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

Plus Two Maths Chapter Wise Questions and Answers Chapter 2 Inverse Trigonometric Functions are part of Plus Two Maths Chapter Wise Questions and Answers. Here we have given Plus Two Maths Chapter Wise Questions and Answers Chapter 2 Inverse Trigonometric Functions.

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^-1x + tan^-1y + tan^-1z = π, show that x + y + z = xyz
Answer:
Given;
tan^-1x + tan^-1y + tan^-1z = π
⇒ tan^-1x + tan^-1y = π – tan^-1z

⇒ 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 – cos² x ⇒ 1 = sin²x
⇒ x = ±\(\frac{\pi}{2}\).

Question 6.
Solve the following

1. 2tan^-1(cosx) = tan^-1(2cosecx)
2. tan^-12x + tan^-13x = \(\frac{\pi}{4}\)

Answer:
1. 2tan^-1(cosx) = tan^-1(2cosecx)

2. tan^-12x + tan^-13x = \(\frac{\pi}{4}\)

⇒ (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 – cos² x
⇒ 1 = sin² 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.

1. Find the principal value of sec^-1\(\left(-\frac{2}{\sqrt{3}}\right)\) (1)
2. 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^-1x + cos^-1y = \(\frac{\pi}{3}\), then
sin^-1x + sin^-1y = ……… (3)
(a) \(\frac{2 \pi}{3}\)
(b) \(\frac{\pi}{3}\)
(c) \(\frac{\pi}{6}\)
(d) \(\frac{\pi}\)
(iii) If tan^-1x + tan^-1y = \(\frac{\pi}{4}\) 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^-1x = 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^-1x (2)
Answer:
(i) Range of cos^-1x 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.

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