The range of \(\sin^{-1}x\) is \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) and \(\sin\frac{\pi}{6}=\frac{1}{2}\). Always remember the principal range in exams.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\pi}{6}\). The range of \(\sin^{-1}x\) is \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) and \(\sin\frac{\pi}{6}=\frac{1}{2}\). Always remember the principal range in exams.
Step 3
Exam Tip
\(\sin^{-1}x\) का परिसर \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) होता है और \(\sin\frac{\pi}{6}=\frac{1}{2}\)। परीक्षा में हमेशा मुख्य मान परिसर याद रखें।
The range of \(\cos^{-1}x\) is \(\left[0,\pi\right]\) and in this range \(\cos\frac{2\pi}{3}=-\frac{1}{2}\). Do not take a negative angle as principal value.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{2\pi}{3}\). The range of \(\cos^{-1}x\) is \(\left[0,\pi\right]\) and in this range \(\cos\frac{2\pi}{3}=-\frac{1}{2}\). Do not take a negative angle as principal value.
Step 3
Exam Tip
\(\cos^{-1}x\) का परिसर \(\left[0,\pi\right]\) है और इस परिसर में \(\cos\frac{2\pi}{3}=-\frac{1}{2}\)। ऋणात्मक उत्तर को मुख्य मान न मानें।
The range of \(\tan^{-1}x\) is (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)) and (\tan\left\(-\frac{\pi}{4}\right\)=-1). Keep the quadrant in mind.
Step 2
Why this answer is correct
The correct answer is B. \(-\frac{\pi}{4}\). The range of \(\tan^{-1}x\) is (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)) and (\tan\left\(-\frac{\pi}{4}\right\)=-1). Keep the quadrant in mind.
Step 3
Exam Tip
\(\tan^{-1}x\) का परिसर (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)) है और (\tan\left\(-\frac{\pi}{4}\right\)=-1)। चतुर्थांश का ध्यान रखें।
The function \(\tan^{-1}x\) is defined for all real (x) but its range is the open interval (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)). The end points are not included.
Step 2
Why this answer is correct
The correct answer is C. (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)). The function \(\tan^{-1}x\) is defined for all real (x) but its range is the open interval (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)). The end points are not included.
Step 3
Exam Tip
\(\tan^{-1}x\) सभी वास्तविक (x) के लिए परिभाषित है लेकिन इसका परिसर खुला अंतराल (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)) है। अंत बिंदु शामिल नहीं होते।
A. \(x=\sin y,\ y\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\)
Step 1
Concept
The meaning of \(\sin^{-1}\) is the angle (y) for which \(\sin y=x\). The principal range must also be included.
Step 2
Why this answer is correct
The correct answer is A. \(x=\sin y,\ y\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). The meaning of \(\sin^{-1}\) is the angle (y) for which \(\sin y=x\). The principal range must also be included.
Step 3
Exam Tip
\(\sin^{-1}\) का अर्थ है वह कोण (y) जिसके लिए \(\sin y=x\)। साथ में मुख्य मान का परिसर भी जरूरी है।
The principal range of \(\cos^{-1}x\) is \(\left[0,\pi\right]\). Both end points are included for \(\cos^{-1}\).
Step 2
Why this answer is correct
The correct answer is B. \(\left[0,\pi\right]\). The principal range of \(\cos^{-1}x\) is \(\left[0,\pi\right]\). Both end points are included for \(\cos^{-1}\).
Step 3
Exam Tip
\(\cos^{-1}x\) का मुख्य परिसर \(\left[0,\pi\right]\) है। \(\cos^{-1}\) में दोनों अंत बिंदु शामिल होते हैं।
When \(\frac{3}{5}\in\left[-1,1\right]\), (\sin\left\(\sin^{-1}x\right\)=x). First check whether the inside value lies in the domain.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{3}{5}\). When \(\frac{3}{5}\in\left[-1,1\right]\), (\sin\left\(\sin^{-1}x\right\)=x). First check whether the inside value lies in the domain.
Step 3
Exam Tip
जब \(\frac{3}{5}\in\left[-1,1\right]\), तब (\sin\left\(\sin^{-1}x\right\)=x)। अंदर का मान प्रांत में है या नहीं पहले देखें।
Since \(\cos\frac{5\pi}{3}=\frac{1}{2}\), \(\cos^{-1}\frac{1}{2}=\frac{\pi}{3}\). Always bring the angle into \(\left[0,\pi\right]\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{\pi}{3}\). Since \(\cos\frac{5\pi}{3}=\frac{1}{2}\), \(\cos^{-1}\frac{1}{2}=\frac{\pi}{3}\). Always bring the angle into \(\left[0,\pi\right]\).
Step 3
Exam Tip
\(\cos\frac{5\pi}{3}=\frac{1}{2}\) और \(\cos^{-1}\frac{1}{2}=\frac{\pi}{3}\)। कोण को हमेशा \(\left[0,\pi\right]\) में लाएं।
Since \(\tan\frac{3\pi}{4}=-1), (\tan^{-1}(-1)=-\frac{\pi}{4}\). The answer must lie in \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\).
Step 2
Why this answer is correct
The correct answer is C. \(-\frac{\pi}{4}\). Since \(\tan\frac{3\pi}{4}=-1), (\tan^{-1}(-1)=-\frac{\pi}{4}\). The answer must lie in \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\).
Step 3
Exam Tip
\(\tan\frac{3\pi}{4}=-1\) और \(\tan^{-1}(-1)=-\frac{\pi}{4}\)। उत्तर \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\) में होना चाहिए।
From \(\csc y=-2\), \(\sin y=-\frac{1}{2}\), so the principal value is \(y=-\frac{\pi}{6}\). Remember the range of \(\csc^{-1}\).
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{\pi}{6}\). From \(\csc y=-2\), \(\sin y=-\frac{1}{2}\), so the principal value is \(y=-\frac{\pi}{6}\). Remember the range of \(\csc^{-1}\).
Step 3
Exam Tip
\(\csc y=-2\) से \(\sin y=-\frac{1}{2}\), इसलिए मुख्य मान \(y=-\frac{\pi}{6}\) है। \(\csc^{-1}\) का परिसर याद रखें।
Since \(\tan\frac{\pi}{3}=\sqrt{3}\) and \(\frac{\pi}{3}\) lies in the principal range. Remember special angles well.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{\pi}{3}\). Since \(\tan\frac{\pi}{3}=\sqrt{3}\) and \(\frac{\pi}{3}\) lies in the principal range. Remember special angles well.
Step 3
Exam Tip
\(\tan\frac{\pi}{3}=\sqrt{3}\) और \(\frac{\pi}{3}\) मुख्य परिसर में है। विशेष कोण याद रखना जरूरी है।
Values of \(\sec y\) do not lie in (\left\(-1,1\right\)), so the domain of \(\sec^{-1}x\) is (\mathbb{R}\setminus\left\(-1,1\right\)). Zero is also not in the domain.
Step 2
Why this answer is correct
The correct answer is B. (\mathbb{R}\setminus\left\(-1,1\right\)). Values of \(\sec y\) do not lie in (\left\(-1,1\right\)), so the domain of \(\sec^{-1}x\) is (\mathbb{R}\setminus\left\(-1,1\right\)). Zero is also not in the domain.
Step 3
Exam Tip
\(\sec y\) के मान (\left\(-1,1\right\)) में नहीं आते, इसलिए \(\sec^{-1}x\) का प्रांत (\mathbb{R}\setminus\left\(-1,1\right\)) है। शून्य भी प्रांत में नहीं है।
Here \(\theta\in\left[0,\pi\right]\) and \(\cos\theta=\frac{8}{17}\), so \(\theta\) is in the first quadrant and \(\sin\theta=\frac{15}{17}\). The triangle method is fast.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{15}{17}\). Here \(\theta\in\left[0,\pi\right]\) and \(\cos\theta=\frac{8}{17}\), so \(\theta\) is in the first quadrant and \(\sin\theta=\frac{15}{17}\). The triangle method is fast.
Step 3
Exam Tip
\(\theta\in\left[0,\pi\right]\) और \(\cos\theta=\frac{8}{17}\), इसलिए \(\theta\) प्रथम चतुर्थांश में है और \(\sin\theta=\frac{15}{17}\)। त्रिभुज विधि तेज है।
From \(\tan\theta=\frac{7}{24}\), the hypotenuse is \(\sqrt{7^2+24^2}=25\), so \(\sin\theta=\frac{7}{25}\). Build a triangle from the ratio.
Step 2
Why this answer is correct
The correct answer is C. \(\frac{7}{25}\). From \(\tan\theta=\frac{7}{24}\), the hypotenuse is \(\sqrt{7^2+24^2}=25\), so \(\sin\theta=\frac{7}{25}\). Build a triangle from the ratio.
Step 3
Exam Tip
\(\tan\theta=\frac{7}{24}\) से कर्ण \(\sqrt{7^2+24^2}=25\), इसलिए \(\sin\theta=\frac{7}{25}\)। अनुपात से त्रिभुज बनाएं।
Let \(\theta=\sin^{-1}\frac{4}{5}\), then \(\sin\theta=\frac{4}{5}\) and \(\cos\theta=\frac{3}{5}\). Therefore \(\tan\theta=\frac{4}{3}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{4}{3}\). Let \(\theta=\sin^{-1}\frac{4}{5}\), then \(\sin\theta=\frac{4}{5}\) and \(\cos\theta=\frac{3}{5}\). Therefore \(\tan\theta=\frac{4}{3}\).
Step 3
Exam Tip
मान लें \(\theta=\sin^{-1}\frac{4}{5}\), तब \(\sin\theta=\frac{4}{5}\) और \(\cos\theta=\frac{3}{5}\)। इसलिए \(\tan\theta=\frac{4}{3}\)।
If \(\theta=\tan^{-1}\frac{12}{5}\), then \(\tan\theta=\frac{12}{5}\) and the hypotenuse is (13). Hence \(\sin\theta=\frac{12}{13}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{12}{13}\). If \(\theta=\tan^{-1}\frac{12}{5}\), then \(\tan\theta=\frac{12}{5}\) and the hypotenuse is (13). Hence \(\sin\theta=\frac{12}{13}\).
Step 3
Exam Tip
यदि \(\theta=\tan^{-1}\frac{12}{5}\), तो \(\tan\theta=\frac{12}{5}\) और कर्ण (13) है। इसलिए \(\sin\theta=\frac{12}{13}\)।
If \(\tan\theta=\frac{15}{8}\), then the hypotenuse is (17). In the principal range \(\theta\) is in the first quadrant, so \(\cos\theta=\frac{8}{17}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{8}{17}\). If \(\tan\theta=\frac{15}{8}\), then the hypotenuse is (17). In the principal range \(\theta\) is in the first quadrant, so \(\cos\theta=\frac{8}{17}\).
Step 3
Exam Tip
यदि \(\tan\theta=\frac{15}{8}\), तो कर्ण (17) है। मुख्य परिसर में \(\theta\) प्रथम चतुर्थांश में है, इसलिए \(\cos\theta=\frac{8}{17}\)।
A. (\mathbb{R},\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\))
Step 1
Concept
The domain of \(\tan^{-1}x\) is \(\mathbb{R}\) and its range is (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)). The endpoints of the open interval are not included.
Step 2
Why this answer is correct
The correct answer is A. (\mathbb{R},\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)). The domain of \(\tan^{-1}x\) is \(\mathbb{R}\) and its range is (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)). The endpoints of the open interval are not included.
Step 3
Exam Tip
\(\tan^{-1}x\) का प्रांत \(\mathbb{R}\) और परिसर (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)) है। खुले अंतराल के अंत बिंदु शामिल नहीं हैं।
By making \(\cos x\) one-one on \(\left[0,\pi\right]\), \(\cos^{-1}x\) becomes decreasing on \(\left[-1,1\right]\). The graph direction also confirms this.
Step 2
Why this answer is correct
The correct answer is A. \(\left[-1,1\right]\). By making \(\cos x\) one-one on \(\left[0,\pi\right]\), \(\cos^{-1}x\) becomes decreasing on \(\left[-1,1\right]\). The graph direction also confirms this.
Step 3
Exam Tip
\(\cos x\) को \(\left[0,\pi\right]\) पर एक-एक बनाने से \(\cos^{-1}x\) \(\left[-1,1\right]\) पर घटता है। ग्राफ की दिशा से भी पुष्टि करें।
The function \(\sin^{-1}x\) is increasing on its whole domain \(\left[-1,1\right]\). This matches the chosen branch of \(\sin x\).
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 whole domain \(\left[-1,1\right]\). This matches the chosen branch of \(\sin x\).
Step 3
Exam Tip
\(\sin^{-1}x\) अपने पूरे प्रांत \(\left[-1,1\right]\) पर बढ़ता है। यह \(\sin x\) की चुनी गई शाखा से मिलता है।
Restricting \(\sin x\) to \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) makes it one-one and gives values \(\left[-1,1\right]\). This defines \(\sin^{-1}x\).
Step 2
Why this answer is correct
The correct answer is A. \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). Restricting \(\sin x\) to \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) makes it one-one and gives values \(\left[-1,1\right]\). This defines \(\sin^{-1}x\).
Step 3
Exam Tip
\(\sin x\) को \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) पर प्रतिबंधित करने से यह एक-एक बनता है और मान \(\left[-1,1\right]\) मिलते हैं। इसी से \(\sin^{-1}x\) परिभाषित होता है।
The function \(\cos x\) is made one-one on \(\left[0,\pi\right]\). Therefore the range of \(\cos^{-1}x\) is \(\left[0,\pi\right]\).
Step 2
Why this answer is correct
The correct answer is B. \(\left[0,\pi\right]\). The function \(\cos x\) is made one-one on \(\left[0,\pi\right]\). Therefore the range of \(\cos^{-1}x\) is \(\left[0,\pi\right]\).
Step 3
Exam Tip
\(\cos x\) को \(\left[0,\pi\right]\) पर एक-एक बनाया जाता है। इसलिए \(\cos^{-1}x\) का परिसर \(\left[0,\pi\right]\) होता है।
Here \(\sin^{-1}1=\frac{\pi}{2}\) and \(\cos^{-1}1=0\), so the sum is \(\frac{\pi}{2}\). Remember endpoint values.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\pi}{2}\). Here \(\sin^{-1}1=\frac{\pi}{2}\) and \(\cos^{-1}1=0\), so the sum is \(\frac{\pi}{2}\). Remember endpoint values.
Step 3
Exam Tip
\(\sin^{-1}1=\frac{\pi}{2}\) और \(\cos^{-1}1=0\), इसलिए योग \(\frac{\pi}{2}\) है। अंत बिंदुओं के मान याद रखें।
Since \(\cos^{-1}0=\frac{\pi}{2}\) and \(\sin^{-1}0=0\), the difference is \(\frac{\pi}{2}\). Inverse trigonometric values at zero are very useful.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{\pi}{2}\). Since \(\cos^{-1}0=\frac{\pi}{2}\) and \(\sin^{-1}0=0\), the difference is \(\frac{\pi}{2}\). Inverse trigonometric values at zero are very useful.
Step 3
Exam Tip
\(\cos^{-1}0=\frac{\pi}{2}\) और \(\sin^{-1}0=0\), इसलिए अंतर \(\frac{\pi}{2}\) है। शून्य के उल्टे त्रिकोणमितीय मान बहुत उपयोगी हैं।
Here \(\tan^{-1}0=0\) and \(\cot^{-1}0=\frac{\pi}{2}\), so the sum is \(\frac{\pi}{2}\). Do not write \(\cot^{-1}0\) as (0).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{\pi}{2}\). Here \(\tan^{-1}0=0\) and \(\cot^{-1}0=\frac{\pi}{2}\), so the sum is \(\frac{\pi}{2}\). Do not write \(\cot^{-1}0\) as (0).
Step 3
Exam Tip
\(\tan^{-1}0=0\) और \(\cot^{-1}0=\frac{\pi}{2}\), इसलिए योग \(\frac{\pi}{2}\) है। \(\cot^{-1}0\) को (0) न लिखें।
From \(\sin^{-1}x=\frac{\pi}{4}\), \(x=\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}\). Convert the inverse form into the normal trigonometric form.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{1}{\sqrt{2}}\). From \(\sin^{-1}x=\frac{\pi}{4}\), \(x=\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}\). Convert the inverse form into the normal trigonometric form.
Step 3
Exam Tip
\(\sin^{-1}x=\frac{\pi}{4}\) से \(x=\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}\)। प्रतिलोम को सामान्य त्रिकोणमितीय रूप में बदलें।
From \(\cos^{-1}x=\frac{2\pi}{3}\), \(x=\cos\frac{2\pi}{3}=-\frac{1}{2}\). Decide the sign from the quadrant of the angle.
Step 2
Why this answer is correct
The correct answer is B. \(-\frac{1}{2}\). From \(\cos^{-1}x=\frac{2\pi}{3}\), \(x=\cos\frac{2\pi}{3}=-\frac{1}{2}\). Decide the sign from the quadrant of the angle.
Step 3
Exam Tip
\(\cos^{-1}x=\frac{2\pi}{3}\) से \(x=\cos\frac{2\pi}{3}=-\frac{1}{2}\)। कोण के चतुर्थांश से चिह्न तय करें।
From \(\tan^{-1}x=-\frac{\pi}{6}\), (x=\tan\left\(-\frac{\pi}{6}\right\)=-\frac{1}{\sqrt{3}}). The range of \(\tan^{-1}\) also includes negative angles.
Step 2
Why this answer is correct
The correct answer is B. \(-\frac{1}{\sqrt{3}}\). From \(\tan^{-1}x=-\frac{\pi}{6}\), (x=\tan\left\(-\frac{\pi}{6}\right\)=-\frac{1}{\sqrt{3}}). The range of \(\tan^{-1}\) also includes negative angles.
Step 3
Exam Tip
\(\tan^{-1}x=-\frac{\pi}{6}\) से (x=\tan\left\(-\frac{\pi}{6}\right\)=-\frac{1}{\sqrt{3}})। \(\tan^{-1}\) का परिसर ऋणात्मक कोण भी लेता है।
For every \(x\in\left[-1,1\right]\), \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\). Here \(x=\frac{2}{3}\) is valid.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\pi}{2}\). For every \(x\in\left[-1,1\right]\), \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\). Here \(x=\frac{2}{3}\) is valid.
Step 3
Exam Tip
हर \(x\in\left[-1,1\right]\) के लिए \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\)। यहां \(x=\frac{2}{3}\) मान्य है।
Since \(\sin\left(-\frac{\pi}{3}\right)=-\frac{\sqrt{3}}{2}\) and it lies in the principal range. Therefore \(-\frac{\pi}{3}\) is correct.
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{\pi}{3}\). Since \(\sin\left(-\frac{\pi}{3}\right)=-\frac{\sqrt{3}}{2}\) and it lies in the principal range. Therefore \(-\frac{\pi}{3}\) is correct.
Step 3
Exam Tip
\(\sin\left(-\frac{\pi}{3}\right)=-\frac{\sqrt{3}}{2}\) और यह मुख्य परिसर में है। इसलिए \(-\frac{\pi}{3}\) सही है।
Since \(\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}\) and \(\frac{\pi}{4}\in\left[0,\pi\right]\). Remember the table of special values.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{\pi}{4}\). Since \(\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}\) and \(\frac{\pi}{4}\in\left[0,\pi\right]\). Remember the table of special values.
Step 3
Exam Tip
\(\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}\) और \(\frac{\pi}{4}\in\left[0,\pi\right]\)। विशेष मानों की तालिका याद रखें।
Since \(\sin\left(-\frac{5\pi}{6}\right)=-\frac{1}{2}\), \(\sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}\). The answer must lie in the principal range.
Step 2
Why this answer is correct
The correct answer is B. \(-\frac{\pi}{6}\). Since \(\sin\left(-\frac{5\pi}{6}\right)=-\frac{1}{2}\), \(\sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}\). The answer must lie in the principal range.
Step 3
Exam Tip
\(\sin\left(-\frac{5\pi}{6}\right)=-\frac{1}{2}\), इसलिए \(\sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}\)। उत्तर मुख्य परिसर में होना चाहिए।
Here \(\beta\in\left[0,\pi\right]\) and \(\cos\beta=-\frac{4}{5}\), so \(\beta\) is in the second quadrant and \(\sin\beta=\frac{3}{5}\). The sign of \(\sin\) is positive here.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{3}{5}\). Here \(\beta\in\left[0,\pi\right]\) and \(\cos\beta=-\frac{4}{5}\), so \(\beta\) is in the second quadrant and \(\sin\beta=\frac{3}{5}\). The sign of \(\sin\) is positive here.
Step 3
Exam Tip
\(\beta\in\left[0,\pi\right]\) और \(\cos\beta=-\frac{4}{5}\), इसलिए \(\beta\) द्वितीय चतुर्थांश में है और \(\sin\beta=\frac{3}{5}\)। \(\sin\) का चिह्न यहां धनात्मक है।
From \(\cot\theta=\frac{3}{4}\), the base is (3), perpendicular is (4), and hypotenuse is (5). Hence \(\cos\theta=\frac{3}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{3}{5}\). From \(\cot\theta=\frac{3}{4}\), the base is (3), perpendicular is (4), and hypotenuse is (5). Hence \(\cos\theta=\frac{3}{5}\).
Step 3
Exam Tip
\(\cot\theta=\frac{3}{4}\) से आधार (3), लम्ब (4) और कर्ण (5) होगा। इसलिए \(\cos\theta=\frac{3}{5}\)।
By the formula, the sum is \(\tan^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\right)=\tan^{-1}1=\frac{\pi}{4}\). While adding \(\tan^{-1}\) terms, check \(ab<1\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{\pi}{4}\). By the formula, the sum is \(\tan^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\right)=\tan^{-1}1=\frac{\pi}{4}\). While adding \(\tan^{-1}\) terms, check \(ab<1\).
Step 3
Exam Tip
सूत्र से योग \(\tan^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\right)=\tan^{-1}1=\frac{\pi}{4}\) है। \(\tan^{-1}\) जोड़ते समय \(ab<1\) जांचें।
From \(\sec y=-\sqrt{2}\), \(\cos y=-\frac{1}{\sqrt{2}}\), so the principal value is \(y=\frac{3\pi}{4}\). For \(\sec^{-1}\), decide the quadrant from the sign.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{3\pi}{4}\). From \(\sec y=-\sqrt{2}\), \(\cos y=-\frac{1}{\sqrt{2}}\), so the principal value is \(y=\frac{3\pi}{4}\). For \(\sec^{-1}\), decide the quadrant from the sign.
Step 3
Exam Tip
\(\sec y=-\sqrt{2}\) से \(\cos y=-\frac{1}{\sqrt{2}}\), इसलिए मुख्य मान \(y=\frac{3\pi}{4}\) है। \(\sec^{-1}\) में चिह्न से चतुर्थांश तय करें।
The domain of \(\cos^{-1}x\) is \(\left[-1,1\right]\) and its range is \(\left[0,\pi\right]\). In domain-range questions, identify intervals first.
Step 2
Why this answer is correct
The correct answer is B. \(\cos^{-1}x\). The domain of \(\cos^{-1}x\) is \(\left[-1,1\right]\) and its range is \(\left[0,\pi\right]\). In domain-range questions, identify intervals first.
Step 3
Exam Tip
\(\cos^{-1}x\) का प्रांत \(\left[-1,1\right]\) और परिसर \(\left[0,\pi\right]\) है। प्रांत-परिसर वाले प्रश्नों में अंतराल तुरंत पहचानें।
Since \(\sin\frac{7\pi}{6}=-\frac{1}{2}\), \(\sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}\). Keep the answer in \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\).
Step 2
Why this answer is correct
The correct answer is C. \(-\frac{\pi}{6}\). Since \(\sin\frac{7\pi}{6}=-\frac{1}{2}\), \(\sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}\). Keep the answer in \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\).
Step 3
Exam Tip
\(\sin\frac{7\pi}{6}=-\frac{1}{2}\), इसलिए \(\sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}\)। उत्तर हमेशा \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) में रखें।
Since \(\cos\left(-\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\), \(\cos^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}). The answer of (\cos^{-1}\) is not negative.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{\pi}{4}\). Since \(\cos\left(-\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\), \(\cos^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}). The answer of (\cos^{-1}\) is not negative.
Step 3
Exam Tip
\(\cos\left(-\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\), इसलिए \(\cos^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4})। (\cos^{-1}\) का उत्तर ऋणात्मक नहीं होता।
Since \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\), if both are equal then each is \(\frac{\pi}{4}\). Hence \(x=\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}\).
Step 2
Why this answer is correct
The correct answer is C. \(\frac{1}{\sqrt{2}}\). Since \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\), if both are equal then each is \(\frac{\pi}{4}\). Hence \(x=\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}\).
Step 3
Exam Tip
क्योंकि \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\), दोनों बराबर होने पर हर एक \(\frac{\pi}{4}\) होगा। इसलिए \(x=\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}\)।
Let \(\theta=\cos^{-1}\frac{5}{13}\), then \(\cos\theta=\frac{5}{13}\) and \(\sin\theta=\frac{12}{13}\). Therefore \(\tan\theta=\frac{12}{5}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{12}{5}\). Let \(\theta=\cos^{-1}\frac{5}{13}\), then \(\cos\theta=\frac{5}{13}\) and \(\sin\theta=\frac{12}{13}\). Therefore \(\tan\theta=\frac{12}{5}\).
Step 3
Exam Tip
मान लें \(\theta=\cos^{-1}\frac{5}{13}\), तब \(\cos\theta=\frac{5}{13}\) और \(\sin\theta=\frac{12}{13}\)। इसलिए \(\tan\theta=\frac{12}{5}\)।
From \(\csc y=\frac{2}{\sqrt{3}}\), \(\sin y=\frac{\sqrt{3}}{2}\), so \(y=\frac{\pi}{3}\). Solve \(\csc^{-1}\) by relating it to \(\sin^{-1}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{\pi}{3}\). From \(\csc y=\frac{2}{\sqrt{3}}\), \(\sin y=\frac{\sqrt{3}}{2}\), so \(y=\frac{\pi}{3}\). Solve \(\csc^{-1}\) by relating it to \(\sin^{-1}\).
Step 3
Exam Tip
\(\csc y=\frac{2}{\sqrt{3}}\) से \(\sin y=\frac{\sqrt{3}}{2}\), इसलिए \(y=\frac{\pi}{3}\) है। \(\csc^{-1}\) को \(\sin^{-1}\) से जोड़कर हल करें।
For \(\sin^{-1}(2x)\), we need \(-1\le 2x\le 1\). Hence \(-\frac{1}{2}\le x\le \frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is B. \(-\frac{1}{2}\le x\le \frac{1}{2}\). For \(\sin^{-1}(2x)\), we need \(-1\le 2x\le 1\). Hence \(-\frac{1}{2}\le x\le \frac{1}{2}\).
Step 3
Exam Tip
\(\sin^{-1}(2x)\) के लिए \(-1\le 2x\le 1\) चाहिए। अतः \(-\frac{1}{2}\le x\le \frac{1}{2}\) मिलेगा।
