NCERT Solutions for Class 12 Maths Chapter 2 Miscellaneous Exercise (Inverse Trigonometric Function)

Class 12 Maths Chapter 2 Miscellaneous Exercise Inverse Trigonometric Function:- The NCERT Solutions for Class 12 Maths Chapter 2 on Inverse Trigonometric Functions provide comprehensive answers to all questions in the Miscellaneous Exercise. These solutions are an essential resource for students aiming to solidify their understanding of fundamental concepts and prepare for competitive exams. Check out the NCERT Solutions for Class 12 Maths Chapter 2 Miscellaneous Exercise (Inverse Trigonometric Function) from the below article.

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NCERT Solutions for Class 12 Maths Chapter 2 Miscellaneous Exercise (Inverse Trigonometric Function)

Find the value of the following of NCERT Solutions for Class 12 Maths Chapter 2 Miscellaneous Exercise (Inverse Trigonometric Function)

Question1.

Solution :

Question2.

Solution :

Question3. Prove that:

Solution :

Question4. Prove that:

Solution :

Question5. Prove that:

Solution :

Question6. Prove that:

Solution :

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Question7. Prove that:

Solution :

Question8. Prove that:

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Question9. Prove that:

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Question10. Prove that:

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Question 11. Prove that:

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Question12. Prove that:

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Question13. Solve the equation:

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Question14. Solve the equation:

Solution :

Question15.

Solution :

Therefore, option (D) is correct.

Question16.

Solution :

Therefore, option (C) is correct.

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Question17.

Solution :

Therefore, option (C) is correct.

NCERT Solutions for Class 12 Maths Chapter 2 Miscellaneous Exercise (Inverse Trigonometric Function) Summary 

1. Inverse Trigonometric Functions:

• These are the inverse functions of the basic trigonometric functions (sine, cosine, tangent, etc.).

• The principal value branch of these functions is typically used, meaning each inverse function has a restricted range to ensure they are one-to-one and hence invertible.

2. Principal Value Branch:

• sin⁡−1x\sin^{-1} xsin−1x (arcsine) has a range of [−π2,π2][- \frac{\pi}{2}, \frac{\pi}{2}] [−2π​,2π​].

• cos⁡−1x\cos^{-1} xcos−1x (arccosine) has a range of [0,π][0, \pi] [0,π].

• tan⁡−1x\tan^{-1} xtan−1x (arctangent) has a range of (−π2,π2)(- \frac{\pi}{2}, \frac{\pi}{2}) (−2π​,2π​).

• cot⁡−1x\cot^{-1} xcot−1x (arccotangent) has a range of (0,π)(0, \pi) (0,π).

• sec⁡−1x\sec^{-1} xsec−1x (arcsecant) and csc⁡−1x\csc^{-1} xcsc−1x (arccosecant) have ranges that exclude zero and are defined appropriately.

3. Properties:

• These functions are continuous and differentiable in their respective domains.

• They are used to solve equations involving trigonometric expressions.

Key Problems and Their Solutions:

1. Simplification of Expressions:

• Problems may involve simplifying expressions using properties of inverse trigonometric functions. For example, simplifying sin⁡(sin⁡−1x)=x\sin(\sin^{-1} x) = xsin(sin−1x)=x.

2. Equation Solving:

• Solving equations such as tan⁡−1(3)=π3\tan^{-1} (\sqrt{3}) = \frac{\pi}{3}tan−1(3​)=3π​ by understanding the corresponding angles and their ranges.

3. Derivatives and Integrals:

• Problems may include finding the derivatives and integrals of inverse trigonometric functions. For example, ddx(tan⁡−1x)=11+x2\frac{d}{dx} (\tan^{-1} x) = \frac{1}{1+x^2}dxd​(tan−1x)=1+x21​.

4. Composite Functions:

• Handling composite functions involving inverse trigonometric functions, such as finding the value of sin⁡−1(cos⁡(π/6))\sin^{-1}(\cos(\pi/6))sin−1(cos(π/6)).

5. Principal Value Determination:

• Determining the principal value of inverse trigonometric functions for given inputs, ensuring the results lie within the defined range.

Example Problems:

1. Example 1:

• Solve for xxx in sin⁡−1x+cos⁡−1x=π2\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}sin−1x+cos−1x=2π​.

• Solution: Using the identity sin⁡−1x+cos⁡−1x=π2\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}sin−1x+cos−1x=2π​, the solution directly follows that the given equation is true for all xxx in [−1,1][-1, 1][−1,1].

2. Example 2:

• Find tan⁡(tan⁡−11+tan⁡−12)\tan(\tan^{-1} 1 + \tan^{-1} 2)tan(tan−11+tan−12).

• Solution: Using the formula for the sum of arctangents, tan⁡−1a+tan⁡−1b=tan⁡−1(a+b1−ab)\tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right)tan−1a+tan−1b=tan−1(1−aba+b​) when ab<1ab < 1ab<1. Here, tan⁡−11+tan⁡−12=tan⁡−1(1+21−1∗2)=tan⁡−1(−3)\tan^{-1} 1 + \tan^{-1} 2 = \tan^{-1} \left( \frac{1+2}{1-1*2} \right) = \tan^{-1} (-3)tan−11+tan−12=tan−1(1−1∗21+2​)=tan−1(−3).

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NCERT Solutions for Class 12 Maths Chapter 2 Miscellaneous Exercise (Inverse Trigonometric Function) FAQs

1. What is the principal value of an inverse trigonometric function?

Ans. The principal value is the value within the restricted range of an inverse trigonometric function that makes it one-to-one and invertible. For example, the principal value of sin⁡−1(x)\sin^{-1}(x)sin−1(x) lies in [−π2,π2][- \frac{\pi}{2}, \frac{\pi}{2}] [−2π​,2π​].

2. How do you solve an equation involving inverse trigonometric functions?

Ans. To solve an equation involving inverse trigonometric functions, use the properties and ranges of these functions. For example, for sin⁡−1(x)=y\sin^{-1}(x) = ysin−1(x)=y, ensure that yyy lies within [−π2,π2][- \frac{\pi}{2}, \frac{\pi}{2}] [−2π​,2π​] and sin⁡(y)=x\sin(y) = xsin(y)=x.

3. What is the range of cos⁡−1(x)\cos^{-1}(x)cos−1(x)?

Ans. The range of cos⁡−1(x)\cos^{-1}(x)cos−1(x) is [0,π][0, \pi][0,π]. This means cos⁡−1(x)\cos^{-1}(x)cos−1(x) will give an angle between 0 and π\piπ.

4. How do you differentiate an inverse trigonometric function?

Ans. Use the standard differentiation formulas. For example, ddx(sin⁡−1(x))=11−x2\frac{d}{dx} (\sin^{-1}(x)) = \frac{1}{\sqrt{1 - x^2}}dxd​(sin−1(x))=1−x2​1​ and ddx(tan⁡−1(x))=11+x2\frac{d}{dx} (\tan^{-1}(x)) = \frac{1}{1 + x^2}dxd​(tan−1(x))=1+x21​.

5. What is the integral of an inverse trigonometric function?

Ans. The integrals of inverse trigonometric functions are derived from their derivatives. For instance, ∫11−x2dx=sin⁡−1(x)+C\int \frac{1}{\sqrt{1 - x^2}} dx = \sin^{-1}(x) + C∫1−x2​1​dx=sin−1(x)+C.