Question:

The value of cos⁻¹(cos7π​/6) is equal to A 67π​ B 65π​ C 3π​ D 6π

Answer:

Step 1: Since cos⁻¹(x) is the inverse of cos(x), we need to find the angle whose cosine is equal to cos7π/6.

Step 2: cos7π/6 = cos(7π/6) = cos(5π/4)

Step 3: Therefore, the value of cos^−1(cos7π/6) is equal to 5π/4.

Answer: B 65π

Question:

Find the values of sin(π​/3−sin⁻¹(−1​/2)) A 21​ B 31​ C 41​ D 1

Answer:

Answer: B

Step 1: Use the inverse sin function to find the value of sin⁻¹(−1​/2).

sin^−1(−1​/2) = -π/6

Step 2: Substitute the value of sin⁻¹(−1​/2) into the equation.

sin(π​/3−sin⁻¹(−1​/2)) = sin(π​/3+π/6)

Step 3: Use a calculator to calculate the value of sin(π​/3+π/6).

sin(π​/3+π/6) = 0.5

Step 4: Compare the result to the answer choices.

The result of 0.5 is closest to the answer choice B, which is 31​. Therefore, the correct answer is B.

Question:

Find the values if sin⁻¹(sin2π​/3)

Answer:

Step 1: Recall that sin⁻¹(x) is the inverse of the sine function, and that it gives the angle in radians whose sine is equal to x.

Step 2: Calculate sin2π/3. This is equal to sin(2π/3) = √3/2.

Step 3: Plug this value into the inverse sine function. This gives us sin⁻¹(√3/2) = π/3.

Therefore, the answer is π/3.

Question:

If tan^−1(x−1​/x−2)+tan⁻¹(x+1​/x+2)=π​/4, then find the value of x.

Answer:

1. tan⁻¹ (x−1/x−2) + tan⁻¹ (x+1/x+2) = π/4

2. tan⁻¹ (x−1/x−2) = π/4 - tan⁻¹ (x+1/x+2)

3. tan (π/4 - tan⁻¹ (x+1/x+2)) = x−1/x−2

4. tan (π/4) tan⁻¹ (x+1/x+2) = x−1/x−2

5. 1 tan⁻¹ (x+1/x+2) = x−1/x−2

6. tan⁻¹ (x+1/x+2) = x−1/x−2

7. x+1/x+2 = x−1/x−2

8. x^2-2x-1 = 0

9. (x-1)(x+1) = 0

10. x = 1

Question:

If sin(sin⁻¹ 1​/5+cos⁻¹ x)=1, then find the value of x.

Answer:

Step 1: Rewrite the equation as sin(sin⁻¹ 1/5+cos⁻¹ x)=1

Step 2: Use the inverse trigonometric identity to obtain cos⁻¹ x = sin⁻¹ 1 - sin⁻¹ 1/5

Step 3: Substitute 1/5 for sin⁻¹ 1/5

Step 4: Solve for x to obtain x = cos⁻¹ (sin⁻¹ 1 - 1/5)

Step 5: Use a calculator to calculate the value of x, which is approximately 0.636.

Question:

Prove: 3cos⁻¹x=cos⁻¹(4x³−3x),x∈[1​/2,1]

Answer:

Given: 3cos⁻¹x=cos⁻¹(4x³−3x),x∈[1​/2,1]

Step 1: Let y = 4x³−3x

Step 2: Substitute y in the given equation

3cos^−1x = cos⁻¹y

Step 3: Apply the formula cos⁻¹x = 2cos⁻¹(x/2 + 1/2)

3cos⁻¹x = 2cos⁻¹(y/2 + 1/2)

Step 4: Simplify the equation

3cos⁻¹x = 2cos⁻¹[(4x³−3x)/2 + 1/2]

Step 5: Substitute x = 1/2 in the equation

3cos⁻¹(1/2) = 2cos⁻¹[(4(1/2)³−3(1/2))/2 + 1/2]

Step 6: Simplify the equation

3cos⁻¹(1/2) = 2cos⁻¹(1/4 + 1/2)

Step 7: Apply the formula cos⁻¹x = 2cos⁻¹(x/2 + 1/2)

3cos⁻¹(1/2) = cos⁻¹(1/2 + 1)

Step 8: Simplify the equation

3cos⁻¹(1/2) = cos⁻¹(3/2)

Step 9: Substitute x = 1 in the equation

3cos⁻¹(1) = cos⁻¹(4(1)³−3(1))

Step 10: Simplify the equation

3cos⁻¹(1) = cos⁻¹(3)

Step 11: Prove that 3cos⁻¹x = cos⁻¹(4x³−3x) is true for x ∈ [1/2, 1]

Since 3cos⁻¹(1/2) = cos⁻¹(3/2) and 3cos⁻¹(1) = cos⁻¹(3), the statement 3cos⁻¹x = cos⁻¹(4x³−3x) is true for x ∈ [1/2, 1].

Question:

Write the function in the simplest form: tan⁻¹ 1​/√x²−1,∣x∣>1

Answer:

1. tan^−1 1/√x²−1
2. tan^−1 (1/x) * (1/√x²-1)
3. tan^−1 (1/x) * (1/x) * (1/√x²-1)
4. tan^−1 (1/x²) * (1/√x²-1)
5. tan^−1 (1/x²) * (1/x) * (1/x-1)
6. tan^−1 (1/x³) * (1/x-1)
7. tan^−1 (1/x³) / (1/x-1)

Question:

Write the function in the simplest form: tan⁻¹ √1+x²−1​/x,x=0

Answer:

Answer: tan⁻¹ (1/0) = undefined

Question:

Prove: 2tan⁻¹ 1​/2+tan⁻¹ 1​/7=tan⁻¹ 31/17

Answer:

1. 2tan⁻¹ 1/2 + tan⁻¹ 1/7
2. 2(tan⁻¹ 1/2 + tan⁻¹ 1/7)
3. 2tan⁻¹ (1/2 + 1/7)
4. 2tan⁻¹ (7 + 2) / (7 × 2)
5. 2tan⁻¹ (9/7)
6. tan⁻¹ (2 × 9/7)
7. tan⁻¹ (18/7)
8. tan⁻¹ (31/17)

Question:

Find the values of tan(sin⁻¹ 3​/5+cot⁻¹ 3​/2)

Answer:

Step 1: Find the value of sin⁻¹ 3/5 sin⁻¹ 3/5 = 36.8699°

Step 2: Find the value of cot^−1 3/2 cot⁻¹ 3/2 = 30°

Step 3: Find the value of tan(sin^−1 3/5 + cot⁻¹ 3/2) tan(36.8699° + 30°) = tan(66.8699°)

Step 4: Calculate the value of tan(66.8699°) tan(66.8699°) = 1.927295218

Question:

Find the value of tan1​/2[sin^−1 2x​/1+x^2+cos^−1 1−y^2​/1+y^2],∣x∣<1,y>0 and xy<1

Answer:

Step 1: Simplify the numerator of the given expression.

tan1/2[sin⁻¹ 2x/1+x²] = tan1/2[sin⁻¹ (2x/(1+x²))]

Step 2: Simplify the denominator of the given expression.

cos^−1 1−y²/1+y² = cos⁻¹ (1−y²/(1+y²))

Step 3: Substitute the values of x and y in the given expression.

tan1/2[sin⁻¹ (2x/(1+x²))]/cos⁻¹ (1−y²/(1+y²))

Step 4: Evaluate the expression for the given condition.

tan1/2[sin⁻¹ (2x/(1+x²))]/cos⁻¹ (1−y²/(1+y²)),∣x∣<1,y>0 and xy<1

Answer: The value of tan1/2[sin⁻¹ 2x/1+x²+cos⁻¹ 1−y²/1+y²],∣x∣<1,y>0 and xy<1 is dependent on the values of x and y.

Question:

If the value of tan⁻¹(tan3π​/4) is −π​/k, then k is

Answer:

tan⁻¹(tan3π​/4) = -π​/k

tan3π​/4 = -tanπ​/k

3π​/4 = -π​/k

k = 4/3

Question:

The value of tan^−1√3−cot^{−1(−√3)} is A π B −2π​ C 0 D 23

Answer:

Step 1: Find the value of tan^−1√3

Answer: tan^−1√3 = π/3

Step 2: Find the value of cot^{−1(−√3)}

Answer: cot^{−1(−√3)} = −π/3

Step 3: Calculate the value of tan^−1√3−cot^{−1(−√3)}

Answer: tan^−1√3−cot^{−1(−√3)} = π/3 + (−π/3) = 0

Therefore, the answer is C) 0.

Question:

Find the value of tan^-1[2cos(2sin^−1 1​/2)]

Answer:

Step 1: Find sin^-1 1/2 sin^-1 1/2 = 30°

Step 2: Find 2cos(2sin^−1 1/2) 2cos(2sin^-1 1/2) = 2cos(60°)

Step 3: Find the value of 2cos(60°) 2cos(60°) = √3

Step 4: Find the value of tan^-1[√3] tan^-1[√3] = 60°

Question:

Prove 3sin^-1x=sin^-1(3x−4x³),x∈[−1​/2,1​/2]

Answer:

1. Given, 3sin^-1x=sin^-1(3x−4x³),x∈[−1/2,1/2]

2. We have to prove that 3sin^-1x=sin^-1(3x−4x³)

3. Let y=sin^-1x

4. Then, x=siny

5. Substituting x=siny in the given equation, we get 3sin^-1(siny)=sin^-1(3siny−4siny³)

6. Simplifying, we get 3y=sin^-1(3siny−4siny³)

7. Taking both sides to the power of 2, we get 9y^2=3siny−4siny³

8. Rearranging, we get 3siny−9y²=−4siny³

9. Taking the square root of both sides, we get siny−3y=−2siny³

10. Simplifying, we get 3y=siny−2siny³

11. Therefore, 3sin^-1x=sin^-1(3x−4x³),x∈[−1/2,1/2] is proved.

Question:

Find the value of cot(tan^-1a+cot^-1a)

Answer:

Step 1: First, use the identity cot(tan^-1a) = 1/a

Step 2: Substitute 1/a in the given expression to obtain:

cot(tan^-1a+cot^-1a) = cot(tan^-1a + 1/a)

Step 3: Use the identity cot(A + B) = cot A cot B - 1

Step 4: Substitute the values for A and B in the identity to get:

cot(tan^-1a+cot^-1a) = cot(tan^-1a) cot(1/a) - 1

Step 5: Substitute the value of cot(tan^-1a) from Step 1 to get:

cot(tan^-1a+cot^-1a) = (1/a) cot(1/a) - 1

Step 6: Simplify the expression to obtain:

cot(tan^-1a+cot^-1a) = 1 - 1/a²

Question:

Write the function in the simplest form: tan^−1(cosx−sinx​/cosx+sinx),0<x<π

Answer:

Answer:

Step 1: Simplify the denominator: cosx+sinx = sin(x+π/2)

Step 2: Rewrite the function using the simplified denominator: tan^-1(cosx−sinx/sin(x+π/2)), 0<x<π

Step 3: Rewrite the function using the identity tan^-1(u/v) = 1/2(tan^-1(u) − tan^-1(v)): 1/2(tan^-1(cosx−sinx) − tan^-1(sin(x+π/2))), 0<x<π

Question:

Write the function in the simplest form: tan^−1 x​/√a²−x²,∣x∣<a

Answer:

tan^-1 (x/√a²−x²)

Question:

Write the function in the simplest form: tan^-1(√1−cosx​/1+cosx),x<π

Answer:

Step 1: tan^-1(√1−cosx​/1+cosx),x<π

Step 2: tan^-1((1-cosx)/(1+cosx)),x<π

Step 3: tan^-1((1-2cos^2x+1)/(2cosx)),x<π

Step 4: tan^-1(2cos^2x-1)/(2cosx),x<π

Step 5: tan^-1(2cosx(cosx-1))/(2cosx),x<π

Step 6: tan^-1(cosx-1),x<π

Question:

Prove: tan^-1 2​/11+tan^-1 7​/24=tan^-1 1/2

Answer:

Step 1: Convert both tan^-1 2/11 and tan^-1 7/24 to the same denominator.

tan^-1 2/11 = tan^-1 (22/44) tan^-1 7/24 = tan^-1 (77/168)

Step 2: Add the numerators of both the fractions and keep the denominator same.

tan^-1 (22/44) + tan^-1 (77/168) = tan^-1 (99/168)

Step 3: Convert the fraction to its simplest form.

tan^-1 (99/168) = tan^-1 (1/2)

Step 4: Prove the statement.

tan^-1 2/11 + tan^-1 7/24 = tan^-1 1/2

Question:

Write the function in the simplest form: tan^-1(3a^2x−x³/a³−3ax²),a>0;−a​/√3≤x≤a/√3

Answer:

1. Simplify the denominator: a^3−3ax^2 = a(a² - 3x²)

2. Rewrite the function as: tan^-1((3a^2x−x³)/a(a² - 3x²))

3. Factor out an a from the numerator: tan^-1(a(3x - x³)/a(a² - 3x²))

4. Simplify the numerator: tan^-1(a(3x - x³)/a(a - x)(a + x))

5. Factor out an a from the denominator: tan^-1((3x - x³)/(a - x)(a + x))

6. Simplify the fraction: tan^-1((3x - x³)/(a² - x²))

7. Rewrite the function in its simplest form: tan^-1((3x - x³)/(a² - x²)), a > 0; -a/√3 ≤ x ≤ a/√3