UP Board Solutions Class 12 Maths Chapter 2 Inverse Trigonometric Functions

प्रश्नावली 2.1

Question 1. \( \sin^{ -1 }\left( -\frac { 1 }{ 2 } \right) \) का मुख्य मान ज्ञात कीजिए।
Answer: हल-
sin\(^{-1}\) की मुख्य मान शाखा परिसर \( \left[ -\frac { \pi }{ 2 }, \frac { \pi }{ 2 } \right] \) है।
माना \( \sin^{-1} \left( -\frac { 1 }{ 2 } \right) = \theta \) तब
\( \sin\theta = -\frac { 1 }{ 2 } = \sin \left( -\frac { \pi }{ 6 } \right) \)
\( \implies \theta = -\frac { \pi }{ 6 } \in \left[ -\frac { \pi }{ 2 }, \frac { \pi }{ 2 } \right] \)
\( \therefore \sin^{-1} \left( -\frac { 1 }{ 2 } \right) \) का मुख्य मान \( -\frac { \pi }{ 6 } \) है।
In simple words: To find the principal value of \( \sin^{-1}(-\frac{1}{2}) \), we set it equal to \( \theta \). Since \( \sin(-\frac{\pi}{6}) = -\frac{1}{2} \) and \( -\frac{\pi}{6} \) lies within the principal value range of \( \sin^{-1}x \), the principal value is \( -\frac{\pi}{6} \).

🎯 Exam Tip: Always remember the principal value branch for each inverse trigonometric function to correctly identify the valid angle.

Question 2. \( \cos^{ -1 }\left( \frac { \sqrt { 3 } }{ 2 } \right) \) का मुख्य मान ज्ञात कीजिए।
Answer: हल- \( \cos^{-1} \) की मुख्य मान शाखा परिसर \( [0, \pi] \) है।
माना \( \cos^{-1} \left( \frac { \sqrt { 3 } }{ 2 } \right) = \theta \), तब
\( \cos\theta = \frac { \sqrt { 3 } }{ 2 } = \cos \left( \frac { \pi }{ 6 } \right) \)
\( \implies \theta = \frac { \pi }{ 6 } \in [0, \pi] \)
\( \therefore \cos^{-1} \left( \frac { \sqrt { 3 } }{ 2 } \right) \) का मुख्य मान \( \frac { \pi }{ 6 } \) है।
In simple words: We find the angle \( \theta \) such that \( \cos\theta = \frac{\sqrt{3}}{2} \). Since \( \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \) and \( \frac{\pi}{6} \) is within the principal value range of \( \cos^{-1}x \), the principal value is \( \frac{\pi}{6} \).

🎯 Exam Tip: Familiarize yourself with common trigonometric values for standard angles like \( \pi/6, \pi/4, \pi/3 \), etc., to quickly solve such problems.

Question 3. \( \operatorname{cosec}^{-1} (2) \) का मुख्य मान ज्ञात कीजिए।
Answer: हल-
\( \operatorname{cosec}^{-1} \) की मुख्य माना शाखा परिसर \( \left[ -\frac { \pi }{ 2 }, \frac { \pi }{ 2 } \right] - \{0\} \) है।
माना \( \operatorname{cosec}^{-1} (2) = \theta \), तब
\( \operatorname{cosec}\theta = 2 = \operatorname{cosec} \left( \frac { \pi }{ 6 } \right) \)
\( \implies \theta = \frac { \pi }{ 6 } \in \left[ -\frac { \pi }{ 2 }, \frac { \pi }{ 2 } \right] - \{0\} \)
अब \( \operatorname{cosec}^{-1} (2) \) का मुख्य मान \( \frac { \pi }{ 6 } \) है।
In simple words: To find the principal value of \( \operatorname{cosec}^{-1}(2) \), we let it be \( \theta \). Since \( \operatorname{cosec}(\frac{\pi}{6}) = 2 \) and \( \frac{\pi}{6} \) falls within the defined principal value range, the principal value is \( \frac{\pi}{6} \).

🎯 Exam Tip: Remember that \( \operatorname{cosec}^{-1}x \) has a restriction at \( x=0 \), hence the removal of \( \{0\} \) from its range.

Question 4. \( \tan^{-1} (-\sqrt{3}) \) का मुख्य मान ज्ञात कीजिए।
Answer: हल-
\( \tan^{-1} \) की मुख्य मान शाखा परिसर \( \left( -\frac { \pi }{ 2 }, \frac { \pi }{ 2 } \right) \) है।
माना \( \tan^{-1} (-\sqrt{3}) = \theta \), तब
\( \tan\theta = -\sqrt{3} = \tan \left( -\frac { \pi }{ 3 } \right) \)
\( \implies \theta = -\frac { \pi }{ 3 } \in \left( -\frac { \pi }{ 2 }, \frac { \pi }{ 2 } \right) \)
\( \therefore \tan^{-1} (-\sqrt{3}) \) का मुख्य मान \( -\frac { \pi }{ 3 } \) है।
In simple words: We are looking for the angle \( \theta \) whose tangent is \( -\sqrt{3} \). Since \( \tan(-\frac{\pi}{3}) = -\sqrt{3} \) and \( -\frac{\pi}{3} \) is in the principal range of \( \tan^{-1}x \), the principal value is \( -\frac{\pi}{3} \).

🎯 Exam Tip: For negative arguments, \( \tan^{-1}(-x) = -\tan^{-1}x \), which simplifies finding the principal value directly.

Question 5. \( \cos^{ -1 }\left( -\frac { 1 }{ 2 } \right) \) का मुख्य मान ज्ञात कीजिए।
Answer: हल- \( \cos^{-1} \) की मुख्य मान शाखा परिसर \( [0, \pi] \) है।
माना \( \cos^{-1} \left( -\frac { 1 }{ 2 } \right) = \theta \), तब
\( \cos\theta = -\frac { 1 }{ 2 } = \cos \left( \pi - \frac { \pi }{ 3 } \right) = \cos \left( \frac { 2\pi }{ 3 } \right) \)
\( \implies \theta = \frac { 2\pi }{ 3 } \in [0, \pi] \)
\( \therefore \cos^{-1} \left( -\frac { 1 }{ 2 } \right) \) का मुख्य मान \( \frac { 2\pi }{ 3 } \) है।
In simple words: When the argument of \( \cos^{-1} \) is negative, we use the identity \( \cos^{-1}(-x) = \pi - \cos^{-1}(x) \). So \( \cos^{-1}(-\frac{1}{2}) = \pi - \cos^{-1}(\frac{1}{2}) = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \).

🎯 Exam Tip: For inverse cosine functions with negative values, express the angle as \( \pi - \alpha \) where \( \alpha \) is the reference angle for the positive value, to ensure the result is in \( [0, \pi] \).

Question 6. \( \tan^{-1}(-1) \) का मुख्य मान ज्ञात कीजिए।
Answer: हल-
\( \tan^{-1} \) का मुख्य मान शाखा परिसर \( \left( -\frac { \pi }{ 2 }, \frac { \pi }{ 2 } \right) \) है।
माना \( \tan^{-1}(-1) = \theta \), तब
\( \tan\theta = -1 = \tan \left( -\frac { \pi }{ 4 } \right) \)
\( \implies \theta = -\frac { \pi }{ 4 } \in \left( -\frac { \pi }{ 2 }, \frac { \pi }{ 2 } \right) \)
\( \therefore \tan^{-1}(-1) \) का मुख्य मान \( -\frac { \pi }{ 4 } \) है।
In simple words: The value of \( \theta \) for which \( \tan\theta = -1 \) and \( \theta \) is in the range \( (-\frac{\pi}{2}, \frac{\pi}{2}) \) is \( -\frac{\pi}{4} \).

🎯 Exam Tip: Direct application of \( \tan^{-1}(-x) = -\tan^{-1}(x) \) makes solving these simple and quick.

Question 7. \( \sec^{ -1 }\left(\frac { 2 }{ \sqrt { 3 } } \right) \) का मुख्य मान ज्ञात कीजिए।
Answer: हल-
\( \sec^{-1} \) की मुख्य मान शाखा परिसर \( [0, \pi] - \left\{ \frac { \pi }{ 2 } \right\} \) है।
माना \( \sec^{-1} \left( \frac { 2 }{ \sqrt { 3 } } \right) = \theta \), तब \( \sec\theta = \frac { 2 }{ \sqrt { 3 } } = \sec \left( \frac { \pi }{ 6 } \right) \)
\( \implies \theta = \frac { \pi }{ 6 } \in [0, \pi] - \left\{ \frac { \pi }{ 2 } \right\} \)
\( \therefore \sec^{-1} \left( \frac { 2 }{ \sqrt { 3 } } \right) \) का मुख्य मान \( \frac { \pi }{ 6 } \) है।
In simple words: We find the angle \( \theta \) such that \( \sec\theta = \frac{2}{\sqrt{3}} \). As \( \sec(\frac{\pi}{6}) = \frac{2}{\sqrt{3}} \) and \( \frac{\pi}{6} \) is within the principal value range for \( \sec^{-1}x \), the principal value is \( \frac{\pi}{6} \).

🎯 Exam Tip: When dealing with \( \sec^{-1}x \), it's often helpful to think of it in terms of \( \cos^{-1}(1/x) \) and then apply the principal value rules for cosine.

Question 8. \( \cot^{-1} (\sqrt{3}) \) का मुख्य मान ज्ञात कीजिए।
Answer: हल-
\( \cot^{-1} \) का मुख्य मान शाखा परिसर \( (0, \pi) \) है।
\( \cot\theta = \sqrt{3} = \cot \left( \frac { \pi }{ 6 } \right) \)
\( \implies \theta = \frac { \pi }{ 6 } \in (0, \pi) \)
\( \therefore \cot^{-1} (\sqrt{3}) \) का मुख्य मान \( \frac { \pi }{ 6 } \) है।
In simple words: The angle \( \theta \) for which \( \cot\theta = \sqrt{3} \) and \( \theta \) lies in the range \( (0, \pi) \) is \( \frac{\pi}{6} \).

🎯 Exam Tip: Similar to tangent, the principal value for positive arguments of \( \cot^{-1}x \) is a direct angle, but for negative arguments, it follows \( \pi - \cot^{-1}|x| \).

Question 9. \( \cos^{ -1 }\left( \frac { -1 }{ \sqrt { 2 } } \right) \) का मान ज्ञात कीजिए।
Answer: हल-
\( \cos^{-1} \) का मुख्य मान शाखा परिसर \( [0, \pi] \) है।
माना \( \cos^{-1} \left( -\frac { 1 }{ \sqrt { 2 } } \right) = \theta \) तब
\( \cos\theta = -\frac { 1 }{ \sqrt { 2 } } = \cos \left( \pi - \frac { \pi }{ 4 } \right) = \cos \left( \frac { 3\pi }{ 4 } \right) \)
\( \implies \theta = \frac { 3\pi }{ 4 } \in [0, \pi] \)
\( \therefore \cos^{-1} \left( -\frac { 1 }{ \sqrt { 2 } } \right) \) का मुख्य मान \( \frac { 3\pi }{ 4 } \) है।
In simple words: Using the property \( \cos^{-1}(-x) = \pi - \cos^{-1}(x) \), we find \( \cos^{-1}(-\frac{1}{\sqrt{2}}) = \pi - \cos^{-1}(\frac{1}{\sqrt{2}}) = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \).

🎯 Exam Tip: Be cautious with negative arguments; ensure the final angle lies within the specific principal value range for each inverse trigonometric function.

Question 10. \( \operatorname{cosec}^{-1} (-\sqrt{2}) \) का मुख्य मान ज्ञात कीजिए।
Answer: हल-
\( \operatorname{cosec}^{-1} \) की मुख्य मान शाखा परिसर \( \left[ -\frac { \pi }{ 2 }, \frac { \pi }{ 2 } \right] - \{0\} \) है।
माना \( \operatorname{cosec}^{-1} (-\sqrt{2}) = \theta \), तब
\( \operatorname{cosec}\theta = -\sqrt{2} = \operatorname{cosec} \left( -\frac { \pi }{ 4 } \right) \)
\( \implies \theta = -\frac { \pi }{ 4 } \in \left[ -\frac { \pi }{ 2 }, \frac { \pi }{ 2 } \right] - \{0\} \)
\( \therefore \operatorname{cosec}^{-1} (-\sqrt{2}) \) का मुख्य मान \( -\frac { \pi }{ 4 } \) है।
In simple words: Since \( \operatorname{cosec}(-\frac{\pi}{4}) = -\sqrt{2} \) and this angle is within the principal value range for \( \operatorname{cosec}^{-1}x \), the principal value is \( -\frac{\pi}{4} \).

🎯 Exam Tip: For inverse cosecant with a negative argument, use \( \operatorname{cosec}^{-1}(-x) = -\operatorname{cosec}^{-1}(x) \).

Question 11. का मान ज्ञात कीजिए।
Answer: हल-
\( \tan^{-1} (1) + \cos^{-1} \left( -\frac { 1 }{ 2 } \right) + \sin^{-1} \left( -\frac { 1 }{ 2 } \right) \)
\( = \frac { \pi }{ 4 } + \frac { 2\pi }{ 3 } - \frac { \pi }{ 6 } \)
\( = \frac { 3\pi + 8\pi - 2\pi }{ 12 } = \frac { 9\pi }{ 12 } = \frac { 3\pi }{ 4 } \)
In simple words: We find the principal values for each inverse trigonometric term: \( \tan^{-1}(1) = \frac{\pi}{4} \), \( \cos^{-1}(-\frac{1}{2}) = \frac{2\pi}{3} \), and \( \sin^{-1}(-\frac{1}{2}) = -\frac{\pi}{6} \). Then, we sum these values to get the final result.

🎯 Exam Tip: Break down complex expressions into individual inverse trigonometric evaluations first, then perform the arithmetic operations carefully, ensuring a common denominator for fractions.

Question 12. का मान ज्ञात कीजिए।
Answer: हल-
माना \( y = \cot (\tan^{-1} a + \cot^{-1} a) \)
\( \because \tan^{-1} a + \cot^{-1} a = \frac { \pi }{ 2 } \)
\( = \cot \frac { \pi }{ 2 } \)
\( \therefore \cot \frac { \pi }{ 2 } = 0 \)
\( = 0 \)
In simple words: Using the identity \( \tan^{-1}x + \cot^{-1}x = \frac{\pi}{2} \), the expression simplifies to \( \cot(\frac{\pi}{2}) \), which equals 0.

🎯 Exam Tip: Recognize fundamental inverse trigonometric identities like \( \tan^{-1}x + \cot^{-1}x = \frac{\pi}{2} \) to simplify expressions rapidly.

Question 13. यदि \( \sin^{-1} x = y \), तो
Answer: हल-
\( \because \sin^{-1} \) का मुख्य मान शाखा परिसर \( \left[ -\frac { \pi }{ 2 }, \frac { \pi }{ 2 } \right] \) है।
\( \therefore -\frac { \pi }{ 2 } \le y \le \frac { \pi }{ 2 } \)
\( \therefore \) विकल्प (b) सही है।
In simple words: The principal value range for \( \sin^{-1}x \) is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \). Therefore, if \( y = \sin^{-1}x \), then \( y \) must lie within this interval.

🎯 Exam Tip: Knowing the principal value ranges of all inverse trigonometric functions is crucial for multiple-choice questions and fundamental understanding.

Question 14. \( \tan^{-1}\sqrt{3} - \sec^{-1}(-2) \) का मान बराबर है
Answer: हल-
माना \( x = \tan^{-1}\sqrt{3} \)
\( \implies \tan x = \sqrt{3} = \tan \frac { \pi }{ 3 } \)
\( \therefore x = \frac { \pi }{ 3 } \)
\( x \in \left( -\frac { \pi }{ 2 }, \frac { \pi }{ 2 } \right) \)
पुनः माना \( y = \sec^{-1}(-2) \)
\( \implies \sec y = -2 \)
\( \sec y = -\sec \frac { \pi }{ 3 } = \sec \left( \pi - \frac { \pi }{ 3 } \right) = \sec \left( \frac { 2\pi }{ 3 } \right) \)
\( \implies y = \frac { 2\pi }{ 3 } \)
\( \because y \in [0, \pi] - \left\{ \frac { \pi }{ 2 } \right\} \)
\( \therefore \tan^{-1}\sqrt{3} - \sec^{-1}(-2) = x - y = \frac { \pi }{ 3 } - \frac { 2\pi }{ 3 } = -\frac { \pi }{ 3 } \)
अतः विकल्प (B) सही है।
In simple words: We calculate \( \tan^{-1}\sqrt{3} = \frac{\pi}{3} \) and \( \sec^{-1}(-2) = \pi - \sec^{-1}(2) = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \). Subtracting these values gives \( \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3} \).

🎯 Exam Tip: Always evaluate each inverse trigonometric term separately, considering its principal value range, before combining them into the final expression.

प्रश्नावली 2.2

निम्नलिखित को सिद्ध कीजिए-

Question 1.
Answer: हल-
\( 3\sin^{-1}x = \sin^{-1}(3x - 4x^3) \)
माना \( \sin^{-1}x = \theta \)
\( \therefore \sin\theta = x \)
पुनः \( \sin 3\theta = 3 \sin^3 \theta - 4 \sin^9 \theta \)
\( \implies \sin 3\theta = 3x - 4x^3 \)
\( \implies 3\theta = \sin^{-1}(3x - 4x^3) \)
\( \therefore 3\sin^{-1}x = \sin^{-1}(3x - 4x^3) \) इति सिद्धम्
In simple words: By substituting \( x = \sin\theta \), the right-hand side becomes \( \sin^{-1}(3\sin\theta - 4\sin^3\theta) \). Recognizing the identity \( 3\sin\theta - 4\sin^3\theta = \sin(3\theta) \), the expression simplifies to \( \sin^{-1}(\sin(3\theta)) = 3\theta \), which is \( 3\sin^{-1}x \), proving the identity.

🎯 Exam Tip: For identities involving \( 3\sin^{-1}x \) or \( 3\cos^{-1}x \), substitutions like \( x=\sin\theta \) or \( x=\cos\theta \) are key to using the triple angle formulas.

Question 2.
Answer: हल-
\( 3\cos^{-1}x = \cos^{-1}(4x^3 - 3x) \)
माना \( \cos^{-1}x = \theta \)
\( \therefore \cos\theta = x \)
\( \therefore \cos 3\theta = 4\cos^3\theta - 3\cos\theta \)
\( \implies \cos 3\theta = 4x^3 - 3x \)
\( \implies 3\theta = \cos^{-1}(4x^3 - 3x) \)
\( \therefore 3\cos^{-1}x = \cos^{-1}(4x^3 - 3x) \) इति सिद्धम्
In simple words: Let \( x = \cos\theta \). The right side becomes \( \cos^{-1}(4\cos^3\theta - 3\cos\theta) \). Using the identity \( 4\cos^3\theta - 3\cos\theta = \cos(3\theta) \), the expression simplifies to \( \cos^{-1}(\cos(3\theta)) = 3\theta \), which is \( 3\cos^{-1}x \), proving the identity.

🎯 Exam Tip: Remember the triple angle formulas for sine and cosine. These are essential for proving inverse trigonometric identities that involve a factor of 3.

Question 3.
Answer: हल-
बायाँ पक्ष \( = \tan^{-1}\frac{2}{11} + \tan^{-1}\frac{7}{24} \)
\( = \tan^{-1} \left( \frac { \frac { 2 }{ 11 } + \frac { 7 }{ 24 } }{ 1 - \frac { 2 }{ 11 } \times \frac { 7 }{ 24 } } \right) \)
\( = \tan^{-1} \left( \frac { 48 + 77 }{ 264 - 14 } \right) \)
\( = \tan^{-1} \left( \frac { 125 }{ 250 } \right) \)
\( = \tan^{-1} \left( \frac { 1 }{ 2 } \right) \) = दायाँ पक्ष
In simple words: We apply the sum formula for inverse tangents, \( \tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right) \), and simplify the fractions to arrive at the desired result of \( \tan^{-1}(\frac{1}{2}) \).

🎯 Exam Tip: The \( \tan^{-1}x + \tan^{-1}y \) formula is frequently tested. Pay close attention to the denominator \( 1-xy \) to avoid errors, especially when \( xy > 1 \).

Question 4.
Answer: हल-
ज्ञात है, \( 2\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{7}\right) = \tan^{-1}\left(\frac{31}{17}\right) \)
\( \therefore 2\tan^{-1}\left(\frac{1}{2}\right) = \tan^{-1}\left(\frac{2 \times \frac{1}{2}}{1 - \left(\frac{1}{2}\right)^2}\right) = \tan^{-1}\left(\frac{1}{1 - \frac{1}{4}}\right) = \tan^{-1}\left(\frac{1}{\frac{3}{4}}\right) = \tan^{-1}\left(\frac{4}{3}\right) \)
\( \left[ \because 2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \right] \)
बायाँ पक्ष \( = 2\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{7}\right) \)
\( = \tan^{-1}\left(\frac{4}{3}\right) + \tan^{-1}\left(\frac{1}{7}\right) \)
\( = \tan^{-1}\left(\frac{\frac{4}{3} + \frac{1}{7}}{1 - \frac{4}{3} \times \frac{1}{7}}\right) \)
\( = \tan^{-1}\left(\frac{28+3}{21-4}\right) \)
\( = \tan^{-1}\left(\frac{31}{17}\right) \) = दायाँ पक्ष
इति सिद्धम्
In simple words: First, convert \( 2\tan^{-1}(\frac{1}{2}) \) into a single \( \tan^{-1} \) term using the formula \( 2\tan^{-1}x = \tan^{-1}(\frac{2x}{1-x^2}) \). This simplifies to \( \tan^{-1}(\frac{4}{3}) \). Then, apply the \( \tan^{-1}A + \tan^{-1}B \) formula to \( \tan^{-1}(\frac{4}{3}) + \tan^{-1}(\frac{1}{7}) \) to reach the right-hand side \( \tan^{-1}(\frac{31}{17}) \).

🎯 Exam Tip: When an expression contains \( 2\tan^{-1}x \), always convert it first to a single \( \tan^{-1} \) form before combining it with other inverse tangent terms.

निम्नलिखित फलनों को सरलतम रूप में लिखिए

Question 5.
Answer: हल-
\( \tan^{-1}\left(\frac{\sqrt{1+x^2} - 1}{x}\right) \)
माना \( x = \tan\theta \)
तब \( \tan^{-1}\left(\frac{\sqrt{1+\tan^2\theta} - 1}{\tan\theta}\right) \)
\( = \tan^{-1}\left(\frac{\sec\theta - 1}{\tan\theta}\right) \)
\( = \tan^{-1}\left(\frac{\frac{1}{\cos\theta} - 1}{\frac{\sin\theta}{\cos\theta}}\right) \)
\( = \tan^{-1}\left(\frac{1 - \cos\theta}{\sin\theta}\right) \)
\( = \tan^{-1}\left(\frac{2\sin^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)}\right) \)
\( = \tan^{-1}\left(\frac{\sin(\theta/2)}{\cos(\theta/2)}\right) \)
\( = \tan^{-1}(\tan(\theta/2)) = \theta/2 = \frac{1}{2}\tan^{-1}x \)
In simple words: Substitute \( x = \tan\theta \) to simplify the expression under the square root and the fraction. Using trigonometric identities, the expression transforms to \( \tan^{-1}(\tan(\frac{\theta}{2})) \), which simplifies to \( \frac{\theta}{2} \), then replace \( \theta \) back with \( \tan^{-1}x \).

🎯 Exam Tip: For expressions involving \( \sqrt{1+x^2} \), a substitution like \( x=\tan\theta \) or \( x=\cot\theta \) is often effective. For \( \sqrt{1-x^2} \), use \( x=\sin\theta \) or \( x=\cos\theta \). For \( \sqrt{x^2-1} \), use \( x=\sec\theta \) or \( x=\operatorname{cosec}\theta \).

Question 6. \( \tan^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right), |x|>1 \)
Answer: हल-
माना \( x = \operatorname{cosec}\theta \)
\( \tan^{-1}\left(\frac{1}{\sqrt{\operatorname{cosec}^2\theta-1}}\right) \)
\( = \tan^{-1}\left(\frac{1}{\sqrt{\cot^2\theta}}\right) \)
\( = \tan^{-1}\left(\frac{1}{\cot\theta}\right) \)
\( = \tan^{-1}(\tan\theta) = \theta \)
\( = \operatorname{cosec}^{-1}x = \frac{\pi}{2} - \sec^{-1}x \)
In simple words: Substitute \( x = \operatorname{cosec}\theta \) (since \( |x|>1 \)). The expression simplifies to \( \tan^{-1}(\frac{1}{\sqrt{\operatorname{cosec}^2\theta-1}}) = \tan^{-1}(\frac{1}{\cot\theta}) = \tan^{-1}(\tan\theta) = \theta \). Since \( x = \operatorname{cosec}\theta \), then \( \theta = \operatorname{cosec}^{-1}x \).

🎯 Exam Tip: Remember the Pythagorean identity \( \operatorname{cosec}^2\theta - 1 = \cot^2\theta \), which is crucial for simplifying expressions involving \( \sqrt{x^2-1} \) when \( x=\operatorname{cosec}\theta \).

Question 7.
Answer: हल-
माना \( y = \tan^{-1}\sqrt{\frac{1-\cos x}{1+\cos x}} \)
\( = \tan^{-1}\sqrt{\frac{1-(1-2\sin^2(x/2))}{1+(2\cos^2(x/2)-1)}} \)
\( = \tan^{-1}\sqrt{\frac{2\sin^2(x/2)}{2\cos^2(x/2)}} \)
\( = \tan^{-1}\sqrt{\tan^2(x/2)} \)
\( = \tan^{-1}(\tan(x/2)) \)
\( = x/2 \)
In simple words: Use the half-angle identities \( 1-\cos x = 2\sin^2(x/2) \) and \( 1+\cos x = 2\cos^2(x/2) \). The expression inside the square root simplifies to \( \tan^2(x/2) \), leading to \( \tan^{-1}(\tan(x/2)) = x/2 \).

🎯 Exam Tip: Identities involving \( 1 \pm \cos x \) are very common. Always simplify them using half-angle formulas to \( 2\sin^2(x/2) \) or \( 2\cos^2(x/2) \).

Question 8.
Answer: हल-
\( \tan^{-1}\left(\frac{\cos x - \sin x}{\cos x + \sin x}\right) \)
अंश व हर में \( \cos x \) से भाग देने पर,
\( = \tan^{-1}\left(\frac{1 - \frac{\sin x}{\cos x}}{1 + \frac{\sin x}{\cos x}}\right) \)
\( = \tan^{-1}\left(\frac{1 - \tan x}{1 + \tan x}\right) \)
\( = \tan^{-1}\left(\tan\left(\frac{\pi}{4} - x\right)\right) \)
\( = \frac{\pi}{4} - x \)
In simple words: Divide the numerator and denominator by \( \cos x \) to convert the expression into terms of \( \tan x \). Recognize the form \( \frac{1-\tan x}{1+\tan x} \) as the expansion of \( \tan(\frac{\pi}{4}-x) \), which simplifies the expression to \( \frac{\pi}{4}-x \).

🎯 Exam Tip: When you see expressions like \( (\cos x - \sin x) / (\cos x + \sin x) \), immediately think of dividing by \( \cos x \) to form \( \tan(\frac{\pi}{4} \pm x) \).

Question 9.
Answer: हल-
\( \tan^{-1}\left(\frac{x}{\sqrt{a^2 - x^2}}\right) \)
माना \( x = a\sin\theta \)
\( \tan^{-1}\left(\frac{a\sin\theta}{\sqrt{a^2 - a^2\sin^2\theta}}\right) \)
\( = \tan^{-1}\left(\frac{a\sin\theta}{\sqrt{a^2(1 - \sin^2\theta)}}\right) \)
\( = \tan^{-1}\left(\frac{a\sin\theta}{\sqrt{a^2\cos^2\theta}}\right) \)
\( = \tan^{-1}\left(\frac{a\sin\theta}{a\cos\theta}\right) \)
\( = \tan^{-1}(\tan\theta) = \theta = \sin^{-1}\left(\frac{x}{a}\right) \)
In simple words: Substitute \( x = a\sin\theta \) to simplify the expression, especially the term \( \sqrt{a^2-x^2} \). The expression then reduces to \( \tan^{-1}(\tan\theta) = \theta \). Finally, replace \( \theta \) with \( \sin^{-1}(\frac{x}{a}) \).

🎯 Exam Tip: For expressions involving \( \sqrt{a^2-x^2} \), the standard substitution is \( x=a\sin\theta \) or \( x=a\cos\theta \). Always choose the substitution that leads to simplification using Pythagorean identities.

Question 10.
Answer: हल-
माना \( y = \tan^{-1}\left(\frac{3a^2x - x^3}{a^3 - 3ax^2}\right) \)
\( x = a\tan\theta \) रखने पर,
\( \therefore \tan\theta = \frac{x}{a} \implies \theta = \tan^{-1}\left(\frac{x}{a}\right) \)
\( y = \tan^{-1}\left(\frac{3a^2(a\tan\theta) - (a\tan\theta)^3}{a^3 - 3a(a\tan\theta)^2}\right) \)
\( = \tan^{-1}\left(\frac{3a^3\tan\theta - a^3\tan^3\theta}{a^3 - 3a^3\tan^2\theta}\right) \)
\( = \tan^{-1}\left(\frac{a^3(3\tan\theta - \tan^3\theta)}{a^3(1 - 3\tan^2\theta)}\right) \)
\( = \tan^{-1}\left(\frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}\right) \)
\( = \tan^{-1}(\tan(3\theta)) \)
\( \left[ \because \tan 3A = \frac{3\tan A - \tan^3 A}{1 - 3\tan^2 A} \right] \)
\( = 3\theta = 3\tan^{-1}\left(\frac{x}{a}\right) \)
In simple words: Substitute \( x = a\tan\theta \) to simplify the complex fractional expression. After cancellation of \( a^3 \) and rearranging terms, the expression matches the formula for \( \tan(3\theta) \). Thus, it simplifies to \( \tan^{-1}(\tan(3\theta)) = 3\theta \), and finally \( 3\tan^{-1}(\frac{x}{a}) \).

🎯 Exam Tip: This form \( \tan^{-1}\left(\frac{3a^2x - x^3}{a^3 - 3ax^2}\right) \) is a classic identity that becomes \( 3\tan^{-1}(\frac{x}{a}) \) after substitution \( x=a\tan\theta \). Memorize this specific transformation.

Question 11.
Answer: हल-
\( \tan^{-1}\left\{2\cos\left(2\sin^{-1}\left(\frac{1}{2}\right)\right)\right\} \)
माना \( \sin^{-1}\left(\frac{1}{2}\right) = \theta \)
तब \( \sin\theta = \frac{1}{2} \implies \theta = \frac{\pi}{6} \)
\( = \tan^{-1}\left\{2\cos\left(2 \times \frac{\pi}{6}\right)\right\} \)
\( = \tan^{-1}\left\{2\cos\left(\frac{\pi}{3}\right)\right\} \)
\( = \tan^{-1}\left\{2 \times \frac{1}{2}\right\} \)
\( = \tan^{-1}(1) = \frac{\pi}{4} \)
In simple words: First, evaluate the innermost expression \( \sin^{-1}(\frac{1}{2}) = \frac{\pi}{6} \). Substitute this into the next layer to get \( 2\cos(2 \times \frac{\pi}{6}) = 2\cos(\frac{\pi}{3}) = 2 \times \frac{1}{2} = 1 \). Finally, calculate \( \tan^{-1}(1) = \frac{\pi}{4} \).

🎯 Exam Tip: Evaluate complex nested inverse trigonometric expressions from the innermost function outwards, step-by-step, to simplify them correctly.

Question 12. cot \( (\tan^{-1} a + \cot^{-1} a) \) का मान ज्ञात कीजिए।
Answer: हल-
माना \( y = \cot (\tan^{-1} a + \cot^{-1} a) \)
\( \because \tan^{-1} a + \cot^{-1} a = \frac{\pi}{2} \)
\( = \cot \frac{\pi}{2} \)
\( = 0 \)
\( \left[ \because \cot \frac{\pi}{2} = 0 \right] \)
In simple words: Using the fundamental identity \( \tan^{-1}x + \cot^{-1}x = \frac{\pi}{2} \), the expression simplifies to \( \cot(\frac{\pi}{2}) \), which has a value of 0.

🎯 Exam Tip: Recognize and apply the complementary inverse trigonometric identities, such as \( \tan^{-1}x + \cot^{-1}x = \frac{\pi}{2} \), to quickly solve problems involving these pairs.

Question 13.
Answer: हल-
माना \( y = \frac{1}{2} \left[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) + \cos^{-1}\left(\frac{1-y^2}{1+y^2}\right) \right] \)
माना \( x = \tan\theta \) तथा \( y = \tan\phi \)
\( \implies \theta = \tan^{-1}x \) तथा \( \phi = \tan^{-1}y \)
\( y = \frac{1}{2} \left[ \sin^{-1}\left(\frac{2\tan\theta}{1+\tan^2\theta}\right) + \cos^{-1}\left(\frac{1-\tan^2\phi}{1+\tan^2\phi}\right) \right] \)
\( = \frac{1}{2} [\sin^{-1}(\sin 2\theta) + \cos^{-1}(\cos 2\phi)] \)
\( = \frac{1}{2} [2\theta + 2\phi] = \frac{1}{2} \times 2(\theta + \phi) = \theta + \phi \)
\( = \tan^{-1}x + \tan^{-1}y \)
\( = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \)
(यह प्रश्न गलत है, यहाँ \( y \) का मान \( x \) के पदों में ज्ञात करना है।)
\( y = \tan^{-1}x + \tan^{-1}y \)
\( y - \tan^{-1}y = \tan^{-1}x \)
In simple words: Substitute \( x = \tan\theta \) and \( y = \tan\phi \). The expressions for \( \sin^{-1} \) and \( \cos^{-1} \) simplify to \( 2\theta \) and \( 2\phi \) respectively using double angle formulas. Summing these gives \( \theta+\phi \), which translates back to \( \tan^{-1}x + \tan^{-1}y \). The problem statement might have a typo as \( y \) is expressed in terms of \( x \) and \( y \) itself.

🎯 Exam Tip: Recognize the standard substitutions for \( \sin^{-1}\left(\frac{2x}{1+x^2}\right) \) and \( \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \), which are \( 2\tan^{-1}x \) (under appropriate domain conditions). These are crucial for simplification.

Question 14.
Answer: हल-
\( \sin \left( \sin^{-1}\left(\frac{1}{5}\right) + \cos^{-1}x \right) = 1 \)
या \( \sin^{-1}\left(\frac{1}{5}\right) + \cos^{-1}x = \sin^{-1}(1) \)
या \( \sin^{-1}\left(\frac{1}{5}\right) + \cos^{-1}x = \frac{\pi}{2} \)
चूँकि \( \sin^{-1}y + \cos^{-1}y = \frac{\pi}{2} \)
अतः \( x = \frac{1}{5} \)
(वैकल्पिक विधि)
या \( \sin \left[ \cos^{-1}\sqrt{1-\left(\frac{1}{5}\right)^2} + \cos^{-1}x \right] = 1 \)
या \( \sin \left[ \cos^{-1}\left(\frac{\sqrt{24}}{5}\right) + \cos^{-1}x \right] = 1 \)
या \( \cos^{-1}\left(\frac{\sqrt{24}}{5}\right) + \cos^{-1}x = \sin^{-1}(1) = \frac{\pi}{2} \)
या \( \cos^{-1}x = \frac{\pi}{2} - \cos^{-1}\left(\frac{\sqrt{24}}{5}\right) \)
या \( \cos^{-1}x = \sin^{-1}\left(\frac{\sqrt{24}}{5}\right) \)
या \( \cos^{-1}x = \cos^{-1}\sqrt{1 - \left(\frac{\sqrt{24}}{5}\right)^2} \)
या \( \cos^{-1}x = \cos^{-1}\sqrt{1 - \frac{24}{25}} \)
या \( \cos^{-1}x = \cos^{-1}\sqrt{\frac{1}{25}} \)
या \( \cos^{-1}x = \cos^{-1}\left(\frac{1}{5}\right) \)
या \( x = \frac{1}{5} \)
In simple words: We are given \( \sin(\sin^{-1}(\frac{1}{5}) + \cos^{-1}x) = 1 \). This implies \( \sin^{-1}(\frac{1}{5}) + \cos^{-1}x = \sin^{-1}(1) = \frac{\pi}{2} \). Since we know \( \sin^{-1}y + \cos^{-1}y = \frac{\pi}{2} \), by comparing the terms, we directly find \( x = \frac{1}{5} \). An alternative, longer method involves converting \( \sin^{-1}(\frac{1}{5}) \) to \( \cos^{-1}(\frac{\sqrt{24}}{5}) \) and then using properties of inverse functions.

🎯 Exam Tip: If an equation is in the form \( \sin(A+B)=1 \), immediately deduce \( A+B = \frac{\pi}{2} \). Then apply inverse trigonometric identities such as \( \sin^{-1}y + \cos^{-1}y = \frac{\pi}{2} \) to simplify and solve for the unknown.

Question 15.
Answer: हल-
प्रश्नानुसार, \( \tan^{-1}\left(\frac{x-1}{x-2}\right) + \tan^{-1}\left(\frac{x+1}{x+2}\right) = \frac{\pi}{4} \)
या \( \tan^{-1}\left(\frac{\frac{x-1}{x-2} + \frac{x+1}{x+2}}{1 - \frac{x-1}{x-2} \frac{x+1}{x+2}}\right) = \frac{\pi}{4} \)
या \( \frac{(x-1)(x+2) + (x+1)(x-2)}{(x-2)(x+2) - (x-1)(x+1)} = \tan\left(\frac{\pi}{4}\right) \)
या \( \frac{(x^2+2x-x-2) + (x^2-2x+x-2)}{(x^2-4) - (x^2-1)} = 1 \)
या \( \frac{x^2+x-2 + x^2-x-2}{x^2-4-x^2+1} = 1 \)
या \( \frac{2x^2-4}{-3} = 1 \)
या \( 2x^2-4 = -3 \)
\( \implies 2x^2 = 1 \)
या \( x^2 = \frac{1}{2} \)
\( \implies x = \pm\frac{1}{\sqrt{2}} \)
In simple words: Use the \( \tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right) \) formula. Simplify the algebraic expression on the left-hand side, then set it equal to \( \tan(\frac{\pi}{4}) = 1 \). Solve the resulting equation for \( x \).

🎯 Exam Tip: When using the \( \tan^{-1}A + \tan^{-1}B \) formula, be meticulous with algebraic simplification of the numerator and denominator to avoid errors in solving the quadratic equation.

Question 16.
Answer: हल-
\( \sin^{-1}\left(\sin\left(\frac{2\pi}{3}\right)\right) \)
\( \sin^{-1} \) की मुख्य मान शाखा परिसर \( \left[ -\frac { \pi }{ 2 }, \frac { \pi }{ 2 } \right] \) है।
\( \therefore \sin^{-1}\left(\sin\left(\frac{2\pi}{3}\right)\right) \neq \frac{2\pi}{3} \)
अब \( \sin^{-1}\left(\sin\left(\frac{2\pi}{3}\right)\right) = \sin^{-1}\left\{\sin\left(\pi - \frac{\pi}{3}\right)\right\} = \sin^{-1}\left(\sin\left(\frac{\pi}{3}\right)\right) \)
\( = \frac{\pi}{3} \)
\( \because \frac{\pi}{3} \in \left[ -\frac { \pi }{ 2 }, \frac { \pi }{ 2 } \right] \)
In simple words: Since \( \frac{2\pi}{3} \) is not in the principal value range of \( \sin^{-1}x \), we use the identity \( \sin(\pi - \theta) = \sin\theta \) to write \( \sin(\frac{2\pi}{3}) = \sin(\pi - \frac{\pi}{3}) = \sin(\frac{\pi}{3}) \). Then \( \sin^{-1}(\sin(\frac{\pi}{3})) = \frac{\pi}{3} \), which is in the correct range.

🎯 Exam Tip: Remember that \( \sin^{-1}(\sin\theta) = \theta \) only if \( \theta \) is in the principal value range \( [-\frac{\pi}{2}, \frac{\pi}{2}] \). If not, use trigonometric identities to find an equivalent angle within the range.

Question 17.
Answer: हल-
\( \tan^{-1}\left(\tan\left(\frac{3\pi}{4}\right)\right) \)
\( \tan^{-1} \) का मुख्य मान शाखा परिसर \( \left( -\frac { \pi }{ 2 }, \frac { \pi }{ 2 } \right) \) है।
\( \therefore \tan^{-1}\left(\tan\left(\frac{3\pi}{4}\right)\right) \neq \frac{3\pi}{4} \)
अब \( \tan^{-1}\left(\tan\left(\frac{3\pi}{4}\right)\right) = \tan^{-1}\left\{\tan\left(\pi - \frac{\pi}{4}\right)\right\} = \tan^{-1}\left(-\tan\left(\frac{\pi}{4}\right)\right) \)
\( = \tan^{-1}\left(\tan\left(-\frac{\pi}{4}\right)\right) \)
\( = -\frac{\pi}{4} \)
\( \because -\frac{\pi}{4} \in \left( -\frac { \pi }{ 2 }, \frac { \pi }{ 2 } \right) \)
In simple words: Since \( \frac{3\pi}{4} \) is outside the principal value range of \( \tan^{-1}x \), we use the identity \( \tan(\pi - \theta) = -\tan\theta \) to write \( \tan(\frac{3\pi}{4}) = \tan(\pi - \frac{\pi}{4}) = -\tan(\frac{\pi}{4}) \). Then, using \( -\tan\theta = \tan(-\theta) \), we get \( \tan(-\frac{\pi}{4}) \). Thus, \( \tan^{-1}(\tan(-\frac{\pi}{4})) = -\frac{\pi}{4} \), which is in the correct range.

🎯 Exam Tip: For \( \tan^{-1}(\tan\theta) \), similar to \( \sin^{-1}(\sin\theta) \), ensure \( \theta \) is within \( (-\frac{\pi}{2}, \frac{\pi}{2}) \). If not, adjust the angle using periodicity or quadrant rules to fit the principal range.

Question 18.
Answer: हल-
\( \tan \left( \sin^{-1}\left(\frac{3}{5}\right) + \cot^{-1}\left(\frac{3}{2}\right) \right) \)
माना \( \sin^{-1}\left(\frac{3}{5}\right) = \theta \) तब \( \sin\theta = \frac{3}{5} \)
\( \cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \)
\( \therefore \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{3/5}{4/5} = \frac{3}{4} \)
\( \implies \theta = \tan^{-1}\left(\frac{3}{4}\right) \)
\( \cot^{-1}\left(\frac{3}{2}\right) = \tan^{-1}\left(\frac{2}{3}\right) \)
\( = \tan \left( \tan^{-1}\left(\frac{3}{4}\right) + \tan^{-1}\left(\frac{2}{3}\right) \right) \)
\( = \tan \left( \tan^{-1}\left(\frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} \times \frac{2}{3}}\right) \right) \)
\( = \tan \left( \tan^{-1}\left(\frac{\frac{9+8}{12}}{\frac{12-6}{12}}\right) \right) \)
\( = \tan \left( \tan^{-1}\left(\frac{17}{6}\right) \right) \)
\( = \frac{17}{6} \)
In simple words: Convert both inverse functions to \( \tan^{-1} \) form. For \( \sin^{-1}(\frac{3}{5}) \), find \( \tan\theta = \frac{3}{4} \). For \( \cot^{-1}(\frac{3}{2}) \), it's simply \( \tan^{-1}(\frac{2}{3}) \). Then apply the \( \tan^{-1}A + \tan^{-1}B \) formula inside the tangent function, which simplifies to \( \tan(\tan^{-1}(\text{value})) = \text{value} \).

🎯 Exam Tip: To evaluate expressions involving a mix of inverse trigonometric functions inside a regular trigonometric function, convert all inverse functions to the same type (usually \( \tan^{-1} \)) and then use the appropriate sum/difference identity.

Question 19. \( \cos^{-1} \left( \cos \left( \frac { 7\pi }{ 6 } \right) \right) \) का मान बराबर है
(A) \( \frac{7\pi}{6} \)
(B) \( \frac{5\pi}{6} \)
(C) \( \frac{\pi}{3} \)
(D) \( \frac{\pi}{6} \)
Answer: हल-
\( \cos^{-1} \) की मुख्य मान शाखा परिसर \( [0, \pi] \) है।
\( \therefore \cos^{-1} \left( \cos \left( \frac { 7\pi }{ 6 } \right) \right) \neq \frac{7\pi}{6} \)
अब \( \cos^{-1} \left( \cos \left( \frac { 7\pi }{ 6 } \right) \right) = \cos^{-1} \left\{ \cos \left( 2\pi - \frac { 7\pi }{ 6 } \right) \right\} \)
\( = \cos^{-1} \left\{ \cos \left( \frac { 12\pi - 7\pi }{ 6 } \right) \right\} = \cos^{-1} \left( \cos \left( \frac { 5\pi }{ 6 } \right) \right) \)
\( = \frac { 5\pi }{ 6 } \)
\( \because \frac { 5\pi }{ 6 } \in [0, \pi] \)
अतः विकल्प (B) सही है।
In simple words: The angle \( \frac{7\pi}{6} \) is outside the principal value range for \( \cos^{-1}x \). Use the identity \( \cos(2\pi - \theta) = \cos\theta \) to find an equivalent angle within the range \( [0, \pi] \). \( \cos(\frac{7\pi}{6}) = \cos(2\pi - \frac{5\pi}{6}) = \cos(\frac{5\pi}{6}) \). Thus, \( \cos^{-1}(\cos(\frac{5\pi}{6})) = \frac{5\pi}{6} \).

🎯 Exam Tip: For \( \cos^{-1}(\cos\theta) \), if \( \theta \) is not in \( [0, \pi] \), use the periodicity and symmetry properties of cosine to find an equivalent angle that is within the principal range, typically \( \cos(2\pi-\theta) \) or \( \cos(-\theta) \).

Question 20. \( \sin \left[ \frac{\pi}{3} - \sin^{-1} \left( -\frac{1}{2} \right) \right] \) का मान है-
(A) \( \frac{1}{2} \)
(B) \( \frac{1}{3} \)
(C) \( \frac{1}{4} \)
(D) 1
Answer: हल-
\( = \sin \left[ \frac{\pi}{3} - \sin^{-1} \left( -\frac{1}{2} \right) \right] \)
\( = \sin \left[ \frac{\pi}{3} - \left( -\frac{\pi}{6} \right) \right] \)
\( = \sin \left[ \frac{\pi}{3} + \frac{\pi}{6} \right] \)
\( = \sin \left[ \frac{2\pi + \pi}{6} \right] \)
\( = \sin \left[ \frac{3\pi}{6} \right] \)
\( = \sin \left[ \frac{\pi}{2} \right] = 1 \)
अतः विकल्प (D) सही है।
In simple words: First, evaluate \( \sin^{-1}(-\frac{1}{2}) = -\frac{\pi}{6} \). Substitute this value back into the expression: \( \sin[\frac{\pi}{3} - (-\frac{\pi}{6})] = \sin[\frac{\pi}{3} + \frac{\pi}{6}] = \sin[\frac{2\pi+\pi}{6}] = \sin[\frac{3\pi}{6}] = \sin[\frac{\pi}{2}] = 1 \).

🎯 Exam Tip: Evaluate nested expressions from the inside out, strictly adhering to the principal value ranges for each inverse trigonometric function. Perform arithmetic operations on angles carefully.

Question 21. \( \tan^{-1}\sqrt{3} - \cot^{-1} (-\sqrt{3}) \) का मान है
(A) \( \pi \)
(B) \( -\frac { \pi }{ 2 } \)
(C) 0
(D) \( 2\sqrt{3} \)
Answer: हल-
\( \tan^{-1}\sqrt{3} = \frac{\pi}{3} \)
फलन \( \cot^{-1} \) की मुख्य मान शाखा का परिसर \( (0, \pi) \) है।
\( \therefore \cot^{-1} (-\sqrt{3}) = \cot^{-1} \left( \cot \left( \pi - \frac{\pi}{6} \right) \right) \)
\( = \cot^{-1} \left( \cot \left( \frac{5\pi}{6} \right) \right) \)
\( = \frac{5\pi}{6} \)
\( \therefore \tan^{-1}\sqrt{3} - \cot^{-1} (-\sqrt{3}) = \frac{\pi}{3} - \frac{5\pi}{6} \)
\( = \frac{2\pi - 5\pi}{6} = -\frac{3\pi}{6} = -\frac{\pi}{2} \)
अतः विकल्प (B) सही है।
In simple words: First, \( \tan^{-1}\sqrt{3} = \frac{\pi}{3} \). For \( \cot^{-1}(-\sqrt{3}) \), since the argument is negative, we use the property \( \cot^{-1}(-x) = \pi - \cot^{-1}(x) = \pi - \cot^{-1}(\sqrt{3}) = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \). Subtracting these values gives \( \frac{\pi}{3} - \frac{5\pi}{6} = -\frac{\pi}{2} \).

🎯 Exam Tip: Be careful with negative arguments for inverse cotangent; use the property \( \cot^{-1}(-x) = \pi - \cot^{-1}(x) \) to ensure the principal value is in the range \( (0, \pi) \).