Question 1

The value of sin⁻¹(sin(5π/6)) is:

Accepted Answer

Step 1: Note the principal value range of sin⁻¹ is [−π/2, π/2].
Step 2: sin(5π/6) = sin(π − π/6) = sin(π/6) = 1/2.
Step 3: Now we need sin⁻¹(1/2). Since 1/2 is in [−1,1] and π/6 ∈ [−π/2, π/2], we get sin⁻¹(1/2) = π/6.
Step 4: Therefore sin⁻¹(sin(5π/6)) = π/6.
Key Rule: sin⁻¹(sin θ) = θ only when θ ∈ [−π/2, π/2]. Otherwise, we first simplify sin θ to a value whose arcsine lies in the principal range.

Question 2

If sin⁻¹(x) + sin⁻¹(y) = π/2, where x, y ∈ [0, 1], then which of the following is correct?

Accepted Answer

We are given sin⁻¹(x) + sin⁻¹(y) = π/2.

Let sin⁻¹(x) = α and sin⁻¹(y) = β, where α, β ∈ [0, π/2] (principal range of sin⁻¹).

Then α + β = π/2, which means β = π/2 − α.

Taking sine of both sides: sin(β) = sin(π/2 − α) = cos(α).

Since sin(β) = y and sin(α) = x, we have:
y = cos(α) = cos(sin⁻¹(x)).

Using the identity cos(sin⁻¹(x)) = √(1 − x²) for x ∈ [0, 1]:
y = √(1 − x²).

Alternatively, squaring: x² + y² = x² + (1 − x²) = 1.

Therefore, x² + y² = 1.

Question 3

The value of tan⁻¹(1) + tan⁻¹(2) + tan⁻¹(3) is:

Accepted Answer

We use the addition formula for inverse tangent:
tan⁻¹(a) + tan⁻¹(b) = tan⁻¹((a + b)/(1 − ab)) when ab < 1,
tan⁻¹(a) + tan⁻¹(b) = π + tan⁻¹((a + b)/(1 − ab)) when ab > 1 and a, b > 0.

Step 1: Compute tan⁻¹(1) + tan⁻¹(2).
Here a = 1, b = 2, so ab = 2 > 1.
tan⁻¹(1) + tan⁻¹(2) = π + tan⁻¹((1 + 2)/(1 − 1·2)) = π + tan⁻¹(3/(−1)) = π + tan⁻¹(−3) = π − tan⁻¹(3).

Step 2: Now add tan⁻¹(3):
tan⁻¹(1) + tan⁻¹(2) + tan⁻¹(3) = [π − tan⁻¹(3)] + tan⁻¹(3) = π.

Therefore, the answer is π.

Question 4

If cos⁻¹(x) − sin⁻¹(x) = π/6, where x ∈ [0, 1], then x equals:

Accepted Answer

We use the identity sin⁻¹(x) + cos⁻¹(x) = π/2 for x ∈ [0, 1].

Let sin⁻¹(x) = α. Then cos⁻¹(x) = π/2 − α.

Given: cos⁻¹(x) − sin⁻¹(x) = π/6.

Substitute: (π/2 − α) − α = π/6.

Simplify: π/2 − 2α = π/6.

Solve for α: 2α = π/2 − π/6 = 3π/6 − π/6 = 2π/6 = π/3.

Therefore: α = π/6, which means sin⁻¹(x) = π/6.

Taking sine: x = sin(π/6) = 1/2.

Verification: cos⁻¹(1/2) = π/3, sin⁻¹(1/2) = π/6.
cos⁻¹(1/2) − sin⁻¹(1/2) = π/3 − π/6 = 2π/6 − π/6 = π/6. ✓

Question 5

The principal value of sin⁻¹(sin(7π/6)) is:

Accepted Answer

The principal value of sin⁻¹ has range [−π/2, π/2].

We need to find the angle in [−π/2, π/2] whose sine equals sin(7π/6).

Step 1: Compute sin(7π/6).
7π/6 is in the third quadrant (between π and 3π/2).
sin(7π/6) = sin(π + π/6) = −sin(π/6) = −1/2.

Step 2: Find the angle in [−π/2, π/2] whose sine is −1/2.
In the range [−π/2, π/2], the angle with sine −1/2 is −π/6.

Verification: sin(−π/6) = −sin(π/6) = −1/2. ✓
Also, −π/6 ∈ [−π/2, π/2]. ✓

Therefore, sin⁻¹(sin(7π/6)) = −π/6.

Question 6

If tan⁻¹(x) + tan⁻¹(y) = π/4 and xy < 1, then x + y equals:

Accepted Answer

We use the addition formula for inverse tangent:
tan⁻¹(x) + tan⁻¹(y) = tan⁻¹((x + y)/(1 − xy)) when xy < 1.

Given: tan⁻¹(x) + tan⁻¹(y) = π/4 and xy < 1.

Apply the formula:
tan⁻¹((x + y)/(1 − xy)) = π/4.

Take tangent of both sides:
(x + y)/(1 − xy) = tan(π/4) = 1.

Solve for x + y:
x + y = 1 · (1 − xy) = 1 − xy.

Rearrange:
x + y + xy = 1.

Alternatively, from (x + y)/(1 − xy) = 1:
x + y = 1 − xy.

Therefore: x + y = 1 − xy, or equivalently, x + y + xy = 1.

Question 7

Find the value of sin⁻¹(sin(7π/6)).

Accepted Answer

The inverse sine function sin⁻¹ has range [-π/2, π/2]. We cannot directly substitute 7π/6 into sin⁻¹(sin x) = x because 7π/6 lies outside this range.

Step 1: Find sin(7π/6).
7π/6 is in the third quadrant (between π and 3π/2).
sin(7π/6) = sin(π + π/6) = -sin(π/6) = -1/2.

Step 2: Apply sin⁻¹ to the result.
sin⁻¹(-1/2) = -π/6 (since -π/6 ∈ [-π/2, π/2] and sin(-π/6) = -1/2).

Therefore, sin⁻¹(sin(7π/6)) = -π/6.

Key theorem: sin⁻¹(sin x) = x only if x ∈ [-π/2, π/2]. For angles outside this range, we must first reduce to an equivalent angle within the principal range using periodicity and symmetry properties.

Question 8

If tan⁻¹(x) + tan⁻¹(y) = π/4, where x, y > 0, then xy + x + y equals:

Accepted Answer

We use the addition formula for inverse tangent:
tan⁻¹(x) + tan⁻¹(y) = tan⁻¹((x + y)/(1 − xy)) when xy < 1.

Step 1: Apply the formula.
tan⁻¹((x + y)/(1 − xy)) = π/4.

Step 2: Take tangent of both sides.
(x + y)/(1 − xy) = tan(π/4) = 1.

Step 3: Solve for the relationship.
x + y = 1 − xy.

Step 4: Rearrange.
xy + x + y = 1.

Key theorem: tan⁻¹(a) + tan⁻¹(b) = tan⁻¹((a+b)/(1−ab)) + nπ, where n depends on the range. For positive x, y with xy < 1, the formula holds directly without the nπ term.

Question 9

The value of cos⁻¹(cos(−2π/3)) is:

Accepted Answer

The inverse cosine function cos⁻¹ has range [0, π]. We cannot directly substitute −2π/3 because it lies outside this range.

Step 1: Use the property cos(−θ) = cos(θ).
cos(−2π/3) = cos(2π/3).

Step 2: Check if 2π/3 is in the range [0, π].
Yes, 2π/3 ∈ [0, π].

Step 3: Apply the inverse cosine.
cos⁻¹(cos(2π/3)) = 2π/3 (since 2π/3 is already in the principal range).

Step 4: Verify.
cos(2π/3) = −1/2, and cos⁻¹(−1/2) = 2π/3 ✓.

Key theorem: cos⁻¹(cos x) = x if and only if x ∈ [0, π]. For angles outside this range, use the even property of cosine (cos(−θ) = cos(θ)) or periodicity to reduce to the principal range.

Question 10

If sin⁻¹(a) + cos⁻¹(a) = π/2 for a ∈ [−1, 1], then a equals:

Accepted Answer

We use the fundamental identity relating inverse sine and inverse cosine.

Step 1: Recall the identity.
For all a ∈ [−1, 1]: sin⁻¹(a) + cos⁻¹(a) = π/2.

Step 2: Verify the given equation.
The equation sin⁻¹(a) + cos⁻¹(a) = π/2 is true for ALL values of a in the domain [−1, 1].

Step 3: Interpret the question.
The question asks: for which value(s) of a is this equation satisfied? Since it is satisfied for all a ∈ [−1, 1], the answer is "all real numbers in [−1, 1]."

However, if the question intends to test whether students know this is an identity, the most reasonable interpretation is that any specific value in [−1, 1] satisfies it. Among the options, we need to identify which is a valid element of [−1, 1].

Key theorem: sin⁻¹(x) + cos⁻¹(x) = π/2 for all x ∈ [−1, 1]. This is a fundamental identity, not a conditional equation.

Question 11

Find the value of tan(sin⁻¹(3/5)).

Accepted Answer

We need to find tan of an angle whose sine is 3/5.

Step 1: Let θ = sin⁻¹(3/5).
Then sin(θ) = 3/5 and θ ∈ [−π/2, π/2].

Step 2: Since sin⁻¹ returns values in [−π/2, π/2] and 3/5 > 0, we have θ ∈ (0, π/2), so θ is in the first quadrant where all trig functions are positive.

Step 3: Find cos(θ) using the Pythagorean identity.
sin²(θ) + cos²(θ) = 1.
(3/5)² + cos²(θ) = 1.
9/25 + cos²(θ) = 1.
cos²(θ) = 16/25.
cos(θ) = 4/5 (positive because θ is in the first quadrant).

Step 4: Calculate tan(θ).
tan(θ) = sin(θ)/cos(θ) = (3/5)/(4/5) = 3/4.

Key theorem: For θ = sin⁻¹(x) with x ∈ [−1, 1], we have sin(θ) = x and θ ∈ [−π/2, π/2]. Use the Pythagorean identity to find cos(θ), then compute other trig ratios.

Question 12

If tan⁻¹(x) + tan⁻¹(y) = π/4, where x, y > 0, then find the value of x + y + xy.

Accepted Answer

We use the addition formula for inverse tangent:
tan⁻¹(x) + tan⁻¹(y) = tan⁻¹((x + y)/(1 - xy)) when xy < 1.

Step 1: Given tan⁻¹(x) + tan⁻¹(y) = π/4.

Step 2: Apply tan to both sides:
tan(tan⁻¹(x) + tan⁻¹(y)) = tan(π/4) = 1.

Step 3: Use the tangent addition formula:
tan(A + B) = (tan A + tan B)/(1 - tan A · tan B).

Here, tan(tan⁻¹(x)) = x and tan(tan⁻¹(y)) = y, so:
(x + y)/(1 - xy) = 1.

Step 4: Solve for the relationship:
x + y = 1 - xy.

Step 5: Rearrange:
x + y + xy = 1.

Therefore, x + y + xy = 1.

Question 13

Evaluate cos⁻¹(cos(2π/3)).

Accepted Answer

The inverse cosine function cos⁻¹ has range [0, π]. We must check if the given angle is in this range.

Step 1: Identify the angle. 2π/3 is in the interval [0, π], so it is already in the principal range of cos⁻¹.

Step 2: Apply the property. Since 2π/3 ∈ [0, π], we have cos⁻¹(cos(2π/3)) = 2π/3 directly.

Step 3: Verify. cos(2π/3) = -1/2 (second quadrant cosine is negative). And cos⁻¹(-1/2) = 2π/3 (the unique angle in [0, π] with cosine -1/2).

Therefore, cos⁻¹(cos(2π/3)) = 2π/3.

Question 14

If sin⁻¹(a) + cos⁻¹(a) = π/2 for a ∈ [0, 1], then which of the following is true?

Accepted Answer

We use the fundamental identity relating inverse sine and inverse cosine.

Step 1: Recall the identity: sin⁻¹(x) + cos⁻¹(x) = π/2 for all x ∈ [-1, 1].

Step 2: This identity holds because if sin⁻¹(x) = θ, then sin(θ) = x where θ ∈ [-π/2, π/2].

Step 3: Since sin(θ) = x, we have cos(π/2 - θ) = x (using the co-function identity).

Step 4: Since θ ∈ [-π/2, π/2], we have π/2 - θ ∈ [0, π], which is the range of cos⁻¹.

Step 5: Therefore, cos⁻¹(x) = π/2 - θ = π/2 - sin⁻¹(x).

Step 6: Rearranging: sin⁻¹(x) + cos⁻¹(x) = π/2.

This identity is true for all a ∈ [0, 1] (and in fact for all a ∈ [-1, 1]).

Question 15

Find the value of tan⁻¹(1) + tan⁻¹(2) + tan⁻¹(3).

Accepted Answer

We use the tangent addition formula for inverse tangent, applied iteratively.

Step 1: Compute tan⁻¹(1) + tan⁻¹(2) using tan⁻¹(x) + tan⁻¹(y) = tan⁻¹((x+y)/(1-xy)) when xy < 1.

Here, x = 1, y = 2, so xy = 2 > 1. This means we must use the modified formula:
tan⁻¹(x) + tan⁻¹(y) = tan⁻¹((x+y)/(1-xy)) + π (when xy > 1 and x, y > 0).

Step 2: Compute (1+2)/(1-1·2) = 3/(1-2) = 3/(-1) = -3.

So tan⁻¹(1) + tan⁻¹(2) = tan⁻¹(-3) + π.

Step 3: Now add tan⁻¹(3):
tan⁻¹(-3) + π + tan⁻¹(3) = tan⁻¹(-3) + tan⁻¹(3) + π.

Step 4: Use the identity tan⁻¹(x) + tan⁻¹(-x) = 0:
tan⁻¹(3) + tan⁻¹(-3) = 0.

Step 5: Therefore:
tan⁻¹(1) + tan⁻¹(2) + tan⁻¹(3) = 0 + π = π.

Alternatively, note that tan(π) is undefined, but we can verify: tan⁻¹(1) = π/4, and tan⁻¹(2) + tan⁻¹(3) should sum to 3π/4.

Question 16

If tan⁻¹(x) + tan⁻¹(y) = π/4, where x, y > 0, then the value of x + y + xy is:

Accepted Answer

We use the addition formula for inverse tangent:
tan⁻¹(x) + tan⁻¹(y) = tan⁻¹((x+y)/(1-xy)) when xy < 1.

Given: tan⁻¹(x) + tan⁻¹(y) = π/4

Step 1: Apply the addition formula.
tan⁻¹((x+y)/(1-xy)) = π/4

Step 2: Take the tangent of both sides.
(x+y)/(1-xy) = tan(π/4) = 1

Step 3: Solve for the relationship between x and y.
x + y = 1 - xy
x + y + xy = 1

Therefore: x + y + xy = 1.

Question 17

If cos⁻¹(x) - sin⁻¹(x) = π/6, then the value of x is:

Accepted Answer

We use two key identities:
1. sin⁻¹(x) + cos⁻¹(x) = π/2 for all x ∈ [-1, 1].
2. This allows us to express cos⁻¹(x) = π/2 - sin⁻¹(x).

Step 1: Substitute the identity into the given equation.
cos⁻¹(x) - sin⁻¹(x) = π/6
(π/2 - sin⁻¹(x)) - sin⁻¹(x) = π/6

Step 2: Simplify.
π/2 - 2sin⁻¹(x) = π/6

Step 3: Solve for sin⁻¹(x).
-2sin⁻¹(x) = π/6 - π/2 = π/6 - 3π/6 = -2π/6 = -π/3
sin⁻¹(x) = π/6

Step 4: Find x.
If sin⁻¹(x) = π/6, then x = sin(π/6) = 1/2.

Question 18

If sin⁻¹(x) + sin⁻¹(y) = π/3 and cos⁻¹(x) + cos⁻¹(y) = 2π/3, then the value of x² + y² is:

Accepted Answer

We use the fundamental identities: sin⁻¹(a) + cos⁻¹(a) = π/2 for any a ∈ [−1, 1].

Given:
  sin⁻¹(x) + sin⁻¹(y) = π/3  ... (1)
  cos⁻¹(x) + cos⁻¹(y) = 2π/3  ... (2)

From the identity sin⁻¹(x) + cos⁻¹(x) = π/2, we have:
  cos⁻¹(x) = π/2 − sin⁻¹(x)
  cos⁻¹(y) = π/2 − sin⁻¹(y)

Substituting into equation (2):
  (π/2 − sin⁻¹(x)) + (π/2 − sin⁻¹(y)) = 2π/3
  π − (sin⁻¹(x) + sin⁻¹(y)) = 2π/3
  π − π/3 = 2π/3  ✓ (Consistent)

Now, let sin⁻¹(x) = α and sin⁻¹(y) = β, where α + β = π/3.

Then x = sin(α) and y = sin(β).

We need: x² + y² = sin²(α) + sin²(β)

Using the constraint α + β = π/3, we have β = π/3 − α.

Therefore:
  x² + y² = sin²(α) + sin²(π/3 − α)

Expanding sin(π/3 − α) = sin(π/3)cos(α) − cos(π/3)sin(α) = (√3/2)cos(α) − (1/2)sin(α):

sin²(π/3 − α) = [(√3/2)cos(α) − (1/2)sin(α)]²
              = (3/4)cos²(α) − (√3/2)sin(α)cos(α) + (1/4)sin²(α)

Therefore:
  x² + y² = sin²(α) + (3/4)cos²(α) − (√3/2)sin(α)cos(α) + (1/4)sin²(α)
          = (5/4)sin²(α) + (3/4)cos²(α) − (√3/2)sin(α)cos(α)
          = (5/4)sin²(α) + (3/4)(1 − sin²(α)) − (√3/2)sin(α)cos(α)
          = (5/4)sin²(α) + 3/4 − (3/4)sin²(α) − (√3/2)sin(α)cos(α)
          = (1/2)sin²(α) + 3/4 − (√3/2)sin(α)cos(α)

Using sin(2α) = 2sin(α)cos(α), so sin(α)cos(α) = sin(2α)/2:
  x² + y² = (1/2)sin²(α) + 3/4 − (√3/4)sin(2α)

Using sin²(α) = (1 − cos(2α))/2:
  x² + y² = (1/4)(1 − cos(2α)) + 3/4 − (√3/4)sin(2α)
          = 1/4 − (1/4)cos(2α) + 3/4 − (√3/4)sin(2α)
          = 1 − (1/4)cos(2α) − (√3/4)sin(2α)
          = 1 − (1/4)[cos(2α) + √3·sin(2α)]
          = 1 − (1/2)[(1/2)cos(2α) + (√3/2)sin(2α)]
          = 1 − (1/2)sin(2α + π/6)

Since α ∈ [0, π/2] and β = π/3 − α ∈ [0, π/2], we have α ∈ [0, π/3].
Thus 2α ∈ [0, 2π/3], so 2α + π/6 ∈ [π/6, 5π/6].

The maximum of sin(2α + π/6) in this range is 1 (at 2α + π/6 = π/2, i.e., α = π/6).
The minimum is sin(π/6) = 1/2 (at α = 0) or sin(5π/6) = 1/2 (at α = π/3).

At α = π/6 (and β = π/6): x² + y² = 1 − (1/2)(1) = 1/2.

Question 19

The value of tan⁻¹(1/2) + tan⁻¹(1/3) is equal to:

Accepted Answer

We use the addition formula for inverse tangent:
  tan⁻¹(a) + tan⁻¹(b) = tan⁻¹((a + b)/(1 − ab))  when ab < 1.

Here, a = 1/2 and b = 1/3.

First, check: ab = (1/2)(1/3) = 1/6 < 1  ✓

Therefore:
  tan⁻¹(1/2) + tan⁻¹(1/3) = tan⁻¹((1/2 + 1/3)/(1 − (1/2)(1/3)))
                            = tan⁻¹((3/6 + 2/6)/(1 − 1/6))
                            = tan⁻¹((5/6)/(5/6))
                            = tan⁻¹(1)
                            = π/4

Verification: tan(π/4) = 1, and the principal value of tan⁻¹(1) is π/4 ∈ (−π/2, π/2). ✓

Question 20

If cos⁻¹(x) − sin⁻¹(x) = π/6, then x is equal to:

Accepted Answer

We use the identity: sin⁻¹(x) + cos⁻¹(x) = π/2 for all x ∈ [−1, 1].

Given: cos⁻¹(x) − sin⁻¹(x) = π/6  ... (1)

From the identity: cos⁻¹(x) = π/2 − sin⁻¹(x)  ... (2)

Substituting (2) into (1):
  (π/2 − sin⁻¹(x)) − sin⁻¹(x) = π/6
  π/2 − 2sin⁻¹(x) = π/6
  −2sin⁻¹(x) = π/6 − π/2
  −2sin⁻¹(x) = π/6 − 3π/6
  −2sin⁻¹(x) = −2π/6
  −2sin⁻¹(x) = −π/3
  sin⁻¹(x) = π/6

Therefore: x = sin(π/6) = 1/2

Verification:
  sin⁻¹(1/2) = π/6
  cos⁻¹(1/2) = π/3
  cos⁻¹(1/2) − sin⁻¹(1/2) = π/3 − π/6 = 2π/6 − π/6 = π/6  ✓

Question 21

The domain of the function f(x) = sin⁻¹(2x − 1) is:

Accepted Answer

The function sin⁻¹(u) is defined only when u ∈ [−1, 1].

For f(x) = sin⁻¹(2x − 1) to be defined, we require:
  −1 ≤ 2x − 1 ≤ 1

Solving the left inequality:
  −1 ≤ 2x − 1
  −1 + 1 ≤ 2x
  0 ≤ 2x
  0 ≤ x

Solving the right inequality:
  2x − 1 ≤ 1
  2x ≤ 2
  x ≤ 1

Combining both: 0 ≤ x ≤ 1

Therefore, the domain is [0, 1].

NDA Inverse Trig Functions

NDA Inverse Trig Functions — Past Exam Questions

Inverse Trig Functions— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Inverse Trig Functions — Practice Questions

60-Second Revision — Inverse Trig Functions