Board | SCERT, Kerala |
Text Book | NCERT Based |
Class | Plus Two |
Subject | Math's Textbook Solution |
Chapter | Chapter 2 |
Exercise | Miscellaneous Exercise |
Chapter Name | Inverse Trigonometric Functions |
Category | Plus Two Kerala |
Kerala Syllabus Plus Two Math's Textbook Solution Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise
Chapter 2 Inverse Trigonometric Functions Solution
Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise
Prove
Consider,
Prove
L.H.S =
Prove
test
Prove
Now, we have:
Prove
Using (1) and (2), we have
Solveis equal to
(A) (B) (C) (D)
sin
Hence option D
Solveis equal to
(A) (B). (C) (D)
Hence, the correct answer is C.
Find the value of
We know that cos−1 (cos x) = x if, which is the principal value branch of cos −1x.
Here,
Now, can be written as:
Find the value of
We know that tan−1 (tan x) = x if, which is the principal value branch of tan −1x.
Here,
Now, can be written as:
Prove
Now, we have:
Prove
Now, we have:
Prove
Now, we will prove that:
Prove
Let Then,
Prove [Hint: putx = cos 2θ]
Put so that , then we have
Solve
Solve
Solve, then x is equal to
(A) (B) (C) 0 (D)
Therefore, from equation (1), we have
Put x = sin y. Then, we have:
But, when, it can be observed that:
is not the solution of the given equation.
Thus, x = 0.
Hence, the correct answer is C.
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Chapter 2: Inverse Trigonometric Functions Miscellaneous Exercise Solution
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Plus Two Math's Chapter Wise Textbook Solution PDF Download
- Chapter 1: Relations and Functions
- Chapter 2: Inverse Trigonometric Functions
- Chapter 3: Matrices
- Chapter 4: Determinants
- Chapter 5: Continuity and Differentiability
- Chapter 6: Application of Derivatives
- Chapter 7: Application of Integrals
- Chapter 8: Integrals
- Chapter 9: Differential Equations
- Chapter 10: Vector Algebra
- Chapter 11: Three Dimensional Geometry
- Chapter 12: Linear Programming
- Chapter 13: Probability
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