Class 12 General Mathematics Introduction To Trigonometry Practice
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Introduction to inverse trigonometric functions
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The value of \(\sin x\) lies only in ([-1,1]) so the domain of \(\sin^{-1}x\) is ([-1,1]). In exams check the domain first.
Step 2
Why this answer is correct
The correct answer is A. अंतराल ([-1,1]) / Interval ([-1,1]). The value of \(\sin x\) lies only in ([-1,1]) so the domain of \(\sin^{-1}x\) is ([-1,1]). In exams check the domain first.
Step 3
Exam Tip
\(\sin x\) का मान केवल ([-1,1]) में आता है इसलिए \(\sin^{-1}x\) का डोमेन ([-1,1]) है। परीक्षा में पहले डोमेन जरूर जांचें।
The value of \(\cos x\) also lies in ([-1,1]) so the domain of \(\cos^{-1}x\) is ([-1,1]). Check the range of (x) before substitution.
Step 2
Why this answer is correct
The correct answer is A. अंतराल ([-1,1]) / Interval ([-1,1]). The value of \(\cos x\) also lies in ([-1,1]) so the domain of \(\cos^{-1}x\) is ([-1,1]). Check the range of (x) before substitution.
Step 3
Exam Tip
\(\cos x\) का मान भी ([-1,1]) में रहता है इसलिए \(\cos^{-1}x\) का डोमेन ([-1,1]) है। मान रखने से पहले (x) की सीमा देखें।
A. अंतराल (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\))/Interval (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\))
Step 1
Concept
The principal value of \(\tan^{-1}x\) lies in (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)). The endpoints \(\pm\frac{\pi}{2}\) are not included.
Step 2
Why this answer is correct
The correct answer is A. अंतराल (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)) / Interval (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)). The principal value of \(\tan^{-1}x\) lies in (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)). The endpoints \(\pm\frac{\pi}{2}\) are not included.
Step 3
Exam Tip
\(\tan^{-1}x\) का प्रधान मान (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)) में होता है। सिरों \(\pm\frac{\pi}{2}\) को शामिल नहीं किया जाता।
Because \(\tan \frac{\pi}{4}=1\) and \(\frac{\pi}{4}\) lies in the principal range. Remembering basic values helps solve quickly.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\pi}{4}\). Because \(\tan \frac{\pi}{4}=1\) and \(\frac{\pi}{4}\) lies in the principal range. Remembering basic values helps solve quickly.
Step 3
Exam Tip
क्योंकि \(\tan \frac{\pi}{4}=1\) और \(\frac{\pi}{4}\) प्रधान सीमा में है। मान याद रखने से मूल प्रश्न जल्दी हल होता है।
A. जब \(x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\)/When \(x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\)
Step 1
Concept
The principal range of \(\sin^{-1}x\) is \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) so (\sin^{-1}\(\sin x\)=x) is directly true there. Without range it can be a mistake.
Step 2
Why this answer is correct
The correct answer is A. जब \(x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) / When \(x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). The principal range of \(\sin^{-1}x\) is \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) so (\sin^{-1}\(\sin x\)=x) is directly true there. Without range it can be a mistake.
Step 3
Exam Tip
\(\sin^{-1}x\) की प्रधान सीमा \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) है इसलिए इसी में (\sin^{-1}\(\sin x\)=x) सीधे सही है। सीमा के बिना यह गलती बन सकती है।
The principal range of \(\cos^{-1}x\) is \([0,\pi]\). Therefore (\cos^{-1}\(\cos x\)=x) applies directly only in this range.
Step 2
Why this answer is correct
The correct answer is A. जब \(x\in[0,\pi]\) / When \(x\in[0,\pi]\). The principal range of \(\cos^{-1}x\) is \([0,\pi]\). Therefore (\cos^{-1}\(\cos x\)=x) applies directly only in this range.
Step 3
Exam Tip
\(\cos^{-1}x\) की प्रधान सीमा \([0,\pi]\) है। इसलिए (\cos^{-1}\(\cos x\)=x) इसी सीमा में सीधे लागू होता है।
A. जब (x\in\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\))/When (x\in\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\))
Step 1
Concept
The principal range of \(\tan^{-1}x\) is (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)). Therefore that interval is the correct choice.
Step 2
Why this answer is correct
The correct answer is A. जब (x\in\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)) / When (x\in\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)). The principal range of \(\tan^{-1}x\) is (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)). Therefore that interval is the correct choice.
Step 3
Exam Tip
\(\tan^{-1}x\) की प्रधान सीमा (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)) है। इसलिए वही अंतराल सही चयन है।
Taking \(\sin\) on both sides gives \(x=\sin\frac{\pi}{6}=\frac{1}{2}\). In inverse functions convert the angle back to the original value.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{2}\). Taking \(\sin\) on both sides gives \(x=\sin\frac{\pi}{6}=\frac{1}{2}\). In inverse functions convert the angle back to the original value.
Step 3
Exam Tip
दोनों तरफ \(\sin\) लेने पर \(x=\sin\frac{\pi}{6}=\frac{1}{2}\) मिलता है। उल्टे फलन में कोण से मूल मान निकालें।
Taking \(\cos\) on both sides gives \(x=\cos\frac{\pi}{3}=\frac{1}{2}\). The angle lies in the principal range so the answer is valid.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{2}\). Taking \(\cos\) on both sides gives \(x=\cos\frac{\pi}{3}=\frac{1}{2}\). The angle lies in the principal range so the answer is valid.
Step 3
Exam Tip
दोनों तरफ \(\cos\) लेने पर \(x=\cos\frac{\pi}{3}=\frac{1}{2}\) है। कोण प्रधान सीमा में है इसलिए उत्तर मान्य है।
Taking \(\tan\) on both sides gives \(x=\tan\frac{\pi}{4}=1\). In \(\tan^{-1}\) questions the value comes directly from tangent.
Step 2
Why this answer is correct
The correct answer is A. (\1). Taking \(\tan\) on both sides gives \(x=\tan\frac{\pi}{4}=1\). In \(\tan^{-1}\) questions the value comes directly from tangent.
Step 3
Exam Tip
दोनों तरफ \(\tan\) लेने पर \(x=\tan\frac{\pi}{4}=1\) है। \(\tan^{-1}\) वाले प्रश्न में मान सीधे टैन से मिलता है।
A. समस्त वास्तविक संख्याएं \(\mathbb{R}\)/All real numbers \(\mathbb{R}\)
Step 1
Concept
Since \(\cot x\) can take all real values the domain of \(\cot^{-1}x\) is \(\mathbb{R}\). Do not mix it with the domain of \(\sin^{-1}x\).
Step 2
Why this answer is correct
The correct answer is A. समस्त वास्तविक संख्याएं \(\mathbb{R}\) / All real numbers \(\mathbb{R}\). Since \(\cot x\) can take all real values the domain of \(\cot^{-1}x\) is \(\mathbb{R}\). Do not mix it with the domain of \(\sin^{-1}x\).
Step 3
Exam Tip
\(\cot x\) सभी वास्तविक मान ले सकता है इसलिए \(\cot^{-1}x\) का डोमेन \(\mathbb{R}\) है। इसे \(\sin^{-1}\) के डोमेन से न मिलाएं।
The principal value of \(\cot^{-1}x\) is usually taken in (\(0,\pi\)). Hence \(\cot^{-1}0=\frac{\pi}{2}\).
Step 2
Why this answer is correct
The correct answer is A. अंतराल (\(0,\pi\)) / Interval (\(0,\pi\)). The principal value of \(\cot^{-1}x\) is usually taken in (\(0,\pi\)). Hence \(\cot^{-1}0=\frac{\pi}{2}\).
Step 3
Exam Tip
\(\cot^{-1}x\) का प्रधान मान प्रायः (\(0,\pi\)) में लिया जाता है। इस कारण \(\cot^{-1}0=\frac{\pi}{2}\) होता है।
The expression \(\sin^{-1}x\) is defined only for \(x\in[-1,1]\), and \(2\notin[-1,1]\). In domain questions check each option quickly.
Step 2
Why this answer is correct
The correct answer is A. \(\sin^{-1}2\). The expression \(\sin^{-1}x\) is defined only for \(x\in[-1,1]\), and \(2\notin[-1,1]\). In domain questions check each option quickly.
Step 3
Exam Tip
\(\sin^{-1}x\) केवल \(x\in[-1,1]\) के लिए परिभाषित है और \(2\notin[-1,1]\)। डोमेन आधारित प्रश्नों में विकल्प तुरंत जांचें।
If \(\cos^{-1}x=0\), then \(x=\cos0=1\). In an inverse trigonometric equation apply the corresponding trigonometric function on both sides.
Step 2
Why this answer is correct
The correct answer is A. (\1). If \(\cos^{-1}x=0\), then \(x=\cos0=1\). In an inverse trigonometric equation apply the corresponding trigonometric function on both sides.
Step 3
Exam Tip
यदि \(\cos^{-1}x=0\) है तो \(x=\cos0=1\)। उल्टे त्रिकोणमितीय समीकरण में दोनों ओर संबंधित त्रिकोणमितीय फलन लगाएं।
A. \(\sin^{-1}x\) का डोमेन ([-1,1]) है/The domain of \(\sin^{-1}x\) is ([-1,1])
Step 1
Concept
For \(\sin^{-1}x\), the value of (x) must lie in ([-1,1]). The other statements are wrong in domain or range.
Step 2
Why this answer is correct
The correct answer is A. \(\sin^{-1}x\) का डोमेन ([-1,1]) है / The domain of \(\sin^{-1}x\) is ([-1,1]). For \(\sin^{-1}x\), the value of (x) must lie in ([-1,1]). The other statements are wrong in domain or range.
Step 3
Exam Tip
\(\sin^{-1}x\) के लिए (x) का मान ([-1,1]) में होना चाहिए। बाकी कथन डोमेन या सीमा में गलत हैं।
Since (\sin^{-1}(-x)=-\sin^{-1}x), \(\sin^{-1}x\) is odd. The function \(\tan^{-1}x\) is also odd but is not in the options.
Step 2
Why this answer is correct
The correct answer is A. \(\sin^{-1}x\). Since (\sin^{-1}(-x)=-\sin^{-1}x), \(\sin^{-1}x\) is odd. The function \(\tan^{-1}x\) is also odd but is not in the options.
Step 3
Exam Tip
(\sin^{-1}(-x)=-\sin^{-1}x) इसलिए \(\sin^{-1}x\) विषम है। \(\tan^{-1}x\) भी विषम होता है पर विकल्प में नहीं है।
The identity is (\cos^{-1}(-x)=\pi-\cos^{-1}x). Do not solve it by treating \(\cos^{-1}x\) as an odd function.
Step 2
Why this answer is correct
The correct answer is A. \(\pi-\cos^{-1}x\). The identity is (\cos^{-1}(-x)=\pi-\cos^{-1}x). Do not solve it by treating \(\cos^{-1}x\) as an odd function.
Step 3
Exam Tip
पहचान (\cos^{-1}(-x)=\pi-\cos^{-1}x) होती है। इसे \(\cos^{-1}x\) को विषम फलन मानकर हल न करें।
The function \(\sin^{-1}x\) is odd, so (\sin^{-1}(-x)=-\sin^{-1}x). This identity is useful in sign-change questions.
Step 2
Why this answer is correct
The correct answer is A. \(-\sin^{-1}x\). The function \(\sin^{-1}x\) is odd, so (\sin^{-1}(-x)=-\sin^{-1}x). This identity is useful in sign-change questions.
Step 3
Exam Tip
\(\sin^{-1}x\) विषम फलन है इसलिए (\sin^{-1}(-x)=-\sin^{-1}x)। चिन्ह बदलने वाले प्रश्नों में यह पहचान उपयोगी है।
The angle \(-\frac{\pi}{6}\) lies in the principal range (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)), so the value is the same. For \(\tan^{-1}\), the interval is open.
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{\pi}{6}\). The angle \(-\frac{\pi}{6}\) lies in the principal range (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)), so the value is the same. For \(\tan^{-1}\), the interval is open.
Step 3
Exam Tip
\(-\frac{\pi}{6}\) प्रधान सीमा (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)) में है इसलिए मान वही है। \(\tan^{-1}\) में सीमा खुली होती है।
A. \(\sin y=x\) और \(y\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\)/\(\sin y=x\) and \(y\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\)
Step 1
Concept
The statement \(y=\sin^{-1}x\) means \(\sin y=x\) and (y) lies in the principal range. Understanding inverse functions by definition is the safest method.
Step 2
Why this answer is correct
The correct answer is A. \(\sin y=x\) और \(y\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) / \(\sin y=x\) and \(y\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). The statement \(y=\sin^{-1}x\) means \(\sin y=x\) and (y) lies in the principal range. Understanding inverse functions by definition is the safest method.
Step 3
Exam Tip
\(y=\sin^{-1}x\) का अर्थ है \(\sin y=x\) और (y) प्रधान सीमा में है। उल्टे फलन को परिभाषा से समझना सबसे सुरक्षित तरीका है।
A. \(\tan y=x\) और (y\in\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\))/\(\tan y=x\) and (y\in\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\))
Step 1
Concept
The expression \(y=\tan^{-1}x\) means \(\tan y=x\). Here (y) stays in the principal range (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)).
Step 2
Why this answer is correct
The correct answer is A. \(\tan y=x\) और (y\in\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)) / \(\tan y=x\) and (y\in\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)). The expression \(y=\tan^{-1}x\) means \(\tan y=x\). Here (y) stays in the principal range (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)).
Step 3
Exam Tip
\(y=\tan^{-1}x\) का मतलब \(\tan y=x\) होता है। यहां (y) प्रधान सीमा (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)) में रहता है।
Because \(\tan\frac{\pi}{3}=\sqrt{3}\) and \(\frac{\pi}{3}\) lies in the principal range. For \(\tan^{-1}\), take the angle in (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\pi}{3}\). Because \(\tan\frac{\pi}{3}=\sqrt{3}\) and \(\frac{\pi}{3}\) lies in the principal range. For \(\tan^{-1}\), take the angle in (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)).
Step 3
Exam Tip
क्योंकि \(\tan\frac{\pi}{3}=\sqrt{3}\) है और \(\frac{\pi}{3}\) प्रधान सीमा में है। \(\tan^{-1}\) में कोण (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)) में लें।
Because \(\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}\). Inverse values are found quickly from basic trigonometric values.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\pi}{6}\). Because \(\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}\). Inverse values are found quickly from basic trigonometric values.
Step 3
Exam Tip
क्योंकि \(\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}\) है। मूल त्रिकोणमितीय मानों से उल्टे मान जल्दी मिलते हैं।
Since \(\frac{\pi}{4}\in[0,\pi]\), (\cos^{-1}\left\(\cos\frac{\pi}{4}\right\)=\frac{\pi}{4}). If the angle lies in the principal range, it remains unchanged.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\pi}{4}\). Since \(\frac{\pi}{4}\in[0,\pi]\), (\cos^{-1}\left\(\cos\frac{\pi}{4}\right\)=\frac{\pi}{4}). If the angle lies in the principal range, it remains unchanged.
Step 3
Exam Tip
\(\frac{\pi}{4}\in[0,\pi]\) है इसलिए (\cos^{-1}\left\(\cos\frac{\pi}{4}\right\)=\frac{\pi}{4})। प्रधान सीमा में कोण हो तो वही उत्तर आता है।
The angle \(\frac{\pi}{3}\) lies in the principal range (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)). Therefore the value of the expression is \(\frac{\pi}{3}\).
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 (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)). Therefore the value of the expression is \(\frac{\pi}{3}\).
Step 3
Exam Tip
\(\frac{\pi}{3}\) प्रधान सीमा (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)) में आता है। इसलिए व्यंजक का मान \(\frac{\pi}{3}\) है।
Because \(\operatorname{cosec}\frac{\pi}{6}=2\). In \(\operatorname{cosec}^{-1}x\), the magnitude of (x) cannot be less than (1).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\pi}{6}\). Because \(\operatorname{cosec}\frac{\pi}{6}=2\). In \(\operatorname{cosec}^{-1}x\), the magnitude of (x) cannot be less than (1).
Step 3
Exam Tip
क्योंकि \(\operatorname{cosec}\frac{\pi}{6}=2\) है। \(\operatorname{cosec}^{-1}x\) में (x) का परिमाण (1) से कम नहीं हो सकता।
Because \(\cot\frac{\pi}{6}=\sqrt{3}\) and (\frac{\pi}{6}\in\(0,\pi\)). Remember (\(0,\pi\)) for the principal value of \(\cot^{-1}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\pi}{6}\). Because \(\cot\frac{\pi}{6}=\sqrt{3}\) and (\frac{\pi}{6}\in\(0,\pi\)). Remember (\(0,\pi\)) for the principal value of \(\cot^{-1}\).
Step 3
Exam Tip
क्योंकि \(\cot\frac{\pi}{6}=\sqrt{3}\) है और (\frac{\pi}{6}\in\(0,\pi\)) है। \(\cot^{-1}\) के प्रधान मान के लिए (\(0,\pi\)) याद रखें।
The expression \(\cos^{-1}x\) is real only for \(x\in[-1,1]\), and \(\frac{7}{6}>1\). Hence it will not give a real value.
Step 2
Why this answer is correct
The correct answer is A. (\cos^{-1}\left\(\frac{7}{6}\right\)). The expression \(\cos^{-1}x\) is real only for \(x\in[-1,1]\), and \(\frac{7}{6}>1\). Hence it will not give a real value.
Step 3
Exam Tip
\(\cos^{-1}x\) केवल \(x\in[-1,1]\) पर वास्तविक है और \(\frac{7}{6}>1\) है। इसलिए यह वास्तविक मान नहीं देगा।
The range of \(\sin^{-1}x\) is \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). The value \(\frac{2\pi}{3}\) lies outside this range.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{2\pi}{3}\). The range of \(\sin^{-1}x\) is \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). The value \(\frac{2\pi}{3}\) lies outside this range.
Step 3
Exam Tip
\(\sin^{-1}x\) की सीमा \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) है। \(\frac{2\pi}{3}\) इस सीमा के बाहर है।
The range of \(\cos^{-1}x\) is \([0,\pi]\), so a negative value is not possible. In such questions check only the output range.
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{\pi}{3}\). The range of \(\cos^{-1}x\) is \([0,\pi]\), so a negative value is not possible. In such questions check only the output range.
Step 3
Exam Tip
\(\cos^{-1}x\) की सीमा \([0,\pi]\) है इसलिए ऋणात्मक मान संभव नहीं है। ऐसे प्रश्नों में केवल आउटपुट रेंज देखें।
The range of \(\tan^{-1}x\) is (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)), which does not include \(\frac{\pi}{2}\). Pay attention to open intervals.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\pi}{2}\). The range of \(\tan^{-1}x\) is (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)), which does not include \(\frac{\pi}{2}\). Pay attention to open intervals.
Step 3
Exam Tip
\(\tan^{-1}x\) की सीमा (\left\(-\frac{\pi}{2},\frac{\pi}{2}\right\)) है जिसमें \(\frac{\pi}{2}\) शामिल नहीं है। खुले अंतराल पर ध्यान दें।
Let \(\theta=\cos^{-1}\frac{3}{5}\), then \(\cos\theta=\frac{3}{5}\) and \(\theta\in[0,\pi]\). Here \(\sin\theta=\frac{4}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{4}{5}\). Let \(\theta=\cos^{-1}\frac{3}{5}\), then \(\cos\theta=\frac{3}{5}\) and \(\theta\in[0,\pi]\). Here \(\sin\theta=\frac{4}{5}\).
Step 3
Exam Tip
मान लें \(\theta=\cos^{-1}\frac{3}{5}\), तब \(\cos\theta=\frac{3}{5}\) और \(\theta\in[0,\pi]\)। यहां \(\sin\theta=\frac{4}{5}\) होगा।
Because \(\cos\frac{3\pi}{4}=-\frac{1}{\sqrt{2}}\) and \(\frac{3\pi}{4}\in[0,\pi]\). For \(\cos^{-1}\), do not take a negative principal value.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{3\pi}{4}\). Because \(\cos\frac{3\pi}{4}=-\frac{1}{\sqrt{2}}\) and \(\frac{3\pi}{4}\in[0,\pi]\). For \(\cos^{-1}\), do not take a negative principal value.
Step 3
Exam Tip
क्योंकि \(\cos\frac{3\pi}{4}=-\frac{1}{\sqrt{2}}\) और \(\frac{3\pi}{4}\in[0,\pi]\) है। \(\cos^{-1}\) में प्रधान मान ऋणात्मक नहीं लेते।
Let \(\theta=\sin^{-1}\frac{5}{13}\), then \(\sin\theta=\frac{5}{13}\) and \(\theta\) is in the principal range. Therefore \(\cos\theta=\frac{12}{13}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{12}{13}\). Let \(\theta=\sin^{-1}\frac{5}{13}\), then \(\sin\theta=\frac{5}{13}\) and \(\theta\) is in the principal range. Therefore \(\cos\theta=\frac{12}{13}\).
Step 3
Exam Tip
मान लें \(\theta=\sin^{-1}\frac{5}{13}\), तब \(\sin\theta=\frac{5}{13}\) और \(\theta\) प्रधान सीमा में है। इसलिए \(\cos\theta=\frac{12}{13}\) होगा।
Let \(\theta=\cos^{-1}\frac{4}{5}\), then \(\cos\theta=\frac{4}{5}\) and \(\sin\theta=\frac{3}{5}\). Hence \(\tan\theta=\frac{3}{4}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{3}{4}\). Let \(\theta=\cos^{-1}\frac{4}{5}\), then \(\cos\theta=\frac{4}{5}\) and \(\sin\theta=\frac{3}{5}\). Hence \(\tan\theta=\frac{3}{4}\).
Step 3
Exam Tip
मान लें \(\theta=\cos^{-1}\frac{4}{5}\), तब \(\cos\theta=\frac{4}{5}\) और \(\sin\theta=\frac{3}{5}\)। इसलिए \(\tan\theta=\frac{3}{4}\) होगा।
For (\sin^{-1}(2x)), we need \(-1\le 2x\le1\). This gives \(x\in\left[-\frac{1}{2},\frac{1}{2}\right]\).
Step 2
Why this answer is correct
The correct answer is A. \(\left[-\frac{1}{2},\frac{1}{2}\right]\). For (\sin^{-1}(2x)), we need \(-1\le 2x\le1\). This gives \(x\in\left[-\frac{1}{2},\frac{1}{2}\right]\).
Step 3
Exam Tip
(\sin^{-1}(2x)) के लिए \(-1\le 2x\le1\) होना चाहिए। इससे \(x\in\left[-\frac{1}{2},\frac{1}{2}\right]\) मिलता है।
Because (\operatorname{cosec}\left\(-\frac{\pi}{6}\right\)=-2) and \(-\frac{\pi}{6}\) lies in the principal range. For negative values choose the correct principal angle.
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{\pi}{6}\). Because (\operatorname{cosec}\left\(-\frac{\pi}{6}\right\)=-2) and \(-\frac{\pi}{6}\) lies in the principal range. For negative values choose the correct principal angle.
Step 3
Exam Tip
क्योंकि (\operatorname{cosec}\left\(-\frac{\pi}{6}\right\)=-2) और \(-\frac{\pi}{6}\) प्रधान सीमा में है। ऋणात्मक मानों में सही प्रधान कोण चुनें।
