Inverse Trigonometric Functions Exercise 02
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:

tan^{−1} (x−1/x−2) + tan^{−1} (x+1/x+2) = π/4

tan^{−1} (x−1/x−2) = π/4  tan^{−1} (x+1/x+2)

tan (π/4  tan^{−1} (x+1/x+2)) = x−1/x−2

tan (π/4) tan^{−1} (x+1/x+2) = x−1/x−2

1 tan^{−1} (x+1/x+2) = x−1/x−2

tan^{−1} (x+1/x+2) = x−1/x−2

x+1/x+2 = x−1/x−2

x^22x1 = 0

(x1)(x+1) = 0

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^{−1}x=cos^{−1}(4x^{3}−3x),x∈[1/2,1]
Answer:
Given: 3cos^{−1}x=cos^{−1}(4x^{3}−3x),x∈[1/2,1]
Step 1: Let y = 4x^{3}−3x
Step 2: Substitute y in the given equation
3cos^−1x = cos^{−1}y
Step 3: Apply the formula cos^{−1}x = 2cos^{−1}(x/2 + 1/2)
3cos^{−1}x = 2cos^{−1}(y/2 + 1/2)
Step 4: Simplify the equation
3cos^{−1}x = 2cos^{−1}[(4x^{3}−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^{−1}x = 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^{−1}x = cos^{−1}(4x^{3}−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^{−1}x = cos^{−1}(4x^{3}−3x) is true for x ∈ [1/2, 1].
Question:
Write the function in the simplest form: tan^{−1} 1/√x^{2}−1,∣x∣>1
Answer:
 tan^−1 1/√x^{2}−1
 tan^−1 (1/x) * (1/√x^{2}1)
 tan^−1 (1/x) * (1/x) * (1/√x^{2}1)
 tan^−1 (1/x^{2}) * (1/√x^{2}1)
 tan^−1 (1/x^{2}) * (1/x) * (1/x1)
 tan^−1 (1/x^{3}) * (1/x1)
 tan^−1 (1/x^{3}) / (1/x1)
Question:
Write the function in the simplest form: tan^{−1} √1+x^{2}−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:
 2tan^{−1} 1/2 + tan^{−1} 1/7
 2(tan^{−1} 1/2 + tan^{−1} 1/7)
 2tan^{−1} (1/2 + 1/7)
 2tan^{−1} (7 + 2) / (7 × 2)
 2tan^{−1} (9/7)
 tan^{−1} (2 × 9/7)
 tan^{−1} (18/7)
 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+x^{2}] = tan1/2[sin^{−1} (2x/(1+x^{2}))]
Step 2: Simplify the denominator of the given expression.
cos^−1 1−y^{2}/1+y^{2} = cos^{−1} (1−y^{2}/(1+y^{2}))
Step 3: Substitute the values of x and y in the given expression.
tan1/2[sin^{−1} (2x/(1+x^{2}))]/cos^{−1} (1−y^{2}/(1+y^{2}))
Step 4: Evaluate the expression for the given condition.
tan1/2[sin^{−1} (2x/(1+x^{2}))]/cos^{−1} (1−y^{2}/(1+y^{2})),∣x∣<1,y>0 and xy<1
Answer: 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 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^{1}x=sin^{1}(3x−4x^{3}),x∈[−1/2,1/2]
Answer:

Given, 3sin^{1}x=sin^{1}(3x−4x^{3}),x∈[−1/2,1/2]

We have to prove that 3sin^{1}x=sin^{1}(3x−4x^{3})

Let y=sin^{1}x

Then, x=siny

Substituting x=siny in the given equation, we get 3sin^{1}(siny)=sin^{1}(3siny−4siny^{3})

Simplifying, we get 3y=sin^{1}(3siny−4siny^{3})

Taking both sides to the power of 2, we get 9y^2=3siny−4siny^{3}

Rearranging, we get 3siny−9y^{2}=−4siny^{3}

Taking the square root of both sides, we get siny−3y=−2siny^{3}

Simplifying, we get 3y=siny−2siny^{3}

Therefore, 3sin^{1}x=sin^{1}(3x−4x^{3}),x∈[−1/2,1/2] is proved.
Question:
Find the value of cot(tan^{1}a+cot^{1}a)
Answer:
Step 1: First, use the identity cot(tan^{1}a) = 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/a^{2}
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/√a^{2}−x^{2},∣x∣<a
Answer:
tan^{1} (x/√a^{2}−x^{2})
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}((1cosx)/(1+cosx)),x<π
Step 3: tan^{1}((12cos^2x+1)/(2cosx)),x<π
Step 4: tan^{1}(2cos^2x1)/(2cosx),x<π
Step 5: tan^{1}(2cosx(cosx1))/(2cosx),x<π
Step 6: tan^{1}(cosx1),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}−x^{3}/a^{3}−3ax^{2}),a>0;−a/√3≤x≤a/√3
Answer:

Simplify the denominator: a^3−3ax^2 = a(a^{2}  3x^{2})

Rewrite the function as: tan^{1}((3a^{2x}−x^{3})/a(a^{2}  3x^{2}))

Factor out an a from the numerator: tan^{1}(a(3x  x^{3})/a(a^{2}  3x^{2}))

Simplify the numerator: tan^{1}(a(3x  x^{3})/a(a  x)(a + x))

Factor out an a from the denominator: tan^{1}((3x  x^{3})/(a  x)(a + x))

Simplify the fraction: tan^{1}((3x  x^{3})/(a^{2}  x^{2}))

Rewrite the function in its simplest form: tan^{1}((3x  x^{3})/(a^{2}  x^{2})), a > 0; a/√3 ≤ x ≤ a/√3