NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions — Exercise 2.20

Find the value of the following:

Question 1: –1 13 cos cos 6 π     

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Question 2: –1 7 tan tan 6 π      Prove that

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Question 3: –1 3 –1 24 2sin tan 5 7 =

Question 4: –1 8 –1 3 –1 77 sin sin tan 17 5 36 + =

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Question 5: –1 4 –1 12 –1 33 cos cos cos 5 13 65 + =

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Question 6: –1 12 –1 3 –1 56 cos sin sin 13 5 65 + =

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Question 7: –1 63 –1 5 –1 3 tan sin cos 16 13 5 = +

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Question 8: –1 1 1 1 1 1 1 1 tan tan tan tan 5 7 3 8 4 − − − p Prove that

Question 9: –1 1 –1 1 tan cos 2 1 x x x − = + , x Î [0, 1]

Question 10:–1 1 sin 1 sin cot 1 sin 1 sin 2 x x x x x + + −   =  + − −  , 0, 4 x π  Î   

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Question 11:–1 1 1 1 –1 tan cos 1 1 4 2 x x x x x + − −  p  = −  + + −  , 1 1 2 − ≤ x ≤ [Hint: Put x = cos 2q]

Find the value of the following: 1. 1 13

Question 12:9 9 1 1 9 1 2 2 sin sin 8 4 3 4 3 π − − − = Solve the following equations:

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Question 13:2tan–1 (cos x) = tan–1 (2 cosec x)

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Question 14: –1 1 1 –1 tan tan ,( 0) 1 2 x x x x − = > +

Question 15:sin (tan–1 x), | x| < 1 is equal to (A) 2 1 x − x (B) 2 1 1− x (C) 2 1 1+ x (D) 2 1 x + x

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Question 16:sin–1 (1 – x) – 2 sin–1 x = 2 p , then x is equal to (A) 0, 1 2 (B) 1, 1 2 (C) 0 (D) 1 2

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Question 17: 1 1 tan tan x x y y x y −  − −   −   + is equal to (A) 2 p (B) 3 p (C) 4 p (D) 3 4 − p