Students can Download Chapter 2 Inverse Trigonometric Functions Questions and Answers, Plus Two Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

## Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 2 Inverse Trigonometric Functions

### Plus Two Maths Inverse Trigonometric Functions Three Mark Questions and Answers

Question 1.

Prove the following

Answer:

Question 2.

Find the value of

Answer:

Question 3.

If tan^{-1}x + tan^{-1}y + tan^{-1}z = π, show that x + y + z = xyz

Answer:

Given;

tan^{-1}x + tan^{-1}y + tan^{-1}z = π

⇒ tan^{-1}x + tan^{-1}y = π – tan^{-1}z

⇒ x + y + z = xyz.

Question 4.

Match the following

Answer:

Question 5.

Solve 2 tan^{-1}(cos x) = tan^{-1}(2 cos x)

Answer:

2 tan^{-1}(cosx) = tan^{-1}(2cosx)

⇒ \(\frac{2 \cos x}{1-\cos ^{2} x}\) = 2cosx

⇒ 1 = 1 – cos^{2} x ⇒ 1 = sin^{2}x

⇒ x = ±\(\frac{\pi}{2}\).

Question 6.

Solve the following

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

Answer:

1. 2tan^{-1}(cosx) = tan^{-1}(2cosecx)

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

⇒ (6x – 1)(x + 1) = 0

⇒ x = \(\frac{1}{6}\), x = – 1

Since x = – 1 does not satisfy the equation, as the LHS becomes negative. So x = \(\frac{1}{6}\).

Question 7.

Solve 2 tan^{-1}(cos x) = tan^{-1}(2 cos x)

Answer:

2 tan^{-1}(cos x) = tan^{-1}(2 cos x)

⇒ \(\frac{2 \cos x}{1-\cos ^{2} x}\) = 2cosx

⇒ 1 = 1 – cos^{2} x

⇒ 1 = sin^{2} x ⇒ x = ±\(\frac{\pi}{2}\)

### Plus Two Maths Inverse Trigonometric Functions Four Mark Questions and Answers

Question 1.

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

Answer:

Question 2.

- Find the principal value of sec
^{-1}\(\left(-\frac{2}{\sqrt{3}}\right)\) (1) - if sin\(\left(\sin ^{-1}\left(\frac{1}{5}\right)+\cos ^{-1}(x)\right)=1\), then find the value of x. (3)

Answer:

1. principal value of:

2. find the value of x:

Question 3.

Solve the following

Answer:

The value x = –\(\frac{\sqrt{3}}{\sqrt{28}}\) makes the LHS negative, so rejected.

Question 4.

(i) Choose the correct answer from the bracket. cos(tan^{-1} x), |x| < 1 is equal to (1)

Answer:

(draw a right triangle to convert ‘tan’ to ‘sin’).

Question 5.

(i) In which quadrants are the graph of cos^{-1} (x) lies, x ∈ [-1,1 ] (1)

(ii) If cos^{-1}x + cos^{-1}y = \(\frac{\pi}{3}\), then

sin^{-1}x + sin^{-1}y = ……… (3)

(a) \(\frac{2 \pi}{3}\)

(b) \(\frac{\pi}{3}\)

(c) \(\frac{\pi}{6}\)

(d) \(\frac{\pi}\)

(iii) If tan^{-1}x + tan^{-1}y = \(\frac{\pi}{4}\) then prove that x + y + xy = 1 (2)

Answer:

(i) First and Second quadrant

⇒ x + y = 1 – xy ⇒ x + y + xy = 1.

Question 6.

(i) sin(tan^{-1}(1)) is equal to

Answer:

### Plus Two Maths Inverse Trigonometric Functions Six Mark Questions and Answers

Question 1.

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

Answer:

(draw a right triangle to convert ‘tan’ to ‘cos’).

Question 2.

(i) Choose the correct answer from the Bracket.

If cos^{-1}x = y, then y is equal to (1)

(a) π ≤ y ≤ π

(b) 0 ≤ y ≤ π

(c) \(-\frac{\pi}{2}\) ≤ y ≤ \(\frac{\pi}{2}\)

(d) 0 ≤ y ≤ π

(ii) Find the value of cos^{-1} cos\(\left(\frac{7 \pi}{3}\right)\) (3)

(iii) Solve for x if, tan^{-1}\(\left(\frac{1+x}{1-x}\right)\) = 2 tan^{-1}x (2)

Answer:

(i) Range of cos^{-1}x is [0, π] ⇒ 0 ≤ y ≤ π

(ii) Here \(\left(\frac{7 \pi}{3}\right)\) lie outside the interval [0, π]. TO make it in the interval proceed as follows.