MCQs on Inverse Trigonometric Functions Class 12 OP Malhotra ISC Maths Solutions Ch-4. In this article you would learn full concept recap of Inverse Trigonometric Functions. Visit official Website CISCE for detail information about ISC Board Class-12 Mathematics.

Inverse Trigonometric Functions Class 12 OP Malhotra MCQ and short answer ISC Maths Solutions Ch-4

Board	ISC
Publications	S Chand
Subject	Maths
Class	12th
Chapter-4	Inverse Trigonometric Functions
Writer	OP Malhotra
Exe	MCQ and Short Answer

MCQs on Inverse Trigonometric Functions

 OP Malhotra ISC Class 12 Maths Solutions

Fill In The Blanks

1. sin^-1(sin π/6) = ______

Ans – sin^-1(sin π/6) = π/6

2. sin^-1(sin 5π/6) = ______

Ans – sin^-1(sin 5π/6) = 5π/6

3. The principal value of sec^-1(-2) is ______

Ans – 2π/3
sec^-1(-2) = cos^-1(1/-2) = cos^-1(-1/2)

We know,
cos (2π/3) = -1/2

∴ Principal value = 2π/3

4. If sin^-1α = tan^-1(3/4), then α = ______

Ans – α = 3/5
sin^-1α = tan^-1(3/4)

Taking sin on both sides,
α = sin(tan^-1(3/4))

Let tan θ = 3/4
Opposite = 3, Adjacent = 4
Hypotenuse = 5

∴ sin θ = 3/5

∴ α = 3/5

5. cos^-1(-1) − sin^-1(1) = ______

Ans – π/2
cos^-1(−1) = π
sin^-1(1) = π/2

So,
cos^-1(−1) − sin^-1(1)
= π − π/2
= π/2

6. tan^-1(1/7) + tan^-1(1/13) = ______

Ans – tan^-1(2/9)
tan^-1(1/7) + tan^-1(1/13)

Using formula,
tan^-1a + tan^-1b =
tan^-1((a + b) / (1 − ab))

Here a = 1/7, b = 1/13
= tan^-1((1/7 + 1/13) / (1 − (1/7 × 1/13)))
= tan^-1((20/91) / (90/91))
= tan^-1(20/90)
= tan^-1(2/9)

7. The value of cos(sin^-1 x) is ______

Ans – √(1 − x²)
Let θ = sin^-1x

So, sin θ = x

In right triangle,
Opposite = x
Hypotenuse = 1
Adjacent = √(1 − x²)

∴ cos(sin^-1x) = cos θ
= √(1 − x²)

8. cos[ cos^-1(1/9) + sin^-1(-1/9) ] = ______

Ans – 8√5/81
cos[ cos^-1(1/9) + sin^-1(-1/9) ]

Let A = cos^-1(1/9)
Let B = sin^-1(-1/9)

Using formula,
cos(A + B) = cosA cosB − sinA sinB

cosA = 1/9
sinA = √(1 − (1/9)²) = 4√5/9

sinB = −1/9
cosB = √(1 − (1/9)²) = 4√5/9
= (1/9 × 4√5/9) − (4√5/9 × −1/9)
= 4√5/81 + 4√5/81
= 8√5/81

9. 3 tan^-1 a is equal to ______

Ans – (3a − a³) / (1 − 3a²)
Let θ = tan^-1a

So, tan θ = a

We know,
tan 3θ = (3 tanθ − tan³θ) / (1 − 3 tan²θ)

∴ tan(3 tan^-1a)
= (3a − a³) / (1 − 3a²)

10. tan^-1x − tan^-1y = tan^-1((x − y)/(1 + xy)) is true when xy is ______

Ans –  xy < 1
tan^-1x − tan^-1y
= tan^-1((x − y) / (1 + xy))

This identity is true when

xy < 1

Multiple Choice Questions

1. cot^-1(1/√3) + 2 sin^-1(1/2) is equal to
(a) π/4   
(b) π/6   
(c) π/3   
(d) 2π/3

Sol: (d)2π/3
cot^-1(1/√3) + 2 sin^-1(1/2)

We know,
cot^-1(1/√3) = π/3
sin^-1(1/2) = π/6

So,
= π/3 + 2 × (π/6)
= π/3 + π/3
= 2π/3

2. cos^-1(-1/2) − 2 sin^-1(1/2) +3 cos^-1(1/√2) − 4 tan^-1(-1) equals
(a) 19π/12   
(b) 13π/12   
(c) 47π/12   
(d) 43π/12

Sol:
cos^-1(-1/2) − 2 sin^-1(1/2)+ 3 cos^-1(1/√2) − 4 tan^-1(-1)

We know,
cos^-1(-1/2) = 2π/3
sin^-1(1/2) = π/6
cos^-1(1/√2) = π/4
tan^-1(-1) = −π/4
= 2π/3 − 2(π/6) + 3(π/4) − 4(−π/4)
= 2π/3 − π/3 + 3π/4 + π
= π/3 + 3π/4 + π

Taking LCM 12,
= 4π/12 + 9π/12 + 12π/12

= 25π/12

3. The value of cos^-1(sin 7π/6) is
(a) -π/3   
(b) π/6   
(c) π/3   
(d) 2π/3

Sol: (d) 2π/3
cos^-1(sin 7π/6)

sin 7π/6 = sin(π + π/6)
= − sin(π/6)
= −1/2

So,
cos^-1(−1/2)

We know,
cos (2π/3) = −1/2

∴ Answer = 2π/3

4. The value of sin^-1(cos 53π/5) is
(a) 3π/5   
(b) -3π/5   
(c) π/10   
(d) -π/10

Sol: (d) −π/10
sin^-1(cos 53π/5)

53π/5 = 50π/5 + 3π/5
= 10π + 3π/5

Since cos has period 2π,
cos(53π/5) = cos(3π/5)

cos θ = sin(π/2 − θ)

cos(3π/5) = sin(π/2 − 3π/5)
= sin(5π/10 − 6π/10)
= sin(−π/10)

∴ sin^-1(sin(−π/10))

Principal value of sin^-1x lies between −π/2 and π/2

∴ Answer = −π/10

5. The value of sin⁻¹[cos(4095°)] is equal to
(a) -π/3   
(b) π/6   
(c) -π/4   
(d) π/4   
(e) π/2

Sol: (c) −π/4
sin^-1[ cos(4095°) ]
4095° = 360° × 11 + 135°

So,
cos(4095°) = cos(135°)
cos(135°) = −√2/2

Now,
sin^-1(−√2/2)

We know,
sin(−π/4) = −√2/2

∴ Answer = −π/4

6. The value of sin[2 sin⁻¹(cos A)] is
(a) sin A   
(b) cos A   
(c) sin 2A   
(d) cos 2A

Sol: (c) sin 2A
sin[ 2 sin^-1(cos A) ]

Let θ = sin^-1(cos A)
So, sin θ = cos A

We know,
sin 2θ = 2 sinθ cosθ
= 2 (cos A) (cos θ)

cos θ = √(1 − cos²A)
= √(sin²A)
= sin A

∴ sin 2θ = 2 cos A sin A
= sin 2A

Answer = sin 2A

7. cos[2 cos⁻¹(1/5) + sin⁻¹(1/5)] is equal to
(a) -√3/2   
(b) 4/5   
(c) -2√6/5   
(d) 2/(3√3)

Sol:
cos[ 2 cos^-1(1/5) + sin^-1(1/5) ]

Let A = cos^-1(1/5)
Let B = sin^-1(1/5)

cos A = 1/5
sin A = √(1 − 1/25) = 2√6/5

sin B = 1/5
cos B = √(1 − 1/25) = 2√6/5

Using formula,
cos(2A + B) = cos2A cosB − sin2A sinB

cos2A = 2 cos²A − 1
= 2(1/25) − 1
= −23/25

sin2A = 2 sinA cosA
= 2(2√6/5)(1/5)
= 4√6/25

Substitute,
= (−23/25)(2√6/5) − (4√6/25)(1/5)
= −46√6/125 − 4√6/125
= −2√6/5

8. If sin⁻¹(x/5) + cosec⁻¹(5/4) = π/2, then x is equal to
(a) 1   
(b) 2   
(c) 3   
(d) 4

Sol: (c) 3
sin^-1(x/5) + cosec^-1(5/4) = π/2

cosec^-1(5/4) means,
cosec θ = 5/4
So, sin θ = 4/5

∴ cosec^-1(5/4) = sin^-1(4/5)

Now equation becomes,
sin^-1(x/5) + sin^-1(4/5) = π/2

We know,
If sin^-1a + sin^-1b = π/2, then
a = √(1 − b²)

Here b = 4/5

√(1 − (4/5)²)
= √(1 − 16/25)
= √(9/25)
= 3/5

So, x/5 = 3/5
∴ x = 3

9. tan⁻¹1 + tan⁻¹2 + tan⁻¹3 is
(a) 0   
(b) tan⁻¹6   
(c) π/4   
(d) π

Sol: (a) 0
tan^-11 + tan^-12 + tan^-13
tan^-11 = π/4

Now,
tan^-12 + tan^-13

Using formula,
tan^-1a + tan^-1b
= tan^-1((a + b) / (1 − ab))
= tan^-1((2 + 3) / (1 − 6))
= tan^-1(5 / −5)
= tan^-1(−1)
= −π/4

Total = π/4 − π/4
= 0

10. The value of cos(2 cos⁻¹ 0.80) is
(a) 0.28   
(b) 0.48   
(c) 0.84   
(d) 0.96

Sol: (a) 0.28
cos(2 cos^-1 0.80)

Let θ = cos^-1(0.80)
So, cos θ = 0.80

We know,
cos 2θ = 2 cos²θ − 1
= 2(0.80)² − 1
= 2(0.64) − 1
= 1.28 − 1
= 0.28

11. The value of cos⁻¹(cos 3π/2) is
(a) π/2   
(b) 3π/2   
(c) 5π/2   
(d) 7π/2

Sol: (a) π/2
cos^-1(cos 3π/2)
cos 3π/2 = 0

Principal value of cos^-1x lies between 0 and π
∴ cos^-1(0) = π/2

12. If tan^-1(1/3) + tan^-1(3/4) − tan^-1(x/3) = 0, then x is equal to
(a) 7/3    
(b) 3    
(c) 11/3    
(d) 13/3

Sol: (d) 13/3
tan^-1(1/3) + tan^-1(3/4) − tan^-1(x/3) = 0

So,
tan^-1(1/3) + tan^-1(3/4)
= tan^-1(x/3)

Using formula,
tan^-1a + tan^-1b
= tan^-1((a + b) / (1 − ab))
= tan^-1((1/3 + 3/4) / (1 − (1/3 × 3/4)))
= tan^-1((13/12) / (3/4))
= tan^-1(13/9)

So,
x/3 = 13/9
∴ x = 13/3

13. The solution of tan^-1x + 2 cot^-1x = 2π/3 is
(a) −1/√3    
(b) 1/√3    
(c) −√3    
(d) √3

Sol: (d) √3
tan^-1x + 2 cot^-1x = 2π/3

We know,
cot^-1x = π/2 − tan^-1x

So,
tan^-1x + 2(π/2 − tan^-1x) = 2π/3
tan^-1x + π − 2 tan^-1x = 2π/3
π − tan^-1x = 2π/3
tan^-1x = π − 2π/3
= π/3

x = tan(π/3)
= √3

14. If sin^-1x + sin^-1y = 2π/3, then the value of cos^-1x + cos^-1y is
(a) 2π/3    
(b) π/3    
(c) π/2    
(d) π

Sol: (b) π/3 
sin^-1x + sin^-1y = 2π/3

We know,
cos^-1x = π/2 − sin^-1x
cos^-1y = π/2 − sin^-1y

Adding both,
cos^-1x + cos^-1y
= (π/2 − sin^-1x) + (π/2 − sin^-1y)
= π − (sin^-1x + sin^-1y)
= π − 2π/3
= π/3

Very Short Answer Type Questions

1. Find the principal value of sin^-1[ cos( sin^-1(1/2) ) ].

Sol: Let θ = sin⁻¹(1/2) ⇒ θ = π/6
cos(sin⁻¹(1/2)) = cos(π/6) = √3/2
sin⁻¹(cos(sin⁻¹(1/2))) = sin⁻¹(√3/2)
= π/3

2. Find the value of tan[ ( sin^-1x + cos^-1x ) / 2 ].

Sol: We know: sin⁻¹x + cos⁻¹x = π/2
So, tan[(sin⁻¹x + cos⁻¹x)/2] = tan(π/4) = 1

3. Find the principal value of cos^-1(−1/2).

Sol:  cos⁻¹(-1/2) = 2π/3

4. If sin^-1(x/5) + cosec^-1(5/4) = π/2, then find x.

Sol:  cosec⁻¹(5/4) = sin⁻¹(4/5)
sin⁻¹(x/5) + sin⁻¹(4/5) = π/2 ⇒ sin⁻¹(x/5) = π/2 – sin⁻¹(4/5)
sin⁻¹(x/5) = cos⁻¹(4/5) ⇒ x/5 = 4/5
⇒ x = 4

5. Find x if 4 sin^-1x + cos^-1x = π.

Sol: cos⁻¹x = π/2 – sin⁻¹x
4 sin⁻¹x + (π/2 – sin⁻¹x) = π ⇒ 3 sin⁻¹x + π/2 = π
3 sin⁻¹x = π/2 ⇒ sin⁻¹x = π/6
⇒ x = 1/2

6. Find x if sin^-1x − cos^-1x = π/6.

Sol: cos⁻¹x = π/2 − sin⁻¹x
sin⁻¹x − (π/2 − sin⁻¹x) = π/6 ⇒ 2 sin⁻¹x − π/2 = π/6
2 sin⁻¹x = 2π/3 ⇒ sin⁻¹x = π/3
⇒ x = √3/2

–: End of MCQs on Inverse Trigonometric Functions Class 12 OP Malhotra ISC Maths Solutions :–

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