Students can Download 2nd PUC Maths Chapter 2 Inverse Trigonometric Functions Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka 2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions

### 2nd PUC Maths Inverse Trigonometric Functions One Mark Questions and Answers

Inverse Function Calculator: With the methods presented, the calculator will find the inverse of the provided function.

Question 1.

Find the principal value of the folowing:

Answer:

Question 2.

Write the domain of f(x) = sec^{-1}x.

Answer:

R – (-1, 1).

Question 3.

Write the set of values of x for which 2 tan^{-1}x = tan^{-1}\(\frac{2 x}{1-x^{2}}\) holds.

Answer:

-1 < x < 1.

Question 4.

Write the range of the principal value branch of the function y = cosec^{-1} x.

Answer:

Question 5.

Write the set of values of x for which 2 tan^{-1} x = Sin^{-1} \(\frac{2 x}{1+x^{2}}\) holds.

Answer:

|x| ≤ 1.

Question 6.

Write the set of values of x for which 2 tan^{-1} = cos^{-1}\(\frac{1-x^{2}}{1+x^{2}}\) holds.

Answer:

x ≥ 0.

Question 7.

Answer:

Let cos^{-1}\(\frac { 1 }{ 2 }\) = θ ⇒ cos θ = \(\frac { 1 }{ 2 }\)

Considering principal branch θ ∈ [0, π ]

Question 8.

Answer:

Question 9.

Answer:

Question 10.

Find the domain of sin^{-1} (2x)

Answer:

Let sin^{-1} (2x) = 0 ⇒ 2x = sin 0

-1 ≤ sin θ ≤ 1 ⇒ -1 ≤ 2x ≤ 1

Question 11.

Write the range of the principal value branch of the function y = sin^{-1} x.

Answer:

### 2nd PUC Maths Inverse Trigonometric Functions Two Marks Questions and Answers

Prove the following:

Question 1.

Answer:

LHS = sin^{-1}(\(\sqrt{1-x^{2}}\)) put x = sin θ ⇒ θ = sin-1 x

= sin^{-1} (2 sin θ \(\sqrt{1-\sin ^{2} \theta}\))

= sin^{-1} (2 sin θ cos θ)

= sin^{-1} (sin 2θ) = 2θ = 2 sin^{-1} x = RHS.

Question 2.

Answer:

LHS = sin^{-1} (2x\(\sqrt{1-x^{2}}\)) putx = cos θ ⇒ θ = cos^{-1} x

= sin^{-1} (2 c0s θ \(\sqrt{1-\cos ^{2} \theta}\))

= sin^{-1} (2 sin θ cos θ)

= sin^{-1} (sin 2θ) = 2θ = 2 cos^{-1} x = RHS.

Question 3.

sin^{-1} (3x – 4x^{3}) = 3 sin^{-1} x, x ∈ \(\left[-\frac{1}{2}, \frac{1}{2}\right]\)

Answer:

LHS = sin^{-1} (3x – 4x^{3})

put x = sin ⇒ θ = sin x

= sin^{-1} (3 sin θ – 4 sin^{3} θ)

= sin^{-1} (sin 3θ) = 3θ = 3 sin^{-1} x = RHS.

Question 4.

cos^{-1}(4x^{2} – 3x) = 3 cos^{-1}x, x ∈ \(\left[\frac{1}{2}, 1\right]\)

Answer:

LHS = cos^{-1}(4x^{3} – 3x)

put x = cos θ ⇒ θ = cos^{-1} x

= cos^{-1} (4 cos^{3} θ – 3 cos θ)

= cos^{-1}(cos 3θ) = 3θ = 3 cos^{-1}x = RHS.

Question 5.

Answer:

Question 6.

Answer:

Question 7.

Answer:

Question 8.

tan^{-1}\(\frac { 1 }{ 2 }\) + tan^{-1}\(\frac { 2 }{ 11 }\) = tan^{-1}\(\frac { 3 }{ 4 }\)

Answer:

Question 9.

tan^{-1}\(\frac { 2 }{ 11 }\) + tan^{-1}\(\frac { 7 }{ 24 }\) = tan^{-1}\(\frac { 1 }{ 2 }\)

Answer:

Question 10.

tan^{-1}\(\frac { 1 }{ 2 }\) + tan^{-1}\(\frac { 1 }{ 3 }\) = \(\frac{\pi}{4}\)

Answer:

Simplify the following

Question 1.

sin^{-1}\(\left(\sin \frac{2 \pi}{3}\right)\) (June 2015)

Answer:

Question 2.

tan^{-1}\(\left(\tan \frac{3 \pi}{4}\right)\)

Answer:

Question 3.

cos^{-1}\(\left(\cos \frac{7 \pi}{6}\right)\)

Answer:

Question 4.

sin^{-1}\(\left(\sin \frac{3 \pi}{5}\right)\)

Answer:

Question 5.

tan^{-1} √3 – sec^{-1} (-2).

Answer:

Question 6.

Answer:

Question 7.

tan\(\left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)\)

Answer:

Question 8.

tan^{-1}\(\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right]\)

Answer:

Question 9.

If sin \(\left\{\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right\}\) = 1, find x.

Answer:

Question 10.

Evaluate sin^{-1}[sin(-600°)]]

Answer:

Question 11.

Evaluate tan(cos^{-1}\(\frac { 3 }{ 5 }\) + tan^{-1}\(\frac { 1 }{ 4 }\))

Answer:

Question 12.

Evaluate sin[2 sin^{-1} 0.6]

Answer:

Let sin^{-1}(0.6) = θ ⇒ sin θ = 0.6

sin 2θ = 2 sinθ cosθ = 2(0.6)(0.8) = 0.96

∴ sin[2sin^{-1} (0.6)] = 0.96

Question 13.

If sin^{-1} x + sin^{-1} y = then find the value of cos^{-1} x + cos^{-1} y.

Answer:

Question 14.

Find the domain of cos^{-1} (x^{2} – 4)

Answer:

cos^{-1} (x^{2} – 4) = θ

Question 15.

Find the domain of sin^{-1}(-x^{2}).

Answer:

Let sin^{-1}(-x^{2}) = θ

Question 16.

Answer:

Question 17.

Answer:

Question 18.

Evaluate tan^{-1}\(\left[\tan \frac{9 \pi}{8}\right]\)

Answer:

Question 19.

Evaluate cos^{-1}[cos(-680)]

Answer:

cos^{-1} [cos (-680)] = cos^{-1} [cos (680)] .

cos^{-1} [cos (720 – 40)] = cos^{-1} [cos(-40)] = cos^{-1} [cos 40] = 40 = \(\frac{2 \pi}{9}\)

Question 20.

Answer:

Question 21.

Answer:

Question 22.

Answer:

Question 23.

Answer:

Question 24.

Answer:

Question 25.

Answer:

Question 26.

Answer:

Let x = a sin θ ⇒ θ = sin^{-1} \(\frac{x}{a}\)

Question 27.

Answer:

Question 28.

Answer:

Question 29.

Answer:

### 2nd PUC Maths Inverse Trigonometric Functions Three Marks Questions With Answers

Question 1.

Prove that tan^{-1}x + tan^{-1}y = tan^{-1}\(\left(\frac{x+y}{1-x y}\right)\) when xy < 1.

Answer:

Let tan^{-1}x = α ⇒ tan α = x

tan^{-1}y = β ⇒ tan β = y

Question 2.

Prove that tan^{-1}x – tan^{-1}y = tan^{-1}\(\left(\frac{x-y}{1+x y}\right)\), xy > – 1.

Answer:

Let tan^{-1}x = α ⇒ tan α = x

Question 3.

Prove that 2 tan^{-1}\(\frac { 1 }{ 2 }\) + tan^{-1}\(\frac { 1 }{ 7 }\) = tan^{-1}\(\frac { 31 }{ 17 }\).

Answer:

Question 4.

Answer:

Question 5.

Solve tan^{-1}2x + tan^{-1}3x = \(\frac{\pi}{4}\)

Answer:

tan^{-1}2x + tan^{-1}3x = \(\frac{\pi}{4}\)

⇒ \(\frac{5 x}{1-6 x^{2}}\) = 1 ⇒ 5x = 1 – 6x^{2} ⇒ 6x^{2} + 5x – 1 = 0 ⇒ x = -1 or x = \(\frac{1}{6}\)

Since x = -1 does not saisíy the equation, x = \(\frac{1}{6}\) is the only solution of the given equation.

Question 6.

Answer:

Let x = tan θ ⇒ θ = tan^{-1}x

= tan^{-1}(tan 3θ) = 3θ = 3 tan^{-1}x = tan^{-1}x + 2tan^{-1}x

= tan^{-1}x + tan^{-1}\(\frac{2 x}{1-x^{2}}\) = LHS.

Question 7.

Show that sin^{-1}\(\frac { 3 }{ 5 }\) – sin^{-1}\(\frac { 8 }{ 17 }\) = cos^{-1}\(\frac { 84 }{ 85 }\).

Answer:

Question 8.

Show that sin^{-1}\(\frac { 12 }{ 13 }\) + cos^{-1}\(\frac { 4 }{ 5 }\) + tan^{-1}\(\frac { 63 }{ 16 }\) = π

Answer:

Question 9.

Answer:

Question 10.

Answer:

Question 11.

Answer:

Question 12.

Answer:

Question 13.

Answer:

Question 14.

Answer:

Question 15.

2 sin^{-1}\(\left(\frac{3}{5}\right)\) = tan^{-1}\(\left(\frac{24}{7}\right)\)

Answer:

Question 16.

Answer:

Question 17.

Answer:

Question 18.

Answer:

Question 19.

Solve for x

2 tan^{-1} (cot x) = tan^{-1} (2 cosec x)

Answer:

Question 20.

Solve for x : tan^{-1} \(\left(\frac{1-x}{1+x}\right)=\frac{1}{2}\) tan^{-1}x (x > 0)

Answer: