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NCERT Class 12 Mathematics Chapter 2 Inverse Trigonometric Functions
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Inverse Trigonometric Functions
Chapter – 2
| Exercise 2.1 |
Find the principal values of the following:
Ans:
Ans:
3. cosec⁻¹ (2).
Ans: Let, cosec⁻¹ (2) = y
Hence,
cosecy = 2
4. tan⁻¹ (-√3)
Ans: Let, tan⁻¹ (- √3) = y
Hence,
tan y = – √3
Ans:
6. tan⁻¹ (-1).
Ans: Let, tan⁻¹ (- 1) = y
Hence,
tan y = – 1
Ans:
8. cot⁻¹ (√3).
Ans: Let, cot⁻¹ (√3) = y
Hence,
cot y = √3
Range of the principal value of cot⁻¹ (x) = (0, π)
Thus, principal value of
Ans: Let,
Hence,
10. cosec⁻¹ (- √2)
Ans: Let, cosec⁻¹ (-√2) = y
Hence,
cosec y = -√2
Find the values of the following:
Ans: Let, tan⁻¹ (1) = x
Hence,
tan x = 1
Ans:
13. If sin⁻¹ x = y, then
(A) 0 ≤ y ≤ π
(C) 0 < y < π
Ans:
14. tan⁻¹ √3 – sec⁻² (- 2) is equal to
Ans: Let tan⁻¹ (√3) = x
Hence, tan x = √3
= tan(π/3)
| EXERCISE 2.2 |
Prove the following:
Ans: Let x = sinθ
⇒ sin⁻¹ x = θ
Thus we have,
R. H. S.= sin⁻¹ (3x – 4x³)
= sin⁻¹ (3 sin θ – 4sin³ θ)
= sin⁻¹ (sin 3 θ)[∵ sin 3A = 3 sin A – 4sin³ A]
= 3 θ[∵ sin⁻¹ (sin A) = A]
= 3sin⁻¹x
= L.H.S.
Therefore, 3sin⁻¹ x = sin⁻¹ (3x – 4x³)
Ans: We have to prove that 3 cos⁻¹ x = cos⁻¹ (4x³ – 3x)
First consider RHS, cos⁻¹ (4x³ – 3x)
Let us take x = cosθ, we know that cos(3θ) = 4cos³ θ – 3 cos θ
⇒ cos⁻¹ (4x³ – 3x) = cos⁻¹ (4cos³ θ – 3cos θ) = cos⁻¹ (cos(3θ) = 3 θ
∵ θ = cos⁻¹ x, ⇒ cos⁻¹ (4x³ – 3x) = 3 cos⁻¹ x
∴ LHS = RHS
0 ≤ 3 cos⁻¹ x ≤ π
0 ≤ cos¹ x ≤ π / 3
∴ χ∈[½,1]
Hence proved
Ans: Let x = tan θ ⇒ θ = tan⁻¹ x
Hence,
Ans:
Ans:
Ans:
Ans:
Find the values of each of the following:
Ans:
Ans:
Find the values of each of the expressions in Exercises 16 to 18.
Ans:
Ans:
Ans:
Ans:
Ans:
Ans: Let tan⁻¹ √3 = x
Hence,
| Miscellaneous Exercise on Chapter 2 |
Find the value of the following:
Ans:
Ans:
Prove that
Ans:
Ans:
Ans:
Ans:
Ans:
Prove that
Ans: Let x = tan² θ
Then,
√x = tanθ
θ = tan⁻¹√x
Therefore,
Ans:
Ans: Let x = cos 2θ ⇒ θ = 1/2 cos⁻¹ x
Thus,
Solve the following equations:
11. 2tan⁻¹ (cos x) = tan⁻¹ (2 cosec x).
Ans: It is given that 2 tan⁻¹ (cos x) = tan⁻¹ (2 cosec x).
since,
Hence,
Ans:
Ans:
Ans: It is given that sin⁻¹ (1 – x) – 2sin⁻¹ x = π/2
⇒ sin⁻¹ (1 – x) – 2sin⁻¹ x = π/2
⇒ – 2sin⁻¹ x = π/2 – sin⁻¹ (1 – x)
⇒ – 2 sin⁻¹ x = cos⁻¹ (1 – x) …(1)
Let sin⁻¹ x = y ⇒ sin y = x
Hence,
cos y = √1 – x²
y = cos⁻¹ (√1 – x²)
sin⁻¹ x = cos⁻¹ √1 – x²
From equation (1), we have
– 2 cos⁻¹ √1 – x² = cos⁻¹ (1 – x)
Put x = sin y
⇒ – 2 cos⁻¹ √1 – sin²y = cos⁻¹ (1 – sin y)
⇒ – 2 cos⁻¹ (cos y) = cos⁻¹ (1 – sin y)
⇒ – 2y = cos⁻¹ (1 – sin y)
⇒ 1 – sin y = cos(- 2y)
⇒ 1 – sin y = cos 2y
⇒ 1 – sin y = 1 – 2sin² y
⇒ 2sin² y – sin y = 0
⇒ sin y (2sin y – 1) = 0
⇒ sin y = 0 , 1/ 2
Therefore,
x = 0, ½
When x = ½, it does not satisfy the equation.
Hence, x = 0 is the only solution
Thus, the correct option is C.
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