Inverse Trigonometric Functions Miscellaneous Exercise
Question:
Prove: cos^{1} 12/13+sin^{1} 3/5=sin^{1} 56/65
Answer:

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

sin^{1} 56/65 = (180/π) * sin^{1} (56/65) = (180/π) * (2/5) = 72

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)/(xy) )
Step 2: Use the inverse tangent value to solve for the answer: tan^{1} ( (x+y)/(xy) ) = π/4
Answer: C. π/4
Question:
Prove: sin^{1} 8/17+sin^{1} 3/5=tan^{1} 77/36
Answer:

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

sin^{1} (8/17 + 3/5) = sin^{1} (77/85) (Simplifying 8/17 + 3/5)

sin^{1} (77/85) = tan^{1} (77/36) (Using the identity sin^{1} x = tan^{1} (x/√(1x^{2})))

Therefore, sin^{1} 8/17+sin^{1} 3/5=tan^{1} 77/36 (Proved)
Question:
Solve : sin^{1}(1−x)−2sin^{1}x=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^{1}x=π/2
Step 2: sin^{1}(1−x) = π/2 + 2sin^{1}x
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:

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

Therefore, tan^{1}(tan7π/6) = 7π/6
Question:
Prove: 2sin^{1} 3/5=tan^{1} 24/7
Answer:

Use the identity sin^{1} x = tan^{1} (x/√(1−x^{2}))

2sin^{1} 3/5 = 2tan^{1} (3/5/√(1−(3/5)^{2}))

2tan^{1} (3/5/√(1−(3/5)^{2})) = 2tan^{1} (24/7/√(1−(24/7)^{2}))

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}))

Use the identity tan^{1} x = tan^{1} (x/√(1+x^{2}))

2tan^{1} (6/7/√(1−(6/7)^2)) = 2tan^{1} (24/7/√(1+ (24/7)^{2}))

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}))
 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:

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

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)))

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)))

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))

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))

We can now use the identity sin^{2} x + cos^{2} x = 1 to simplify the equation: tan^{1}((sinx√1+sinx)/(√1+sinx/√1−sinx))=tan^{1}((sinx√1+sinx)/(cosx))

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))

Finally, we can use the identity tan (2x) = 2 tan x/(1tan^{2} x) to simplify the equation: tan^{1}(2sin(x/2)cos(x/2))=tan^{1}(2tan(x/2)/(1tan^{2}(x/2)))

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:

Let x = cos2θ

Substitute x = cos2θ into the given equation:
tan^{1}((√1+cos2θ−√1−cos2θ)/(√1+cos2θ+√1−cos2θ)=π/4−1/2cos^{1}cos2θ,−1/√2≤cos2θ≤1
 Simplify:
tan^{1}((2sinθ)/(2cosθ))=π/4−1/2cos^{1}cos2θ,−1/√2≤cos2θ≤1
 Use the identity tan^{1}(a/b) = tan^{1}(a) − tan^{1}(b):
tan^{1}(sinθ)−tan^{1}(cosθ)=π/4−1/2cos^{1}cos2θ,−1/√2≤cos2θ≤1
 Use the identity tan^{1}(a) = a for a small angle:
sinθ−cosθ=π/4−1/2cos^{1}cos2θ,−1/√2≤cos2θ≤1
 Simplify:
sinθ=π/4−1/2cos^{1}cos2θ+cosθ,−1/√2≤cos2θ≤1
 Use the identity sinθ = 2sin(θ/2)cos(θ/2):
2sin(θ/2)cos(θ/2)=π/4−1/2cos^{1}cos2θ+cosθ,−1/√2≤cos2θ≤1
 Simplify:
sin(θ/2)=π/4−1/2cos^{1}cos2θ+cos(θ/2),−1/√2≤cos2θ≤1
 Use the identity cos(θ/2) = √(1+cosθ)/2:
sin(θ/2)=π/4−1/2cos^{1}cos2θ+√(1+cosθ)/2,−1/√2≤cos2θ≤1
 Simplify:
sin(θ/2)=π/4−1/2cos^{1}cos2θ+√(1+cos2θ)/2,−1/√2≤cos2θ≤1
 Prove:
tan^{1}((√1+x−√1−x)/(√1+x+√1−x)=π/4−1/2cos^{1}x,−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−x^{2} B 1/√1−x^{2} C 1/√1+x^{2} D x/√1+x^{2}
Answer:
Answer: A x/√1−x^{2}
Question:
Prove: 9π/8−9/4sin^{1} 1/3=9/4sin^{1} 2√2/3
Answer:

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

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

Simplify: 9π/8  3π/8

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

Simplify: 9/4(π/3) = 9/4sin^{1}(2√2/3)

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^{1}x,(x>0)
Answer:
 tan^{1}(1−x)/(1+x) = 1/2tan^{1x}
 (1−x)/(1+x) = 1/2x
 (1−x) = 1/2x(1+x)
 1−x = 1/2x^2 + 1/2x
 1−x = 1/2x(x+1)
 1 = 1/2x^2 + 1/2x + x
 2 = x^{2} + x + 2
 0 = x^{2} + x  1
 0 = (x+1)(x1)
 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:
 cos^{1} 4/5+cos^{1} 12/13
 cos^{1} (4/5 + 12/13) (Using the inverse cosine addition formula)
 cos^{1} (16/65 + 156/65) (Simplifying 4/5 + 12/13)
 cos^{1} (172/65) (Adding 16/65 and 156/65)
 cos^{1} 33/65 (Simplifying 172/65)
Therefore, cos^{1} 4/5+cos^{1} 12/13=cos^{1} 33/65