SOLVING INVERSE TRIGONOMETRIC EQUATIONS PROBLEMS

Problem 1 :

sin^-1(cos x) = π/2 - x is valid for 

1)  -π ≤ x ≤ 0   b)  0 ≤ x ≤ π    3) -π/2 ≤ x ≤ π/2

4)  -π/4 ≤ x ≤ 3π/4

Solution :

sin^-1(cos x) = π/2 - x

cos x = sin(π/2 - x)

cos x = cos x

So, it is valid for 0 ≤ x ≤ π.

Problem 2 :

If cot^-1 x = 2π/5 for some x ∈ R, the value of tan^-1 x is 

1)  -π/10   b)  π/5    3) π/10     4)  -π/5

Solution :

cot^-1 x = 2π/5 ----(1)

We know that tan^-1 x + cot^-1 x = π/2

cot^-1 x = π/2 - tan^-1 x ----(2)

(1) = (2)

π/2 - tan^-1 x = 2π/5

π/2 - 2π/5 = tan^-1 x

(5π - 4π)/10 = tan^-1 x

tan^-1 x = π/10

Problem 3 :

The domain of the function is defined by f(x) = sin^-1 √(x -1) is 

1)  [1, 2]   b)  [-1, 1]    3) [0, 1]     4)  [-1, 0]

Solution :

-1 ≤ sin^-1 x ≤ 1

Since it is square root function negative values can be ignored.

0 ≤ √(x -1) x ≤ 1

By combining both, we get [1, 2]. So, the required domain is [1, 2].

Problem 4 :

If x = 1/5, the value of cos (cos^-1x + 2 sin^-1x) is 

1)  √(24/25)    b)  √24/25    3) (1/5)    4) (-1/5)

Solution :

cos (cos^-1x + 2 sin^-1x)

Problem 5 :

tan^-1(1/4) + tan^-1(2/9) is equal to

1)  √(24/25)    b)  √24/25    3) (1/2) tan^-1 (3/5)     4) tan^-1 (1/2)

Solution :

Problem 6 :

If the function f(x) = sin^-1(x² - 3), then x belongs to

1)  [-1, 1]    b)  [√2, 2]    3) [-2, -√2] U [√2, 2]     4) [-2, -√2]

Solution :

-1 ≤ sin^-1 x ≤ 1

-1 ≤ (x² - 3) ≤ 1

-1 + 3 ≤ x²  ≤ 1 + 3

2 ≤ x²  ≤ 4

√2 ≤ √x²  ≤ √4

√2 ≤ √x ≤ 2

So, the solution is [-2, -√2] U [√2, 2].

Problem 7 :

If cot^-1 2 and cot^-1 3 are two angles of a triangle, then the third angle is 

1)  π/4   b)  3π/4    3) π/6     4)  π/3

Solution :

Problem 8 :

If sin^-1(tan π/4) - sin^-1√(3/x) = π/6. Then x is a root of the equation.

1) x² - x - 6 = 0   2)  x² - x - 12 = 0    3) x² + x - 12 = 0

     4)  x² + x - 6 = 0

Solution :

sin^-1(tan π/4) - sin^-1√(3/x) = π/6

sin^-11 - sin^-1√(3/x) = π/6

1) x² - x - 6 = 0

(x - 3)(x + 2) = 0

x = 3, x = -2

b)  x² - x - 12 = 0

(x - 4)(x + 3) = 0

x = 4, x = -3

3) x² + x - 12 = 0

(x + 4)(x - 3) = 0

x = -4, x = 3

4)  x² + x - 6 = 0

(x + 3)(x - 2) = 0

x = -3, x = 2

So, the answer is option 2)

Problem 9 :

Solution :

Problem 10 :

Solution :

Problem 11 :

If |x| ⩽1, then

Solution :

Problem 12 :

The equation

1) no solution    2) Unique solution

3) two solution      4) infinite number of solutions

Solution :

Problem 13 :

Solution :

Problem 14 :

Solution :

Problem 15 :

sin(tan^-1x), |x| < 1 is equal to

Solution :

Opposite side = x, adjacent side = 1

Hypotenuse = √x²+1²

sin(sin^-1(x/√(x²+1²))

= x/√(x²+1)

So, the answer is 4).