Self Assesment and Revision 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.

 

Self Assesment and Revision on Inverse Trigonometric Functions Class 12 OP Malhotra ISC Maths Solutions Ch-4

Board	ISC
Publications	S Chand
Subject	Maths
Class	12th
Chapter-4	Inverse Trigonometric Functions
Writer	OP Malhotra
Exe	Self assesment and revision

Self Assesment and Revision on Inverse Trigonometric Functions

Inverse Trigonometric Functions Class 12 OP Malhotra Self assesment and revision Solutions

1. Prove that sin^-1(x / √(1 + x²))+ cos^-1((x + 1) / √(x² + 2x + 2)) = tan^-1 (x² + x + 1)

Sol: To prove : sin^-1(x / √(1 + x²)) + cos^-1((x + 1) / √(x² + 2x + 2)) = tan^-1(x² + x + 1)

Proof : sin^-1(x / √(1 + x²)) = tan^-1x

Since if sinθ = x / √(1 + x²), then tanθ = x.
x² + 2x + 2 = (x + 1)² + 1
cos^-1((x + 1) / √(x² + 2x + 2)) = tan^-1(1 / (x + 1))
LHS = tan^-1x + tan^-1(1 / (x + 1))

Using formula:
tan^-1a + tan^-1b = tan^-1((a + b) / (1 − ab))
= tan^-1((x + 1/(x + 1)) / (1 − x/(x + 1)))
= tan^-1(x² + x + 1)

Hence Proved.

2. Solve the equation tan^-1(x + 2) + tan^-1(2 − x) = tan^-1(2/3)

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

Let a = x + 2 and b = 2 − x

a + b = (x + 2) + (2 − x) = 4
ab = (x + 2)(2 − x) = 4 − x²
1 − ab = 1 − (4 − x²) = x² − 3
LHS = tan^-1(4 / (x² − 3))

Comparing both sides
tan^-1(4 / (x² − 3)) = tan^-1(2/3)
4 / (x² − 3) = 2/3

12 = 2(x² − 3)
12 = 2x² − 6
2x² = 18
x² = 9
x = ±3

∴ x = 3 or x = −3

3. Show that sin^-1(√3 / 2) + 2 tan^-1(1 / √3) = 2π / 3

Sol: We know that,
sin(π/3) = √3 / 2

Therefore,
sin^-1(√3 / 2) = π/3

tan(π/6) = 1 / √3

Therefore,
tan^-1(1 / √3) = π/6

2 tan^-1(1 / √3) = 2 × (π/6) = π/3

LHS = π/3 + π/3 = 2π/3

Hence Proved.

4. Show that sin^-1(4 / 5) + cos^-1(2 / √5) = cot^-1(2 / 11)

Sol: Let α = sin^-1(4/5) ⇒ tan α = 4/3

Let β = cos^-1(2/√5) ⇒ tan β = 1/2
α + β = tan^-1(4/3) + tan^-1(1/2)

Using formula:
tan^-1a + tan^-1b = tan^-1((a + b) / (1 − ab))
= tan^-1((4/3 + 1/2) / (1 − (4/3 × 1/2)))
= tan^-1(11/2)
= cot^-1(2/11)

Hence Proved.

5. Prove that 2 [ tan^-11 + tan^-1(1/2) + tan^-1(1/3) ] = π

Sol: tan^-1(1/2) + tan^-1(1/3)

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

LHS = 2 [ π/4 + π/4 ] = 2 × (π/2)
= π

Hence Proved.

6. Prove that cot ( π/4 − 2 cot^-13 ) = 7

Sol: Let θ = cot^-13 ⇒ tanθ = 1/3

tan2θ = (2tanθ)/(1 − tan²θ)
= (2/3)/(1 − 1/9)
= 3/4

tan(π/4 − 2θ) = (1 − 3/4)/(1 + 3/4) = 1/7

∴ cot(π/4 − 2θ) = 7

7. Show that sin^-1(1/√17) + cos^-1(9/√85) = tan^-1(1/2)

Sol: sin^-1(1/√17) + cos^-1(9/√85) = tan^-1(1/2)

Let α = sin^-1(1/√17) ⇒ tanα = 1/4
Let β = cos^-1(9/√85) ⇒ tanβ = 1/2

α + β = tan^-1((1/4 + 1/2)/(1 − (1/4 × 1/2)))
= tan^-1(1/2)

Hence Proved

8. Prove that 2 tan^-1(1/3) + cot^-1(4) = tan^-1(16/13)

Sol: 2 tan^-1(1/3) + cot^-14 = tan^-1(16/13)

Let tanθ = 1/3
tan2θ = (2/3)/(1 − 1/9) = 3/4

cot^-14 = tan^-1(1/4)
tan^-1(3/4) + tan^-1(1/4)
= tan^-1(16/13)

Hence Proved

9. Solve for x: tan^-1(x − 1) + tan^-1x + tan^-1(x + 1) = tan^-1(3x)

Sol: tan^-1(x − 1) + tan^-1x + tan^-1(x + 1) = tan^-1(3x)

tan^-1(x − 1) + tan^-1(x + 1)

Using formula:
tan^-1a + tan^-1b = tan^-1((a + b)/(1 − ab))
= tan^-1((2x)/(1 − (x² − 1)))
= tan^-1((2x)/(2 − x²))

Now add tan^-1x: = tan^-1((2x)/(2 − x²)) + tan^-1x

Again using formula:
= tan^-1(( (2x)/(2 − x²) + x ) /(1 − x·(2x/(2 − x²)) ))
= tan^-1(3x)

x(x² − 1) = 0

So,
x = 0, ±1
∴ x = 0

10. If sin^-1x + sin^-1y + sin^-1z = π, prove that x² − y² − z² + 2yz √(1 − x²) = 0.

Sol: sin^-1x = π − (sin^-1y + sin^-1z)

Taking sine on both sides:
x = sin(sin^-1y + sin^-1z)

Using formula: sin(A + B) = sinA cosB + cosA sinB
x = y√(1 − z²) + z√(1 − y²)

Squaring both sides:
x² = y²(1 − z²) + z²(1 − y²)
+ 2yz√(1 − y²)√(1 − z²)

Simplifying,
x² = y² + z² − 2y²z²
+ 2yz√[(1 − y²)(1 − z²)]

Using given condition again,
√[(1 − y²)(1 − z²)] = √(1 − x²)

Substituting and rearranging:
x² − y² − z²+ 2yz √(1 − x²) = 0

Hence Proved

11. Solve: cos^-1 [ sin (cos^-1x) ] = π/3

Sol: Let cos^-1x = θ
⇒ x = cosθ

Then,
sin(cos^-1x) = sinθ = √(1 − cos²θ) = √(1 − x²)

So equation becomes:
cos^-1(√(1 − x²)) = π/3

Taking cosine on both sides:
√(1 − x²) = 1/2

Squaring:
1 − x² = 1/4
x² = 3/4
x = √3/2

∴x = √3/2

12. Prove that sin [ 2 tan^-1(3/5) − sin^-1(7/25) ] = 304/425

Sol: Let tanθ = 3/5
sin2θ = 2tanθ / (1 + tan²θ)
= (2 × 3/5) / (1 + 9/25)
= (6/5) / (34/25)
= 15/17

cos2θ = (1 − tan²θ)/(1 + tan²θ)
= (1 − 9/25)/(34/25)
= 8/17

Let sinα = 7/25
cosα = √(1 − 49/625)
= 24/25

sin(2θ − α) = sin2θ cosα − cos2θ sinα
= (15/17)(24/25) − (8/17)(7/25)
= 360/425 − 56/425
= 304/425

Hence Proved.

13. Solve for x: sin (2 tan^-1x) = 1

Sol: We know that,
sin(2 tan^-1x) = 2x / (1 + x²)

So,
2x / (1 + x²) = 1
2x = 1 + x²
x² − 2x + 1 = 0
(x − 1)² = 0
x = 1

∴x = 1

14. Prove that 2 tan^-1(1/5) + cos^-1(7/(5√2)) + 2 tan^-1(1/8) = π/4

Sol: Let tanθ = 1/5
tan2θ = (2 × 1/5) / (1 − 1/25)= (2/5) / (24/25)
= 5/12
∴ 2 tan^-1(1/5) = tan^-1(5/12)

Let tanφ = 1/8
tan2φ = (2 × 1/8) / (1 − 1/64)
= (1/4) / (63/64)
= 16/63

∴ 2 tan^-1(1/8) = tan^-1(16/63)

cos^-1(7/(5√2))
Let angle = α
cosα = 7/(5√2)
sinα = √(1 − 49/50) = 1/√50 = 1/(5√2)
tanα = 1/7

∴ α = tan^-1(1/7)

LHS = tan^-1(5/12) + tan^-1(1/7) + tan^-1(16/63)
Using tan^-1a + tan^-1b formula step by step,
= tan^-1(1)
= π/4

Hence Proved.

15. Prove that sec²(tan^-12) + cosec²(cot^-13) = 15

Sol: Let θ = tan^-12
tanθ = 2
sec²θ = 1 + tan²θ
= 1 + 4
= 5

Let φ = cot^-13
cotφ = 3
cosec²φ = 1 + cot²φ
= 1 + 9
= 10

LHS = 5 + 10 = 15

Hence Proved.

16. Prove that cos^-1(63/65) + 2 tan^-1(1/5) = sin^-1(3/5)

Sol: Let tanθ = 1/5
tan2θ = (2 × 1/5) / (1 − 1/25)
= (2/5) / (24/25)
= 5/12
∴ 2 tan^-1(1/5) = tan^-1(5/12)

Let cosα = 63/65
sinα = √(1 − 3969/4225)
= 16/65
tanα = 16/63
∴ α = tan^-1(16/63)

LHS = tan^-1(16/63) + tan^-1(5/12)
= tan^-1((16/63 + 5/12) / (1 − (16/63 × 5/12)))
= tan^-1(3/4)= sin^-1(3/5)

Hence Proved.

17. Prove that tan^-1(1/4) + tan^-1(2/9) = (1/2) sin^-1(4/5)

Sol: Let A = tan^-1(1/4), B = tan^-1(2/9)

Using formula:
tan^-1a + tan^-1b
= tan^-1((a + b)/(1 − ab))
= tan^-1((1/4 + 2/9)/(1 − (1/4 × 2/9)))
= tan^-1((17/36)/(17/18))
= tan^-1(1/2)

Let θ = sin^-1(4/5)
sinθ = 4/5 ⇒ cosθ = 3/5
tan(θ/2) = sinθ / (1 + cosθ)
= (4/5)/(1 + 3/5)
= (4/5)/(8/5)
= 1/2
∴ θ/2 = tan^-1(1/2)
= (1/2) sin^-1(4/5)

Hence Proved.

18. Solve for x: sin^-1x + sin^-1(1 − x) = cos^-1x, x ≠ 0

Sol: We know that,
cos^-1x = π/2 − sin^-1x

So equation becomes:
sin^-1x + sin^-1(1 − x)
= π/2 − sin^-1x

2 sin^-1x + sin^-1(1 − x)
= π/2

Checking x = 1/2,
sin^-1(1/2) = π/6

LHS = π/6 + π/6 = π/3
RHS = cos^-1(1/2) = π/3

Hence satisfied.

∴ x = 1/2

19. Evaluate: tan [ 2 tan^-1(1/2) − cot^-13 ]

Sol: Let tanθ = 1/2tan2θ
= (2tanθ)/(1 − tan²θ)
= (2 × 1/2)/(1 − 1/4)
= 1/(3/4)
= 4/3
∴ 2 tan^-1(1/2) = tan^-1(4/3)

cot^-13 = tan^-1(1/3)

tan [ tan^-1(4/3) − tan^-1(1/3) ]
Using formula:
tan(A − B) = (tanA − tanB)/(1 + tanA tanB)
= (4/3 − 1/3)/(1 + (4/3 × 1/3))
= (1)/(1 + 4/9)
= 1/(13/9)
= 9/13

20. If cos^-1x + cos^-1y + cos^-1z = π, prove that x² + y² + z² + 2xyz = 1.

Sol: Let α = cos^-1x,
β = cos^-1y,
γ = cos^-1z

Given: α + β + γ = π
⇒ γ = π − (α + β)

Taking cosine on both sides:
cos γ = cos [π − (α + β)]

We know: cos(π − θ) = −cos θ
⇒ cos γ = −cos(α + β)

Using formula:
cos(α + β) = cosα cosβ − sinα sinβ
⇒ z = −[xy − √(1 − x²) √(1 − y²)]
⇒ z = −xy + √[(1 − x²)(1 − y²)]

Squaring both sides:
z² = (−xy + √[(1 − x²)(1 − y²)] )²
z² = x²y² + (1 − x²)(1 − y²) − 2xy√[(1 − x²)(1 − y²)]

Expanding:
z² = x²y² + 1 − x² − y² + x²y² − 2xy√[(1 − x²)(1 − y²)]
z² = 1 − x² − y² + 2x²y² − 2xy√[(1 − x²)(1 − y²)]But √[(1 − x²)(1 − y²)] = z + xy

Substitute:
z² = 1 − x² − y² + 2x²y² − 2xy(z + xy)
z² = 1 − x² − y² + 2x²y² − 2xyz − 2x²y²
z² = 1 − x² − y² − 2xyz

Rearranging:
x² + y² + z² + 2xyz = 1

Hence Proved.

21. Solve: cos^-1 ( sin cos^-1x ) = π/6

Sol: Let θ = cos^-1x
⇒ cos θ = x

Now the equation becomes:
cos^-1 ( sin θ ) = π/6

Taking cosine on both sides:
sin θ = cos (π/6)

We know:
cos (π/6) = √3 / 2
⇒ sin θ = √3 / 2

But sin θ = √(1 − cos²θ)
⇒ sin θ = √(1 − x²)

So,
√(1 − x²) = √3 / 2

Squaring both sides:
1 − x² = 3/4
x² = 1/4
x = ±1/2

∴x = ± 1/2

22. Solve the equation for x: sin^-1(5/x) + sin^-1(12/x) = π/2 , x ≠ 0

Sol: Let A = sin^-1(5/x)
Let B = sin^-1(12/x)

Given:
A + B = π/2
Taking sine on both sides:
sin(A + B) = sin(π/2)

We know:
sin(A + B) = sinA cosB + cosA sinB
sin(π/2) = 1

So,
sinA cosB + cosA sinB = 1

Now,
sinA = 5/x
sinB = 12/x
cosA = √(1 − 25/x²)
cosB = √(1 − 144/x²)

Substituting:
(5/x) √(1 − 144/x²) + (12/x) √(1 − 25/x²) = 1

Since A + B = π/2,
cosA = sinB
⇒ √(1 − 25/x²) = 12/x

Squaring both sides:
1 − 25/x² = 144/x²
1 = 169/x²
x² = 169
x = ±13

Checking domain:

|5/x| ≤ 1 and |12/x| ≤ 1
⇒ |x| ≥ 12

Both x = 13 and x = −13 satisfy the condition.

∴x = ±13

23. Solve for x, if tan (cos^-1x) = 2/√5

Sol: Let θ = cos^-1x
⇒ cos θ = x

Given:
tan θ = 2/√5

We know:
tan θ = sin θ / cos θ
⇒ sin θ / x = 2/√5
⇒ sin θ = (2x)/√5

But,
sin²θ + cos²θ = 1
⇒ sin²θ = 1 − x²

Now substitute:
( (2x)/√5 )² = 1 − x²
4x² / 5 = 1 − x²

Multiply by 5:
4x² = 5 − 5x²
9x² = 5
x² = 5/9
x = ± √5 / 3

Now check range:
Since θ = cos^-1x lies in [0, π],
tan θ is positive only in first quadrant.
So θ is acute ⇒ x > 0

∴x = √5 / 3

24. If sin^-1x + tan^-1x = π/2 , prove that 2x² + 1 = √5

Sol: Let A = sin^-1x
Let B = tan^-1x

Given:
A + B = π/2
⇒ A = π/2 − B

Taking sine on both sides:
sin A = sin (π/2 − B)

We know:
sin (π/2 − B) = cos B

So,
x = cos (tan^-1x)
Let θ = tan^-1x
⇒ tan θ = x

Draw a right triangle:
Opposite = x
Adjacent = 1
Hypotenuse = √(1 + x²)

So,
cos θ = 1 / √(1 + x²)

Hence,
x = 1 / √(1 + x²)

Squaring both sides:
x² = 1 / (1 + x²)
x²(1 + x²) = 1
x⁴ + x² − 1 = 0

Let y = x²
y² + y − 1 = 0

Using quadratic formula:
y = (−1 + √5)/2   (since y ≥ 0)

Thus,
2y + 1 = √5
⇒ 2x² + 1 = √5

Hence Proved.

25. Prove that (1/2) cos^-1[(1 − x)/(1 + x)] = tan^-1√x

Sol: Let θ = (1/2) cos^-1[(1 − x)/(1 + x)]
⇒ 2θ = cos^-1[(1 − x)/(1 + x)]

Taking cosine on both sides:
cos 2θ = (1 − x)/(1 + x)

We know the identity:
cos 2θ = (1 − tan²θ) / (1 + tan²θ)

Substitute:
(1 − tan²θ) / (1 + tan²θ) = (1 − x)/(1 + x)

Since denominators are equal, compare numerators:
1 − tan²θ = 1 − x
⇒ tan²θ = x
⇒ tan θ = √x

Hence,
θ = tan^-1(√x)

Therefore,
(1/2) cos^-1[(1 − x)/(1 + x)] = tan^-1(√x)

Hence Proved.

–: End of Self assessment and Revision Inverse Trigonometric Functions Class 12 OP Malhotra ISC Maths Solutions :–

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