Inverse Trigonometric Functions Important Questions

---

Very Short Answer Questions (1 mark)

Ques. Find the value of \(sin^{-1}\bigg(sin \frac{π}{4}\bigg)\)?

Ans. From identity \(sin^{-1}(sin \;x) = x\), we have

∴ \(sin^{-1}\bigg(sin \frac{π}{4}\bigg) = \frac{π}{4}\)

Ques. How many inverse trigonometric functions are used?

Ans. There are six inverse trigonometric functions that are used in mathematics.

Ques. What is the range of arcsin function?

Ans. The range of arcsin function is \(\bigg[-\frac{π}{2},\frac{π}{2}\bigg]\).

Ques. What is the domain of arcsec function?

Ans. The domain of arcsec function is (-∞ ,-1] ∪ [1,∞ ).

Ques. Find the value of x for sin(x) = 6?

Ans. Given, sin (x) = 6

⇒ x =sin^-1(6), which is not possible.

∴ There is no required value of x for which sin x = 6, so the domain of sin^-1x is -1 to 1 for the required values of x.

Ques. Determine the principal value of tan^-1 (-√3)?

Ans. The principal value of tan-1(-√3) is -π/3.

Ques. Find the value of \(sin^{-1}\bigg(sin \frac{π}{3}\bigg)\)?

Ans. From identity sin^-1(sin x) = x, we have

∴ \(sin^{-1}\bigg(sin \frac{π}{3}\bigg) = \frac{π}{3}\)

---

Short Answer Question (2 marks)

Ques. Determine the value of tan^-1(√3) - cot^-1(-√3)?

Ans. Using the inverse trigonometric functions:

⇒ tan^-1(√3) - cot^-1(-√3)

⇒ tan^-1(√3) - (π - cot^-1(√3))

⇒ tan^-1(√3) - π + cot^-1(√3)

⇒ \(\frac{π}{3} - π + \frac{π}{6}\)

⇒ \(\frac{π}{2} - π\)

∴ required value is \(-\frac{π}{2}\)

Ques. What are the formulas for complementary functions in inverse trigonometric functions?

Ans. The formulas for complementary functions in inverse trigonometric functions are as follows:

• sin^-1x = cosec^-11/x, x ∈ R - (-1,1)
• cos^-1x = sec^-11/x, x ∈ R - (-1,1)
• tan^-1x = cot^-11/x, x > 0

Ques. Determine the value of sin^-1(1/2) - sec^-1(-2)?

Ans. Using the inverse trigonometric functions:

⇒ sin^-1 (1/2) - sec^-1(-2) = \(\frac{π}{6} - (π - sec^{-1}2)\)

 ⇒ π/6 - (π - π/3) (Since the range of cos⁻¹(x) is [0,π])

⇒ π/6 - π + π/3

⇒ π/6 + π/3 - π

⇒ π/2 - π

∴ required value is -π/2

Ques. Determine the value of the value of tan^-1(1) + cos^-1(-1/2) + sin^-1(-1/2)?

Ans. Using the inverse trigonometric functions:

⇒ tan^-1(1) + cos^-1(-1/2) + sin^-1(-1/2)

⇒ π/4 + π - cos^-1(1/2) - sin^-1(1/2)

⇒ π/4 + π - π/3 - π/6

⇒ π/4 + π - π/2

⇒ π/4 + π/2

∴ required value is 3π/4

Ques. Determine the value of sin (cos^-1 4/5)?

Ans. Let cos-1 4/5 = x 

⇒ cos x = ⅘

⇒ It is given that sin x = √(1 – cos² x)

⇒ sin x = √(1 – 16/15) = ⅗

∴ sin x = sin (cos-1 4/5) = ⅗

Read More: 

---

Long Answer Question (3 marks)

Ques. Determine value of cot (tan^-1 α + cot^-1 α)?

Ans. Since it is given that cot \((tan^{-1} \alpha + cot^{-1} \alpha)\)

⇒ \(cot (\frac{\pi}{2}) \)(since, tan^-1 x + cot^-1 x = \(\frac{\pi}{2}\))

⇒ cot \((\frac{180°}{2})\) ( as we all know that cot 90° = 0 )

⇒ cot (90°)

∴ required value is 0

Ques. Determine the value of sin (cos^-1 \(\frac{5}{13}\))?

Ans. Let cos^-1 \(\frac{5}{13}\) = x 

⇒ cos x = \(\frac{5}{12}\)

⇒ It is given that sin x = √(1 – cos² x)

⇒ sin x = √(1 – 25/169) 

⇒sin x = sin (cos^-1 12/13) 

∴ required value is 12/13

Ques. What are the sum and difference formulas for inverse trigonometric functions?

Ans. The sum and difference formulas for inverse trigonometric functions are as follows:

• sin^-1x + sin^-1y = sin^-1(x.√(1 - y²) + y√(1 - x²))
• sin^-1x - sin-1y = sin^-1(x.√(1 - y²) - y√(1 - x²))
• cos^-1x + cos^-1y = cos^-1(xy - √(1 - x²).√(1 - y²))
• cos^-1x - cos^-1y = cos^-1(xy + √(1 - x²).√(1 - y²))
• tan^-1x + tan^-1y = tan^-1(x + y)/(1 - xy), if xy < 1
• tan^-1x + tan^-1y = tan^-1(x - y)/(1 + xy), if xy > - 1

Ques. Determine the value of sin-1(sin 7)?

Ans. As it is know that sin x = sin(π – x) 

⇒sin 7 = sin(π – 7) 

⇒ sin(2π + 7) 

⇒ sin(-π – 7) 

⇒ sin(-2π – 7) 

Since sin-1(sin 7)

⇒ sin^-1(sin (-2π + 7)) 

∴ required value is -2π + 7

Ques. Determine the function tan(cos^-1 x)?

Ans. Suppose cos^-1 x = y

⇒ cos y = x where base = x and hypotenuse = 1 

⇒ As a result sin y = √(1 – x²)/1

⇒ tan y = sin y/ cos y

⇒ tan y = √(1 – x²)/x

⇒ y = tan^-1 √(1 – x²)/x

⇒ cos^-1 x = tan-1 √(1 – x²)/x

⇒ tan(cos^-1 x) = tan(tan^-1 √(1 – x²)/x ) 

∴ required value is √(1 – x²)/x.

---

Very Long Answer Question (5 marks)

Ques. Prove that tan^-1 2/11 + tan^-1 7/4 = tan^-1 17/6?

Ans. By using the formula tan-1 x + tan^-1 y = tan^-1 (x + y) / (1 - xy)

⇒ We get, LHS = tan^-1 2/11 + tan^-1 7/24

⇒ tan^-1 [(2/11 + 7/4) / (1 - (2/11). (7/4)]

⇒ tan^-1 [((8 + 77) / 44) / ((44 - 14) / 44)]

⇒ On simplifying the terms,

⇒ tan^-1 (85 / 30)

⇒ tan^-1(17 / 6)

Therefore LHS = RHS

Ques. Find the value of tan (sin^-1 4/5 + cot^-1 3/2)?

Ans. Use the basic trigonometric function of sin x = y can be changed to x = sin-1 y

⇒ Let sin^-1 4/5 = x

⇒ sin x = 4/5

Then, cos x = √ 1 - sin² x = 3/5

⇒ sec x = 5/3

Therefore, tan x = √ sec² x - 1

⇒ values goes as: √ ( 25/9) - 1

⇒ required value is: 4 / 3

⇒ x = tan^-1 4/3

⇒ sin^-1 4 / 5 = tan^-1 4 / 3 ....(1)

⇒ Now, cot^-1 3 / 2 = tan^-1 2 / 3 ....(2)

⇒ By using equation (1) and (2)

⇒ tan (sin^-1 3/5 + cot^-1 2/3) = tan (tan^-1 4/3 + tan^-1 2/3)

⇒ tan [tan^-1 (4/3 + 2/3) / (1 - (4/3).(2/3)]

⇒ tan (tan^-1 6/ 5)

∴ required value is 6 / 5

Ques. Determine the values of tan^-1 1/√3 - cot^-1 (- √3)?

Ans. Use the basic trigonometric function of sinx = y, can be changed to x = sin-1 y

⇒ Consider tan^-1 1/√3 = x

⇒ tan x = 1/√3

⇒ tan π / 6 where π / 6 ∈ (- π/2, π/2)

So, tan^-1 1/√3 = π/6

⇒Let us assume cot^-1 (- √3) = y

Hence, cot y = (- √3)

⇒ x = - cot (π / 6)

⇒ x = cot (π - π / 6)

⇒ x = cot (5π / 6)

⇒ It is given that range of principal value of cot^-1 x = (0, π)

Therefore, cot^-1(- √3) = (5π / 6)

⇒ Then, tan-1 √3 - cot^-1 (- √3)

⇒ required value is π / 6 - 5π / 6

∴ required value is - 2π / 3

Ques. Prove that sin^-1 (4/5) – sin^-1 (6/10) = cos^-1 (48/50)?

Ans. Assume sin^-1 (4/5) = a and sin^-1 (6/10) = b

⇒ We can write sin a = 4/5 and sin b = 6/10

⇒ Now, find the value of cos a and cos b

⇒ First find the value of cos a: Cos a = √[1 – sin² a]

⇒cos a = √[1 – (4/5)² ]

⇒ cos a = √[1 – (16/25)]

⇒ cos a = √[(25-16)/25]

⇒ cos a = 3/5

∴ The obtained value of cos a = 3/5

⇒ Now find the value of cos b: Cos b= √[1 – sin² b]

⇒ cos b = √[1 – (6/10)2 ]

⇒ cos b = √[1 – (36/100)]

⇒ cos b = √[(100-36)/100]

⇒ cos b = 8/10

∴ The obtained value of cos b = 8/10

⇒ As you all know the trigonometric formula: cos (a- b) = cos a cos b + sin a sin b

⇒ Substitute the required value of cos a, cos b, sin a and sin b in the equation above:

⇒ cos (a – b) = (3/5)x (8/10) + (4/5)x(6/10)

⇒ cos (a – b) = (24 + 24)/(10 x 5)

⇒ cos (a – b) = 48/50

⇒ (a – b) = cos-1 (48/50)

⇒ Substituting the values of a and b sin^-1 (4/5)- sin^-1 (6/10) = cos^-1 (48/50)

Ques. Prove that tan^-1 4/9 + tan^-1 7/4 = tan^-1 79/8?

Ans. By using the formula tan-1 x + tan^-1 y = tan^-1 (x + y) / (1 - xy)

⇒ We get, LHS = tan^-1 4/9 + tan^-1 7/24

⇒ tan^-1 [(4/9 + 7/4) / (1 - (4/9). (7/4)]

⇒ tan^-1 [((16 + 63) / 44) / ((36 - 28) / 44)]

⇒ On simplifying the terms,

⇒ tan^-1 (79 / 8)

Thus proving LHS = RHS

For Latest Updates on Upcoming Board Exams, Click Here: https://t.me/class_10_12_board_updates

---

Check-Out: