Students can access the NCERT MCQ Questions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use MCQ Questions for Class 12 Maths with Answers during preparation and score maximum marks in the exam. Students can download the Inverse Trigonometric Functions Class 12 MCQs Questions with Answers from here and test their problem-solving skills. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 12 Maths Chapter 2 Inverse Trigonometric Functions Objective Questions.
Inverse Trigonometric Functions Class 12 MCQs Questions with Answers
Students are advised to solve the Inverse Trigonometric Functions Multiple Choice Questions of Class 12 Maths to know different concepts. Practicing the MCQ Questions on Inverse Trigonometric Functions Class 12 with answers will boost your confidence thereby helping you score well in the exam.
Explore numerous MCQ Questions of Inverse Trigonometric Functions Class 12 with answers provided with detailed solutions by looking below.
Question 1.
If sin-1 x + sin-1 y = \(\frac { 2π }{3}\), then the value of cos-1 x + cos-1 y is
(a) \(\frac { 2π }{3}\)
(b) \(\frac { π }{3}\)
(c) \(\frac { π }{2}\)
(d) π
Answer
Answer: (b) \(\frac { π }{3}\)
Question 2.
tan-1 (√3) – sec-1(-2) is equal to:
(a) π
(b) –\(\frac { π }{3}\)
(c) \(\frac { π }{3}\)
(d) \(\frac { 2π }{3}\)
Answer
Answer: (b) –\(\frac { π }{3}\)
Question 3.
cos-1 (cos \(\frac { 7π }{6}\)) is equal to
(a) \(\frac { 7π }{6}\)
(b) –\(\frac { 5π }{6}\)
(c) \(\frac { π }{3}\)
(d) \(\frac { π }{6}\)
Answer
Answer: (b) –\(\frac { 5π }{6}\)
Question 4.
sin(\(\frac { π }{3}\) – sin-1(-\(\frac { 1 }{2}\))) is equal to
(a) \(\frac { 1 }{2}\)
(b) \(\frac { 1 }{3}\)
(c) \(\frac { 1 }{4}\)
(d) 1
Answer
Answer: (d) 1
Question 5.
tan-1 √3 – cot-1(-√3) is equal to
(a) π
(b) –\(\frac { π }{2}\)
(c) 0
(d) 2√3
Answer
Answer: (b) –\(\frac { π }{2}\)
Question 6.
sin (tan-1 x), |x| < 1, is equal to
(a) \(\frac { x }{\sqrt{1-x^2}}\)
(b) \(\frac { 1 }{\sqrt{1-x^2}}\)
(c) \(\frac { x }{\sqrt{1+x^2}}\)
(d) \(\frac { x }{\sqrt{1+x^2}}\)
Answer
Answer: (d) \(\frac { x }{\sqrt{1+x^2}}\)
Question 7.
sin-1 (1 – x) – 2 sin-1 x = \(\frac { π }{2}\), then x is equal to
(a) 0, \(\frac { 1 }{2}\)
(b) 1, \(\frac { 1 }{2}\)
(c) 0
(d) \(\frac { 1 }{2}\)
Answer
Answer: (c) 0
Question 8.
tan-1 (\(\frac { x }{y}\)) – tan-1 \(\frac { x-y }{x+y}\) is equal to
(a) \(\frac { π }{2}\)
(b) \(\frac { π }{3}\)
(c) \(\frac { π }{4}\)
(d) –\(\frac { 3π }{4}\)
Answer
Answer: (c) \(\frac { π }{4}\)
Question 9.
The value of sin-1(cos(\(\frac { 43π }{5}\))) is
(a) \(\frac { 3π }{5}\)
(b) \(\frac { -7π }{5}\)
(c) \(\frac { π }{10}\)
(d) –\(\frac { -π }{10}\)
Answer
Answer: (d) –\(\frac { -π }{10}\)
Question 10.
The principal value of the expression
cos-1 [cos (-680°)] is
(a) \(\frac { 2π }{9}\)
(b) \(\frac { -2π }{9}\)
(c) \(\frac {34π }{9}\)
(d) –\(\frac { π }{9}\)
Answer
Answer: (a) \(\frac { 2π }{9}\)
Question 11.
The value of cot (sin-1x) is
(a) \(\frac { \sqrt{1+x^2} }{x}\)
(b) \(\frac { x }{\sqrt{1+x^2}}\)
(c) \(\frac {1}{x}\)
(d) \(\frac { \sqrt{1-x^2} }{x}\)
Answer
Answer: (d) \(\frac { \sqrt{1-x^2} }{x}\)
Question 12.
The domain of sin-1 2x is
(a) [0, 1]
(b) [-1, 1]
(c) [\(\frac {-1}{2}\), \(\frac {1}{2}\)]
(d) [-2, 2]
Answer
Answer: (c) [\(\frac {-1}{2}\), \(\frac {1}{2}\)]
Question 13.
The greatest and least values of (sin-1 x)² + (cos-1x)² are respectively
(a) \(\frac { 5π^2 }{4}\) and \(\frac { π^2 }{8}\)
(b) \(\frac { π }{2}\) and \(\frac { -π }{2}\)
(c) \(\frac { π^2 }{4}\) and \(\frac { -π^2 }{4}\)
(d) –\(\frac { π^2 }{4}\) and 0
Answer
Answer: (a) \(\frac { 5π^2 }{4}\) and \(\frac { π^2 }{8}\)
Question 14.
If cos-1 x – cos-1 — = α, then 4x² – 4xy cos α + y² is equal to:
(a) 4
(b) 2 sin² α
(c) -4 sin² α
(d) 4 sin² α.
Answer
Answer: (d) 4 sin² α.
Hint:
Squaring, 4x² + y² cos² α – 4xy cos α
= 4 sin² α – y² sin² α
⇒ 4x² – 4xy cos α + y² = 4 sin² α.
Question 15.
If sin-1 (\(\frac {5}{4}\)) = \(\frac {π}{2}\), then the value of x is
(a) 3
(b) 4
(c) 5
(d) 1
Answer
Answer: (a) 3
Hint:
Question 16.
The value of cot (cosec-1\(\frac {5}{3}\) + tan-1\(\frac {2}{3}\)) is
(a) \(\frac { 5 }{17}\)
(b) \(\frac { 6 }{17}\)
(c) \(\frac { 3 }{17}\)
(d) \(\frac { 4 }{17}\)
Answer
Answer: (b) \(\frac { 6 }{17}\)
Hint:
Question 17.
If tan-1 y = tan-1 x + tan-1(\(\frac { 2x }{1-x^2}\)) when |x| < \(\frac { 1 }{√3}\), then the value of y is:
(a) \(\frac { 3x-x^3 }{1-3x^2}\)
(b) \(\frac { 3x+x^3 }{1-3x^2}\)
(c) \(\frac { 3x-x^3 }{1+3x^2}\)
(d) \(\frac { 3x+x^3 }{1+3x^2}\)
Answer
Answer: (a) \(\frac { 3x-x^3 }{1-3x^2}\)
Hint:
Fill in the blanks
Question 1.
Principal value of sin-1 (-\(\frac {1}{2}\)) is ………………
Answer
Answer: –\(\frac {π}{6}\)
Question 2.
Principal value of sin-1 (-\(\frac {1}{√2}\)) is ……………….
Answer
Answer: \(\frac {-π}{4}\)
Question 3.
Principal value of cos-1 (\(\frac {-1}{2}\)) is ………………
Answer
Answer: \(\frac {2π}{3}\)
Question 4.
Principal value of tan-1 (-√3) is …………………
Answer
Answer: –\(\frac {π}{3}\)
Question 5.
Principal value of tan-1 (-1) is ……………..
Answer
Answer: –\(\frac {π}{4}\)
Question 6.
Principal value of cot-1 (√3) is ……………….
Answer
Answer: \(\frac {π}{6}\)
Question 7.
Principal value of cosec-1 (-√2) is ……………..
Answer
Answer: –\(\frac {π}{4}\)
Question 8.
sin-1 x + cos-1 x = ………………
Answer
Answer: \(\frac {π}{2}\)
Question 9.
tan-1 x + cot-1 x = ………………..
Answer
Answer: \(\frac {π}{2}\)
Question 10.
sec-1 x + cosec-1 x = ……………….
Answer
Answer: \(\frac {π}{2}\)
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