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## Sunday, September 12, 2021

Plus Two Maths Chapter 2 Inverse Trigonometric Functions Chapter Wise Question and Answers PDF Download: Students of Standard 12 can now download Plus Two Maths Chapter 2 Inverse Trigonometric Functions chapter wise question and answers pdf from the links provided below in this article. Plus Two Maths Chapter 2 Inverse Trigonometric Functions Question and Answer pdf will help the students prepare thoroughly for the upcoming Plus Two Maths Chapter 2 Inverse Trigonometric Functions exams.

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

Plus Two Maths Chapter 2 Inverse Trigonometric Functions question and answers consists of questions asked in the previous exams along with the solutions for each question. To help them get a grasp of chapters, frequent practice is vital. Practising these questions and answers regularly will help the reading and writing skills of students. Moreover, they will get an idea on how to answer the questions during examinations. So, let them solve Plus Two Maths Chapter 2 Inverse Trigonometric Functions chapter wise questions and answers to help them secure good marks in class tests and exams.

 Board Kerala Board Study Materials Chapter wise Question and Answers For Year 2021 Class 12 Subject Mathematics Chapters Maths Chapter 2 Inverse Trigonometric Functions Format PDF Provider Spandanam Blog

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Question 1.
Prove the following:

Question 2.
Find the value of

Question 3.
If tan-1x + tan-1y + tan-1z = π, show that x + y + z = xyz
Given;
tan-1x + tan-1y + tan-1z = π
⇒ tan-1x + tan-1y = π – tan-1z

⇒ x + y + z = xyz.

Question 4.
Match the following

Question 5.
Solve 2 tan-1(cos x) = tan-1(2 cos x)
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 = $$\frac{\pi}{4}$$

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

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)

1. principal value of:

2. find the value of x:

Question 3.
Solve the following

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)

(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)

⇒ x + y = 1 – xy ⇒ x + y + xy = 1.

Question 6.
(i) sin(tan-1(1)) is equal to

### 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$$\frac{84}{85}$$.

(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)
(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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