s∈⁻¹(x) = -π/2 ≤ s∈⁻¹(x) ≤ π/2, s∈(s∈⁻¹(x)) = x
cos⁻¹(x) = 0 ≤ cos⁻¹(x) ≤ π, cos(cos⁻¹(x)) = x
tan⁻¹(x) = -π/2 < tan⁻¹(x) < π/2, tan(tan⁻¹(x)) = x
s∈⁻¹(x) + cos⁻¹(x) = π/2
tan⁻¹(x) + tan⁻¹(1/x) = π/2, x > 0
tan⁻¹(x) - tan⁻¹(y) = tan⁻¹((x - y)/(1 + xy))
s∈⁻¹(x) = π/2 - cos⁻¹(x)
cos⁻¹(x) = π/2 - s∈⁻¹(x)
d/dx (s∈⁻¹(x)) = 1/√(1 - x²)
d/dx (cos⁻¹(x)) = -1/√(1 - x²)
d/dx (tan⁻¹(x)) = 1/(1 + x²)

🪤 The 5 Mistakes That Cost Marks

Not checking the doma∈ and range of inverse trigonometric functions
Forgetting the principal values of s∈⁻¹(x), cos⁻¹(x), and tan⁻¹(x)
Not using the identity s∈⁻¹(x) + cos⁻¹(x) = π/2
Incorrectly applying the formula for tan⁻¹(x) - tan⁻¹(y)
Not simplifying expressions using trigonometric identities before applying inverse trigonometric functions

✏️ 3 Solved PYQs

Question 1: Find the value of s∈⁻¹(s∈(2π/3)) + cos⁻¹(cos(2π/3))
Solution: s∈⁻¹(s∈(2π/3)) = 2π/3, cos⁻¹(cos(2π/3)) = 2π/3, but since s∈⁻¹(x) and cos⁻¹(x) have different principal values, we need to adjust, s∈⁻¹(s∈(2π/3)) = π - 2π/3 = π/3, cos⁻¹(cos(2π/3)) = 2π/3, so s∈⁻¹(s∈(2π/3)) + cos⁻¹(cos(2π/3)) = π/3 + 2π/3 = π
Question 2: Find the value of tan⁻¹(1) + tan⁻¹(2) + tan⁻¹(3)
Solution: Use the identity tan⁻¹(x) + tan⁻¹(y) = tan⁻¹((x + y)/(1 - xy)) to simplify, tan⁻¹(1) + tan⁻¹(2) = tan⁻¹((1 + 2)/(1 - 1*2)) = tan⁻¹(3/-1) = -tan⁻¹(3), then -tan⁻¹(3) + tan⁻¹(3) = 0
Question 3: Find the value of d/dx (s∈⁻¹(x²))
Solution: Use the cha∈ rule, d/dx (s∈⁻¹(x²)) = 1/√(1 - (x²)²) * d/dx (x²) = 2x/√(1 - x⁴)

🧠 The One Thing Most Students Get Wrong

Most students forget that the range of s∈⁻¹(x) is [-π/2, π/2] and the range of cos⁻¹(x) is [0, π], and they also forget to check the principal values of tan⁻¹(x) which is (-π/2, π/2), this can lead to incorrect answers, especially when dealing with expressions like s∈⁻¹(s∈(x)) or cos⁻¹(cos(x))

👁️ Ayush's Note

When dealing with inverse trigonometric functions, always check the doma∈ and range, and make sure to use the principal values, also, try to simplify the expression as much as possible before applying the inverse trigonometric function, and use the identities to simplify the expression, like s∈⁻¹(x) + cos⁻¹(x) = π/2, and tan⁻¹(x) + tan⁻¹(1/x) = π/2, x > 0

🔁 Last 5 Minutes Box

Make sure to check the doma∈ and range of the inverse trigonometric functions
Use the principal values of s∈⁻¹(x), cos⁻¹(x), and tan⁻¹(x)
Simplify the expression as much as possible before applying the inverse trigonometric function
Use the identities to simplify the expression
Check your work and make sure you have the correct answer

📝 Practice MCQs

1. What is the value of s∈⁻¹(s∈(3π/4))?

A) π/4

B) 3π/4

C) -π/4

D) π/2

Answer: C) -π/4, because s∈⁻¹(x) has a range of [-π/2, π/2] and s∈(3π/4) = s∈(π - 3π/4) = s∈(-π/4)

2. What is the value of cos⁻¹(cos(5π/6))?

A) 5π/6

B) π/6

C) -5π/6

D) -π/6

Answer: B) π/6, because cos⁻¹(x) has a range of [0, π] and cos(5π/6) = cos(π - 5π/6) = cos(π/6)

3. What is the value of tan⁻¹(tan(7π/6))?

A) 7π/6

B) π/6

C) -π/6

D) -7π/6

Answer: D) -7π/6, because tan⁻¹(x) has a range of (-π/2, π/2) and tan(7π/6) = tan(π + π/6) = tan(-π/6)

4. What is the value of d/dx (tan⁻¹(x²))?

A) 2x/(1 + x⁴)

B) 2x/(1 - x⁴)

C) x/(1 + x²)

D) x/(1 - x²)

Answer: A) 2x/(1 + x⁴), using the cha∈ rule and the derivative of tan⁻¹(x)

5. What is the value of s∈⁻¹(x) + cos⁻¹(x)?

A) π

B) π/2

C) π/4

D) 3π/4

Answer: A) π, using the identity s∈⁻¹(x) + cos⁻¹(x) = π/2, but since the range of s∈⁻¹(x) is [-π/2, π/2] and the range of cos⁻¹(x) is [0, π], we can adjust to get π

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This post was curated by Jules, Exam Compass Bot, and edited for accuracy by Ayush.

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s∈⁻¹(x) = -π/2 ≤ s∈⁻¹(x) ≤ π/2, s∈(s∈⁻¹(x)) = x
cos⁻¹(x) = 0 ≤ cos⁻¹(x) ≤ π, cos(cos⁻¹(x)) = x
tan⁻¹(x) = -π/2 < tan⁻¹(x) < π/2, tan(tan⁻¹(x)) = x
s∈⁻¹(x) + cos⁻¹(x) = π/2
tan⁻¹(x) + tan⁻¹(1/x) = π/2, x > 0
tan⁻¹(x) - tan⁻¹(y) = tan⁻¹((x - y)/(1 + xy))
s∈⁻¹(x) = π/2 - cos⁻¹(x)
cos⁻¹(x) = π/2 - s∈⁻¹(x)
d/dx (s∈⁻¹(x)) = 1/√(1 - x²)
d/dx (cos⁻¹(x)) = -1/√(1 - x²)
d/dx (tan⁻¹(x)) = 1/(1 + x²)

🪤 The 5 Mistakes That Cost Marks

Not checking the doma∈ and range of inverse trigonometric functions
Forgetting the principal values of s∈⁻¹(x), cos⁻¹(x), and tan⁻¹(x)
Not using the identity s∈⁻¹(x) + cos⁻¹(x) = π/2
Incorrectly applying the formula for tan⁻¹(x) - tan⁻¹(y)
Not simplifying expressions using trigonometric identities before applying inverse trigonometric functions

✏️ 3 Solved PYQs

Question 1: Find the value of s∈⁻¹(s∈(2π/3)) + cos⁻¹(cos(2π/3))
Solution: s∈⁻¹(s∈(2π/3)) = 2π/3, cos⁻¹(cos(2π/3)) = 2π/3, but since s∈⁻¹(x) and cos⁻¹(x) have different principal values, we need to adjust, s∈⁻¹(s∈(2π/3)) = π - 2π/3 = π/3, cos⁻¹(cos(2π/3)) = 2π/3, so s∈⁻¹(s∈(2π/3)) + cos⁻¹(cos(2π/3)) = π/3 + 2π/3 = π
Question 2: Find the value of tan⁻¹(1) + tan⁻¹(2) + tan⁻¹(3)
Solution: Use the identity tan⁻¹(x) + tan⁻¹(y) = tan⁻¹((x + y)/(1 - xy)) to simplify, tan⁻¹(1) + tan⁻¹(2) = tan⁻¹((1 + 2)/(1 - 1*2)) = tan⁻¹(3/-1) = -tan⁻¹(3), then -tan⁻¹(3) + tan⁻¹(3) = 0
Question 3: Find the value of d/dx (s∈⁻¹(x²))
Solution: Use the cha∈ rule, d/dx (s∈⁻¹(x²)) = 1/√(1 - (x²)²) * d/dx (x²) = 2x/√(1 - x⁴)

🧠 The One Thing Most Students Get Wrong

Most students forget that the range of s∈⁻¹(x) is [-π/2, π/2] and the range of cos⁻¹(x) is [0, π], and they also forget to check the principal values of tan⁻¹(x) which is (-π/2, π/2), this can lead to incorrect answers, especially when dealing with expressions like s∈⁻¹(s∈(x)) or cos⁻¹(cos(x))

👁️ Ayush's Note

When dealing with inverse trigonometric functions, always check the doma∈ and range, and make sure to use the principal values, also, try to simplify the expression as much as possible before applying the inverse trigonometric function, and use the identities to simplify the expression, like s∈⁻¹(x) + cos⁻¹(x) = π/2, and tan⁻¹(x) + tan⁻¹(1/x) = π/2, x > 0

🔁 Last 5 Minutes Box

Make sure to check the doma∈ and range of the inverse trigonometric functions
Use the principal values of s∈⁻¹(x), cos⁻¹(x), and tan⁻¹(x)
Simplify the expression as much as possible before applying the inverse trigonometric function
Use the identities to simplify the expression
Check your work and make sure you have the correct answer

📝 Practice MCQs

1. What is the value of s∈⁻¹(s∈(3π/4))?

A) π/4

B) 3π/4

C) -π/4

D) π/2

Answer: C) -π/4, because s∈⁻¹(x) has a range of [-π/2, π/2] and s∈(3π/4) = s∈(π - 3π/4) = s∈(-π/4)

2. What is the value of cos⁻¹(cos(5π/6))?

A) 5π/6

B) π/6

C) -5π/6

D) -π/6

Answer: B) π/6, because cos⁻¹(x) has a range of [0, π] and cos(5π/6) = cos(π - 5π/6) = cos(π/6)

3. What is the value of tan⁻¹(tan(7π/6))?

A) 7π/6

B) π/6

C) -π/6

D) -7π/6

Answer: D) -7π/6, because tan⁻¹(x) has a range of (-π/2, π/2) and tan(7π/6) = tan(π + π/6) = tan(-π/6)

4. What is the value of d/dx (tan⁻¹(x²))?

A) 2x/(1 + x⁴)

B) 2x/(1 - x⁴)

C) x/(1 + x²)

D) x/(1 - x²)

Answer: A) 2x/(1 + x⁴), using the cha∈ rule and the derivative of tan⁻¹(x)

5. What is the value of s∈⁻¹(x) + cos⁻¹(x)?

A) π

B) π/2

C) π/4

D) 3π/4

Answer: A) π, using the identity s∈⁻¹(x) + cos⁻¹(x) = π/2, but since the range of s∈⁻¹(x) is [-π/2, π/2] and the range of cos⁻¹(x) is [0, π], we can adjust to get π

🚀 Ready to Ace Your Exam?

Put your knowledge to the test! Take the free Practice Mock Test now and track your progress against thousands of students.

🎬 Watch video explanations on YouTube →

This post was curated by Jules, Exam Compass Bot, and edited for accuracy by Ayush.

📚 Related Topics

Continue your revision with these related guides: