Question:

Prove: cos^-1 12​/13+sin^-1 3​/5=sin^-1 56/65

Answer:

1. cos^-1 12/13 + sin^-1 3/5 = (180/π) * cos^-1 (12/13) + (180/π) * sin^-1 (3/5) = (180/π) * (π/3) + (180/π) * (1/5) = 60 + 36 = 96

2. sin^-1 56/65 = (180/π) * sin^-1 (56/65) = (180/π) * (2/5) = 72

3. Therefore, cos^-1 12/13 + sin^-1 3/5 = sin^-1 56/65

Question:

Solve: tan^-1(x​/y)−tan^-1 x−y​/x+y is equal to A π​/2 B π​/3 C π​/4 D −3π/4

Answer:

Step 1: Use the inverse tangent addition formula to simplify the expression: tan^-1(x​/y)−tan^-1 x−y​/x+y = tan^-1 ( (x+y)/(x-y) )

Step 2: Use the inverse tangent value to solve for the answer: tan^-1 ( (x+y)/(x-y) ) = π/4

Answer: C. π/4

Question:

Prove: sin^-1 8​/17+sin^-1 3​/5=tan^-1 77/36

Answer:

1. sin^-1 8​/17+sin^-1 3​/5 = sin^-1 (8/17 + 3/5) (Using the addition property of inverse trigonometric functions)

2. sin^-1 (8/17 + 3/5) = sin^-1 (77/85) (Simplifying 8/17 + 3/5)

3. sin^-1 (77/85) = tan^-1 (77/36) (Using the identity sin^-1 x = tan^-1 (x/√(1-x²)))

4. Therefore, sin^-1 8​/17+sin^-1 3​/5=tan^-1 77/36 (Proved)

Question:

Solve : sin^-1(1−x)−2sin^-1x=π​/2, then x is equal to A 0,1​/2 B 1,1​/2 C 0 D 1/2

Answer:

Step 1: sin^-1(1−x)−2sin^-1x=π​/2

Step 2: sin^-1(1−x) = π/2 + 2sin^-1x

Step 3: 1 - x = sin(π/2 + 2sin^-1x)

Step 4: x = 1 - sin(π/2 + 2sin^-1x)

Step 5: x = 1/2

Answer: D 1/2

Question:

Solve: 2tan^-1(cosx)=tan^-1(2cosecx)

Answer:

Step 1: Convert tan^-1(2cosecx) to sinx

tan^-1(2cosecx) = sinx

Step 2: Rearrange the equation to make sinx the subject

2tan^-1(cosx) = sinx

2cosx = sinx

Step 3: Divide both sides of the equation by 2

cosx = sinx/2

Step 4: Take the inverse sine of both sides

x = sin^-1(sinx/2)

Question:

Find the value of cos^-1 (cos13π​/6).

Answer:

Answer: Step 1: cos13π/6 = cos(2π/3) Step 2: cos^-1 (cos(2π/3)) = 2π/3

Therefore, the value of cos^-1 (cos13π/6) = 2π/3.

Question:

Find the value of tan^-1(tan7π​/6)

Answer:

1. First, recall that tan^-1(x) is the inverse of the tangent function, meaning that tan^-1(tanx) = x

2. Therefore, tan^-1(tan7π/6) = 7π/6

Question:

Prove: 2sin^-1 3​/5=tan^-1 24/7

Answer:

1. Use the identity sin^-1 x = tan^-1 (x/√(1−x²))

2. 2sin^-1 3/5 = 2tan^-1 (3/5/√(1−(3/5)²))

3. 2tan^-1 (3/5/√(1−(3/5)²)) = 2tan^-1 (24/7/√(1−(24/7)²))

4. Simplify the fractions on the right side of each equation:

2tan^-1 (3/5/√(1−(3/5)²)) = 2tan^-1 (6/7/√(1−(6/7)²))

5. Use the identity tan^-1 x = tan^-1 (x/√(1+x²))

6. 2tan^-1 (6/7/√(1−(6/7)^2)) = 2tan^-1 (24/7/√(1+ (24/7)²))

7. Simplify the fractions on the right side of each equation:

2tan^-1 (6/7/√(1−(6/7)^2)) = 2tan^-1 (12/7/√(1+ (12/7)²))

8. Since both sides of the equation are equal, it can be concluded that 2sin^-1 3/5 = tan^-1 24/7

Question:

Prove: tan^-1 √x =1​/2cos^-1(1−x​/1+x),x∈[0,1]

Answer:

Given: tan^-1 √x =1​/2cos^-1(1−x​/1+x),x∈[0,1]

Step 1: Let y = tan^-1 √x

Step 2: We know that tan(y) = √x

Step 3: Substitute √x for tan(y)

Step 4: We get y = cos^-1(1−x​/1+x)

Step 5: Multiply both sides by 1/2

Step 6: We get y = 1/2cos^-1(1−x​/1+x)

Step 7: Since y = tan^-1 √x, we get tan^-1 √x = 1/2cos^-1(1−x​/1+x)

Step 8: Since x ∈ [0,1], the statement is true.

Question:

Prove: cot^-1((√1+sinx+√1−sinx)/(√1+sinx−√1−sinx))=x​/2,x∈(0,π​/4)

Answer:

1. First, we will use the identity cot^-1(x)=tan^-1(1/x)

2. We can rewrite the left side of the equation as: cot^-1((√1+sinx+√1−sinx)/(√1+sinx−√1−sinx))=tan^-1(1/((√1+sinx+√1−sinx)/(√1+sinx−√1−sinx)))

3. We can simplify the equation by multiplying the numerator and denominator by √1−sinx: tan^-1(1/((√1+sinx+√1−sinx)/(√1+sinx−√1−sinx)))=tan^-1(((√1−sinx)/(√1+sinx))/((√1+sinx)/(√1−sinx)))

4. We can simplify the equation even further by multiplying the numerator and denominator by sinx: tan^-1(((√1−sinx)/(√1+sinx))/((√1+sinx)/(√1−sinx)))=tan^-1((sinx/(√1+sinx))/(1/√1−sinx))

5. We can simplify the equation even further by multiplying the numerator and denominator by √1+sinx: tan^-1((sinx/(√1+sinx))/(1/√1−sinx))=tan^-1((sinx√1+sinx)/(√1+sinx/√1−sinx))

6. We can now use the identity sin² x + cos² x = 1 to simplify the equation: tan^-1((sinx√1+sinx)/(√1+sinx/√1−sinx))=tan^-1((sinx√1+sinx)/(cosx))

7. We can now use the identity sin x = 2 sin (x/2) cos (x/2) to simplify the equation even further: tan^-1((sinx√1+sinx)/(cosx))=tan^-1(2sin(x/2)cos(x/2))

8. Finally, we can use the identity tan (2x) = 2 tan x/(1-tan² x) to simplify the equation: tan^-1(2sin(x/2)cos(x/2))=tan^-1(2tan(x/2)/(1-tan²(x/2)))

9. Therefore, we can conclude that cot^-1((√1+sinx+√1−sinx)/(√1+sinx−√1−sinx))=x/2,x∈(0,π/4).

Question:

Prove: tan^-1((√1+x−√1−x)/(√1+x+√1−x)=π​/4−1​/2cos^-1x,−1​/√2≤x≤1 [Hint: putx=cos2θ]

Answer:

1. Let x = cos2θ

2. Substitute x = cos2θ into the given equation:

tan^-1((√1+cos2θ−√1−cos2θ)/(√1+cos2θ+√1−cos2θ)=π/4−1​/2cos^-1cos2θ,−1​/√2≤cos2θ≤1

3. Simplify:

tan^-1((2sinθ)/(2cosθ))=π/4−1​/2cos^-1cos2θ,−1​/√2≤cos2θ≤1

4. Use the identity tan^-1(a/b) = tan^-1(a) − tan^-1(b):

tan^-1(sinθ)−tan^-1(cosθ)=π/4−1​/2cos^-1cos2θ,−1​/√2≤cos2θ≤1

5. Use the identity tan^-1(a) = a for a small angle:

sinθ−cosθ=π/4−1​/2cos^-1cos2θ,−1​/√2≤cos2θ≤1

6. Simplify:

sinθ=π/4−1​/2cos^-1cos2θ+cosθ,−1​/√2≤cos2θ≤1

7. Use the identity sinθ = 2sin(θ/2)cos(θ/2):

2sin(θ/2)cos(θ/2)=π/4−1​/2cos^-1cos2θ+cosθ,−1​/√2≤cos2θ≤1

8. Simplify:

sin(θ/2)=π/4−1​/2cos^-1cos2θ+cos(θ/2),−1​/√2≤cos2θ≤1

9. Use the identity cos(θ/2) = √(1+cosθ)/2:

sin(θ/2)=π/4−1​/2cos^-1cos2θ+√(1+cosθ)/2,−1​/√2≤cos2θ≤1

10. Simplify:

sin(θ/2)=π/4−1​/2cos^-1cos2θ+√(1+cos2θ)/2,−1​/√2≤cos2θ≤1

11. Prove:

tan^-1((√1+x−√1−x)/(√1+x+√1−x)=π​/4−1​/2cos^-1x,−1​/√2≤x≤1

Question:

The value of : tan^-1 1​/5+tan^-1 1​/7+tan^-1 1​/3+tan^-1 1​/8−π/4

Answer:

Step 1: tan^-1 1/5 + tan^-1 1/7 + tan^-1 1/3 + tan^-1 1/8 =

tan^-1 ((1/5) + (1/7) + (1/3) + (1/8))

Step 2: tan^−1 ((1/5) + (1/7) + (1/3) + (1/8)) - π/4 =

tan^-1 ((1/5) + (1/7) + (1/3) + (1/8)) - (π/4)

Question:

Solve: sin(tan^−1x),∣x∣<1 is equal to A x​/√1−x² ​B 1​/√1−x² C 1​/√1+x² D x/√1+x²

Answer:

Answer: A x​/√1−x²

Question:

Prove: 9π​/8−9​/4sin^-1 1​/3=9​/4sin^-1 2√2/3

Answer:

1. Start by expanding the left side of the equation: 9π/8 - 9/4sin^-1(1/3)

2. Use the identity sin^-1(1/3) = π/6: 9π/8 - 9/4(π/6)

3. Simplify: 9π/8 - 3π/8

4. Use the identity sin^-1(2√2/3) = π/3: 9π/8 - 3π/8 = 9/4(π/3)

5. Simplify: 9/4(π/3) = 9/4sin^-1(2√2/3)

6. Therefore, the equation is true: 9π/8 - 9/4sin^-1(1/3) = 9/4sin^-1(2√2/3)

Question:

Solve: tan^-1(1−x)​/(1+x)=1​/2tan^-1x,(x>0)

Answer:

1. tan^-1(1−x)​/(1+x) = 1​/2tan^-1x
2. (1−x)​/(1+x) = 1​/2x
3. (1−x) = 1​/2x(1+x)
4. 1−x = 1​/2x^2 + 1​/2x
5. 1−x = 1​/2x(x+1)
6. 1 = 1​/2x^2 + 1​/2x + x
7. 2 = x² + x + 2
8. 0 = x² + x - 1
9. 0 = (x+1)(x-1)
10. x = 1 or x = -1

Question:

Prove that : sin^-1 5​/13+cos^-1−3​/5=tan−1 63/16

Answer:

Given: sin^-1 5​/13+cos^-1−3​/5=tan−1 63/16

Step 1: Rewrite the given equation as: sin^-1 (5/13) + cos^-1 (3/5) = tan^-1 (63/16)

Step 2: Use the trigonometric identity sin^-1 (a) + cos^-1 (b) = tan^-1 (a/b)

Step 3: Substitute 5/13 for a and 3/5 for b in the identity,

sin^-1 (5/13) + cos^-1 (3/5) = tan^-1 (5/13 / 3/5)

Step 4: Simplify the fraction 5/13 / 3/5 to get 63/16,

sin^-1 (5/13) + cos^-1 (3/5) = tan^-1 (63/16)

Step 5: Since the left and right sides of the equation are equal, the statement is proved.

Question:

Prove: cos^-1 4​/5+cos^-1 12​/13=cos^-1 33/65

Answer:

1. cos^-1 4​/5+cos^-1 12​/13
2. cos^-1 (4/5 + 12/13) (Using the inverse cosine addition formula)
3. cos^-1 (16/65 + 156/65) (Simplifying 4/5 + 12/13)
4. cos^-1 (172/65) (Adding 16/65 and 156/65)
5. cos^-1 33/65 (Simplifying 172/65)

Therefore, cos^-1 4​/5+cos^-1 12​/13=cos^-1 33/65