ में मूल मान उपयोगी होते हैं।
As \(x\to\infty\), \(\tan^{-1}x\to\frac{\pi}{2}\). The value \(\frac{\pi}{2}\) is a limit, not an attained value.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\pi}{2}\). As \(x\to\infty\), \(\tan^{-1}x\to\frac{\pi}{2}\). The value \(\frac{\pi}{2}\) is a limit, not an attained value.
Step 3
Exam Tip
जब \(x\to\infty\), तब \(\tan^{-1}x\to\frac{\pi}{2}\)। \(\frac{\pi}{2}\) सीमा है, पर वास्तविक मान नहीं बनता।
Since \(\cos^{-1}1=0\), the graph passes through \(\left(1,0\right)\). Remember standard values for graph questions.
Step 2
Why this answer is correct
The correct answer is A. \(\left(1,0\right)\). Since \(\cos^{-1}1=0\), the graph passes through \(\left(1,0\right)\). Remember standard values for graph questions.
Step 3
Exam Tip
\(\cos^{-1}1=0\), इसलिए ग्राफ \(\left(1,0\right)\) से गुजरता है। ग्राफ वाले प्रश्नों में मानक मान याद रखें।
