Class 12 Maths Exemplar Chapter 2 PDF
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This document provides an overview of inverse trigonometric functions, including their domains and ranges. It defines inverse functions and explains their properties. It's a great resource for high school students.
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Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Chapter 2 INVERSE TRIGONOMETRIC FUNCTIONS 2.1 Overview 2.1.1 Inverse function Inverse...
Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com Chapter 2 INVERSE TRIGONOMETRIC FUNCTIONS 2.1 Overview 2.1.1 Inverse function Inverse of a function ‘f ’ exists, if the function is one-one and onto, i.e, bijective. Since trigonometric functions are many-one over their domains, we restrict their domains and co-domains in order to make them one-one and onto and then find their inverse. The domains and ranges (principal value branches) of inverse trigonometric functions are given below: Functions Domain Range (Principal value branches) – y = sin–1x [–1,1] , 2 2 y = cos–1x [–1,1] [0,π] – y = cosec–1x R– (–1,1) , – {0} 2 2 y = sec–1x R– (–1,1) [0,π] – 2 – y = tan–1x R , 2 2 y = cot–1x R (0,π) Notes: (i) The symbol sin–1x should not be confused with (sinx)–1. Infact sin–1x is an angle, the value of whose sine is x, similarly for other trigonometric functions. (ii) The smallest numerical value, either positive or negative, of θ is called the principal value of the function. 20/04/2018 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com INVERSE TRIGONOMETRIC FUNCTIONS 19 (iii) Whenever no branch of an inverse trigonometric function is mentioned, we mean the principal value branch. The value of the inverse trigonometic function which lies in the range of principal branch is its principal value. 2.1.2 Graph of an inverse trigonometric function The graph of an inverse trigonometric function can be obtained from the graph of original function by interchanging x-axis and y-axis, i.e, if (a, b) is a point on the graph of trigonometric function, then (b, a) becomes the corresponding point on the graph of its inverse trigonometric function. It can be shown that the graph of an inverse function can be obtained from the corresponding graph of original function as a mirror image (i.e., reflection) along the line y = x. 2.1.3 Properties of inverse trigonometric functions –π π 1. sin–1 (sin x) = x : x∈ , 2 2 cos–1(cos x) = x : x ∈[0, π ] – tan–1(tan x) = x : x ∈ , 2 2 cot–1(cot x) = x : x ∈ (0, ) sec–1(sec x) = x : x ∈[0, ] – 2 – cosec–1(cosec x) = x : x∈ , – {0} 2 2 2. sin (sin–1 x) = x : x ∈[–1,1] cos (cos–1 x) = x : x ∈[–1,1] tan (tan–1 x) = x : x ∈R cot (cot–1 x) = x : x ∈R sec (sec–1 x) = x : x ∈R – (–1,1) cosec (cosec–1 x) = x : x ∈R – (–1,1) 1 3. sin –1 = cosec –1 x : x ∈R – (–1,1) x 1 cos –1 = sec –1 x : x ∈R – (–1,1) x 20/04/2018 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com 20 MATHEMATICS 1 tan –1 = cot –1 x : x>0 x = – π + cot–1x : x –1 tan–1x – tan–1y = tan–1 1 + xy 2x 7. 2tan–1x = sin–1 : –1 ≤ x ≤ 1 1 + x2 1 – x2 2tan–1x = cos–1 : x≥0 1 + x2 2x 2tan–1x = tan–1 : –1 < x < 1 1 – x2 2.2 Solved Examples Short Answer (S.A.) 3 Example 1 Find the principal value of cos–1x, for x =. 2 20/04/2018 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com INVERSE TRIGONOMETRIC FUNCTIONS 21 3 3 2 = θ , then cos θ = –1 Solution If cos. 2 3 Since we are considering principal branch, θ ∈ [0, π]. Also, since > 0, θ being in 2 3 the first quadrant, hence cos–1 =. 2 6 – Example 2 Evaluate tan–1 sin. 2 – Solution tan–1 sin = tan–1 − sin = tan–1(–1) = −. 2 2 4 13 Example 3 Find the value of cos–1 cos. 6 13 π –1 Solution cos–1 cos = cos–1 cos (2π + ) = cos cos 6 6 6 π =. 6 9 Example 4 Find the value of tan–1 tan. 8 9 π Solution tan–1 tan = tan–1 tan π + 8 8 –1 π = tan tan = 8 8 –1 Example 5 Evaluate tan (tan (– 4)). Solution Since tan (tan–1x) = x, ∀ x ∈ R, tan (tan–1(– 4) = – 4. Example 6 Evaluate: tan–1 3 – sec–1 (–2). 20/04/2018 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com 22 MATHEMATICS Solution tan–1 3 – sec–1 (– 2) = tan–1 3 – [π – sec–12] π 1 2π π π = − π+ cos –1 =− + = −. 3 2 3 3 3 –1 –1 3 Example 7 Evaluate: sin cos sin 2. –1 –1 3 1 Solution sin cos sin = sin –1 cos = sin –1 =. 2 3 2 6 Example 8 Prove that tan(cot–1x) = cot (tan–1x). State with reason whether the equality is valid for all values of x. Solution Let cot–1x = θ. Then cot θ = x or, tan – = x ⇒ tan –1 x = – 2 2 –1 So tan(cot x ) = tan = cot – = cot − cot x = cot(tan x ) –1 –1 2 2 The equality is valid for all values of x since tan–1x and cot–1x are true for x ∈ R. –1 y Example 9 Find the value of sec tan . 2 y y Solution Let tan –1 = , where ∈ − , . So, tanθ = , 2 2 2 2 4 + y2 which gives sec=. 2 y 4 + y2 Therefore, sec tan –1 = sec =. 2 2 –1 8 Example 10 Find value of tan (cos–1x) and hence evaluate tan cos. 17 Solution Let cos–1x = θ, then cos θ = x, where θ ∈ [0,π] 20/04/2018 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com INVERSE TRIGONOMETRIC FUNCTIONS 23 1 – cos 2 1 – x2 Therefore, tan(cos–1x) = tan = =. cos x 2 8 1– Hence 8 17 15. tan cos –1 = = 17 8 8 17 –1 –5 Example 11 Find the value of sin 2cot 12 –5 −5 Solution Let cot–1 = y. Then cot y =. 12 12 –1 –5 Now sin 2cot = sin 2y 12 12 –5 = 2siny cosy = 2 since cot y < 0, so y ∈ 2 , 13 13 –120 = 169 –1 1 4 Example 12 Evaluate cos sin + sec –1 4 3 1 4 –1 1 3 Solution cos sin –1 + sec –1 = cos sin + cos –1 4 3 4 4 –1 1 3 1 3 = cos sin cos cos –1 – sin sin –1 sin cos –1 4 4 4 4 2 2 3 1 1 3 = 4 1– 4 – 4 1– 4 3 15 1 7 3 15 – 7 = 4 4 –4 4 = 16. 20/04/2018 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com 24 MATHEMATICS Long Answer (L.A.) 3 17 π Example 13 Prove that 2sin–1 – tan–1 = 5 31 4 3 3 −π π Solution Let sin–1 = θ, then sinθ = , where θ ∈ , 5 5 2 2 3 3 Thus tan θ = , which gives θ = tan–1. 4 4 3 17 Therefore, 2sin–1 – tan–1 5 31 17 3 17 = 2θ – tan–1 = 2 tan–1 – tan–1 31 4 31 3 2. 17 tan –1 4 – tan –1 24 17 = 31 = tan–1 − tan –1 1– 9 7 31 16 24 17 − tan –1 7 31 π = = 1 + 24. 17 4 7 31 Example 14 Prove that cot–17 + cot–18 + cot–118 = cot–13 Solution We have cot–17 + cot–18 + cot–118 1 1 1 1 = tan–1 + tan–1 + tan–1 (since cot–1 x = tan–1 , if x > 0) 7 8 18 x 1 1 + 1 tan 7 8 + tan –1 –1 1 1. < 1) = 1 1 18 (since x. y = 1− × 7 8 7 8 20/04/2018 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com INVERSE TRIGONOMETRIC FUNCTIONS 25 3 1 + –1 3 –1 1 tan –1 11 18 = tan + tan = (since xy < 1) 11 18 1− 3 × 1 11 18 –1 65 –1 1 = tan = tan = cot–1 3 195 3 Example 15 Which is greater, tan 1 or tan–1 1? Solution From Fig. 2.1, we note that tan x is an increasing function in the interval −π π π π , , since 1 > ⇒ tan 1 > tan. This gives 2 2 4 4 Y tan x tan 1 > 1 π ⇒ tan 1 > 1 > 4 O ⇒ tan 1 > 1 > tan–1 (1). X – /2 /4 /2 Example 16 Find the value of 2 sin 2 tan –1 + cos(tan –1 3). 3 2 2 Solution Let tan–1 = x and tan–1 3 = y so that tan x = and tan y = 3. 3 3 2 Therefore, sin 2 tan –1 + cos(tan –1 3) 3 = sin (2x) + cos y 2 2. 2 tan x 1 3 + 1 = 1 + tan x + = ( ) 2 4 2 1+ tan 2 y 1+ 1+ 3 9 12 1 37 = + =. 13 2 26 20/04/2018 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com 26 MATHEMATICS Example 17 Solve for x 1− x 1 tan –1 = tan x, x > 0 –1 1 + x 2 1− x From given equation, we have 2 tan –1 = tan x –1 Solution 1+ x ⇒ 2 tan –1 1 − tan –1 x = tan –1 x π π ⇒ 2 = 3tan –1 x ⇒ = tan –1 x 4 6 1 ⇒ x= 3 Example 18 Find the values of x which satisfy the equation sin–1 x + sin–1 (1 – x) = cos–1 x. Solution From the given equation, we have sin (sin–1 x + sin–1 (1 – x)) = sin (cos–1x) ⇒ sin (sin–1 x) cos (sin–1 (1 – x)) + cos (sin–1 x) sin (sin–1 (1 – x) ) = sin (cos–1 x) ⇒ x 1– (1– x) 2 + (1− x ) 1 − x 2 = 1− x 2 ⇒ x 2 x – x 2 + 1− x 2 (1− x −1) = 0 ⇒x ( ) 2 x – x2 − 1− x2 = 0 ⇒x = 0 or 2x – x2 = 1 – x2 1 ⇒x = 0 or x=. 2 π Example 19 Solve the equation sin–16x + sin–1 6 3 x = − 2 π Solution From the given equation, we have sin–1 6x = − − sin 6 3 x –1 2 20/04/2018 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com INVERSE TRIGONOMETRIC FUNCTIONS 27 π sin (sin–1 6x) = sin − − sin 6 3 x –1 ⇒ 2 ⇒ 6x = – cos (sin–1 6 3 x) ⇒ 6x = – 1−108x 2. Squaring, we get 36x2 = 1 – 108x2 1 ⇒ 144x2 = 1 ⇒ x= ± 12 1 1 Note that x = – is the only root of the equation as x = does not satisfy it. 12 12 Example 20 Show that α π β –1 sin α cos β 2 tan–1 tan.tan − = tan 2 4 2 cos α + sin β α π β 2 tan.tan − 2 4 2 –1 2 x since 2 tan x = tan –1 –1 Solution L.H.S. = tan α π β 1− x 2 1− tan 2 tan 2 − 2 4 2 β 1− tan α 2 2 tan 2 1 + tan β –1 2 = tan 2 β 1 − tan α 2 1− tan 2 2 1+ tan β 2 α β 2 tan. 1− tan 2 = tan –1 2 2 2 2 β 2 α β 1 + tan − tan 1 − tan 2 2 2 20/04/2018 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com 28 MATHEMATICS α 2 β 1− tan 2 tan = tan –1 2 2 2β 2 α β 2 α 1+ tan 1− tan + 2 tan 1+ tan 2 2 2 2 α β 2 tan 1− tan 2 2 2 2 α 2β 1 + tan 1+ tan = tan –1 2 2 2 α β 1− tan 2 tan 2+ 2 2 α 2β 1 + tan 1+ tan 2 2 sin α cos β = tan –1 = R.H.S. cos α + sin β Objective type questions Choose the correct answer from the given four options in each of the Examples 21 to 41. Example 21 Which of the following corresponds to the principal value branch of tan–1? π π π π (A) − , (B) − 2 , 2 2 2 π π (C) − , – {0} (D) (0, π) 2 2 Solution (A) is the correct answer. Example 22 The principal value branch of sec–1 is π π π (A) − 2 , 2 − {0} (B) [0, π] − 2 π π (C) (0, π) (D) − , 2 2 20/04/2018 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com INVERSE TRIGONOMETRIC FUNCTIONS 29 Solution (B) is the correct answer. Example 23 One branch of cos–1 other than the principal value branch corresponds to π 3π 3π (A) , (B) [π , 2π]− 2 2 2 (C) (0, π) (D) [2π, 3π] Solution (D) is the correct answer. –1 43π Example 24 The value of sin cos is 5 3π −7 π π π (A) (B) (C) (D) – 5 5 10 10 –1 40π+ 3π 3π = sin cos 8π+ –1 Solution (D) is the correct answer. sin cos 5 5 –1 3π –1 π 3π = sin cos = sin sin − 5 2 5 –1 π π = sin sin − = −. 10 10 Example 25 The principal value of the expression cos–1 [cos (– 680°)] is 2π − 2π 34π π (A) (B) (C) (D) 9 9 9 9 Solution (A) is the correct answer. cos–1 (cos (680°)) = cos–1 [cos (720° – 40°)] 2π = cos–1 [cos (– 40°)] = cos–1 [cos (40°)] = 40° =. 9 Example 26 The value of cot (sin–1x) is 1+ x2 x (A) (B) x 1+ x 2 20/04/2018 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com 30 MATHEMATICS 1 1− x 2 (C) (D). x x Solution (D) is the correct answer. Let sin–1 x = θ, then sinθ = x 1 1 ⇒ cosec θ = ⇒ cosec2θ = x x2 1 1− x2 ⇒ 1 + cot2 θ = ⇒ cotθ =. x2 x π Example 27 If tan–1x = for some x ∈ R, then the value of cot–1x is 10 π 2π 3π 4π (A) (B) (C) (D) 5 5 5 5 π Solution (B) is the correct answer. We know tan –1x + cot –1x =. Therefore 2 π π cot–1x = – 2 10 π π 2π ⇒ cot–1x = – =. 2 10 5 Example 28 The domain of sin–1 2x is (A) [0, 1] (B) [– 1, 1] 1 1 (C) − 2 , 2 (D) [–2, 2] Solution (C) is the correct answer. Let sin–12x = θ so that 2x = sin θ. 1 1 Now – 1 ≤ sin θ ≤ 1, i.e.,– 1 ≤ 2x ≤ 1 which gives − ≤x≤. 2 2 − 3 Example 29 The principal value of sin–1 2 is 20/04/2018 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com INVERSE TRIGONOMETRIC FUNCTIONS 31 2π π 4π 5π (A) − (B) − (C) (D). 3 3 3 3 Solution (B) is the correct answer. − 3 –1 π –1 π π sin –1 = sin – sin = – sin sin = –. 2 3 3 3 Example 30 The greatest and least values of (sin–1x)2 + (cos–1x)2 are respectively 5π 2 π2 π −π (A) and (B) and 4 8 2 2 π2 −π2 π2 (C) and (D) and 0. 4 4 4 Solution (A) is the correct answer. We have (sin–1x)2 + (cos–1x)2 = (sin–1x + cos–1x)2 – 2 sin–1x cos–1 x π2 π = − 2sin –1 x − sin –1 x 4 2 π2 ( ) 2 = − π sin –1 x + 2 sin –1 x 4 2 π π2 = 2 ( sin –1 x ) − 2 sin –1 x + 8 –1 2 π π2 = 2 sin x − + . 4 16 π2 π 2 −π π 2 π2 Thus, the least value is 2 i.e. and the Greatest value is 2 − 4 + 16 , 2 16 8 5π2 i.e.. 4 Example 31 Let θ = sin–1 (sin (– 600°), then value of θ is 20/04/2018 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com 32 MATHEMATICS π π 2π − 2π (A) (B) (C) (D). 3 2 3 3 Solution (A) is the correct answer. π −10π sin –1 sin − 600 × = sin sin –1 180 3 –1 2π –1 2π = sin − sin 4 π − = sin sin 3 3 –1 π –1 π π = sin sin π − = sin sin =. 3 3 3 Example 32 The domain of the function y = sin–1 (– x2) is (A) [0, 1] (B) (0, 1) (C) [–1, 1] (D) φ Solution (C) is the correct answer. y = sin–1 (– x2) ⇒ siny = – x2 i.e. – 1 ≤ – x2 ≤ 1 (since – 1 ≤ sin y ≤ 1) ⇒ 1 ≥ x2 ≥ – 1 ⇒ 0 ≤ x2 ≤ 1 ⇒ x ≤ 1 i.e. − 1 ≤ x ≤ 1 Example 33 The domain of y = cos–1 (x2 – 4) is (A) [3, 5] (B) [0, π] (C) − 5, − 3 ∩ − 5, 3 (D) − 5, − 3 ∪ 3, 5 Solution (D) is the correct answer. y = cos–1 (x2 – 4 ) ⇒ cosy = x2 – 4 i.e. – 1 ≤ x2 – 4 ≤ 1 (since – 1 ≤ cos y ≤ 1) ⇒ 3 ≤ x2 ≤ 5 ⇒ 3≤ x ≤ 5 ⇒ x∈ − 5, − 3 ∪ 3, 5 Example 34 The domain of the function defined by f (x) = sin–1x + cosx is 20/04/2018 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com INVERSE TRIGONOMETRIC FUNCTIONS 33 (A) [–1, 1] (B) [–1, π + 1] (C) ( – ∞, ∞ ) (D) φ Solution (A) is the correct answer. The domain of cos is R and the domain of sin–1 is [–1, 1]. Therefore, the domain of cosx + sin–1x is R ∩ [ –1,1] , i.e., [–1, 1]. Example 35 The value of sin (2 sin–1 (.6)) is (A).48 (B).96 (C) 1.2 (D) sin 1.2 Solution (B) is the correct answer. Let sin–1 (.6) = θ, i.e., sin θ =.6. Now sin (2θ) = 2 sinθ cosθ = 2 (.6) (.8) =.96. π Example 36 If sin–1 x + sin–1 y = , then value of cos–1 x + cos–1 y is 2 π 2π (A) (B) π (C) 0 (D) 2 3 π Solution (A) is the correct answer. Given that sin–1 x + sin–1 y =. 2 π –1 π –1 π Therefore, – cos x + – cos y = 2 2 2 π ⇒ cos–1x + cos–1y =. 2 –1 3 1 Example 37 The value of tan cos + tan –1 is 5 4 19 8 19 3 (A) (B) (C) (D) 8 19 12 4 –1 3 1 –1 4 1 Solution (A) is the correct answer. tan cos + tan –1 = tan tan + tan –1 5 4 3 4 20/04/2018 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com 34 MATHEMATICS 4 1 3+4 19 19 –1 = tan tan –1 =. = tan tan 4 1 8 8 1− × 3 4 Example 38 The value of the expression sin [cot–1 (cos (tan–1 1))] is 1 2 (A) 0 (B) 1 (C) (D). 3 3 Solution (D) is the correct answer. π 1 –1 2 2 sin [cot–1 (cos )] = sin [cot–1 ]= sin sin = 4 2 3 3 1 Example 39 The equation tan–1x – cot–1x = tan–1 has 3 (A) no solution (B) unique solution (C) infinite number of solutions (D) two solutions Solution (B) is the correct answer. We have π π tan–1x – cot–1x = and tan–1x + cot–1x = 6 2 2π Adding them, we get 2tan–1x = 3 π ⇒ tan–1x = i.e., x = 3. 3 Example 40 If α ≤ 2 sin–1x + cos–1x ≤β , then −π π (A) α = , β= (B) α = 0, β = π 2 2 −π 3π (C) α = , β= (D) α = 0, β = 2π 2 2 20/04/2018 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com INVERSE TRIGONOMETRIC FUNCTIONS 35 −π π Solution (B) is the correct answer. We have ≤ sin–1 x ≤ 2 2 −π π π π π ⇒ + ≤ sin–1x + ≤ + 2 2 2 2 2 ⇒ 0 ≤ sin x + (sin x + cos x) ≤ π –1 –1 –1 ⇒ 0 ≤ 2sin–1x + cos–1x ≤ π Example 41 The value of tan2 (sec–12) + cot2 (cosec–13) is (A) 5 (B) 11 (C) 13 (D) 15 Solution (B) is the correct answer. tan2 (sec–12) + cot2 (cosec–13) = sec2 (sec–12) – 1 + cosec2 (cosec–13) – 1 = 22 × 1 + 32 – 2 = 11. 2.3 EXERCISE Short Answer (S.A.) 5 –1 13 + cos cos –1 1. Find the value of tan tan . 6 6 – 3 π 2. Evaluate cos cos –1 +. 2 6 π 3. Prove that cot – 2 cot –1 3 = 7. 4 1 1 –π 4. Find the value of tan –1 – + cot –1 + tan –1 sin. 3 3 2 2 5. Find the value of tan–1 tan . 3 –π –1 –4 6. Show that 2tan–1 (–3) = + tan . 2 3 20/04/2018 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com 36 MATHEMATICS 7. Find the real solutions of the equation tan –1 x ( x + 1) + sin –1 x 2 + x + 1 =. 2 –1 1 8. Find the value of the expression sin 2 tan 3 ( + cos tan 2 2. –1 ) 9. If 2 tan–1 (cos θ) = tan–1 (2 cosec θ), then show that θ = , 4 where n is any integer. –1 1 –1 1 10. Show that cos 2 tan = sin 4 tan . 7 3 3 11. ( ) Solve the following equation cos tan –1 x = sin cot –1 . 4 Long Answer (L.A.) 1 + x 2 + 1– x 2 π 1 12. Prove that tan –1 = + cos –1 x 2 1 + x – 1– x 2 2 4 2 –1 3 4 –3π π 13. Find the simplified form of cos cos x + sin x , where x ∈ ,. 5 5 4 4 8 3 77 14. Prove that sin –1 + sin –1 = sin –1. 17 5 85 5 3 63 15. Show that sin –1 + cos –1 = tan –1. 13 5 16 1 2 1 16. Prove that tan –1 + tan –1 = sin −1. 4 9 5 –1 1 1 17. Find the value of 4 tan – tan –1. 5 239 20/04/2018 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com INVERSE TRIGONOMETRIC FUNCTIONS 37 1 –1 3 4– 7 4+ 7 18. Show that tan sin = and justify why the other value 2 4 3 3 is ignored? 19. If a1, a2, a3,...,an is an arithmetic progression with common difference d, then evaluate the following expression. d –1 d –1 d –1 d tan tan –1 + tan + tan +... + tan . 1 + a1 a2 1 + a2 a3 1 + a3 a4 1 + an –1 an Objective Type Questions Choose the correct answers from the given four options in each of the Exercises from 20 to 37 (M.C.Q.). 20. Which of the following is the principal value branch of cos–1x? – (A) 2 , 2 (B) (0, π) (C) [0, π] (0, π) – (D) 2 21. Which of the following is the principal value branch of cosec–1x? – (A) , (B) [0, π] – 2 2 2 – – (C) 2 , 2 (D) 2 , 2 – {0} 22. If 3tan–1 x + cot–1 x = π, then x equals 1 (A) 0 (B) 1 (C) –1 (D). 2 33π 23. The value of sin–1 cos is 5 3 –7π π –π (A) (B) (C) (D) 5 5 10 10 20/04/2018 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com 38 MATHEMATICS 24. The domain of the function cos–1 (2x – 1) is (A) [0, 1] (B) [–1, 1] (C) ( –1, 1) (D) [0, π] 25. The domain of the function defined by f (x) = sin–1 x –1 is (A) [1, 2] (B) [–1, 1] (C) [0, 1] (D) none of these 2 26. If cos sin –1 + cos –1 x = 0 , then x is equal to 5 1 2 (A) (B) (C) 0 (D) 1 5 5 27. The value of sin (2 tan–1 (.75)) is equal to (A).75 (B) 1.5 (C).96 (D) sin 1.5 –1 3π 28. The value of cos cos is equal to 2 π 3π 5π 7π (A) (B) (C) (D) 2 2 2 2 1 29. The value of the expression 2 sec–1 2 + sin–1 is 2 5 7 (A) (B) (C) (D) 1 6 6 6 4 30. If tan–1 x + tan–1y = , then cot–1 x + cot–1 y equals 5 2 3π (A) (B) (C) (D) π 5 5 5 2 2a –1 1– a 2x 31. If sin –1 2 + cos 2 = tan –1 , where a, x ∈ ]0, 1, then 1+ a 1+ a 1– x 2 the value of x is a 2a (A) 0 (B) (C) a (D) 2 1– a 2 20/04/2018 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com INVERSE TRIGONOMETRIC FUNCTIONS 39 –1 7 32. The value of cot cos is 25 25 25 24 7 (A) (B) (C) (D) 24 7 25 24 1 2 33. The value of the expression tan cos –1 is 2 5 (A) 2+ 5 (B) 5–2 5+2 (C) (D) 5+ 2 2 θ 1– cos θ Hint :tan = 2 1+ cos θ 2x 34. If | x | ≤ 1, then 2 tan–1 x + sin–1 is equal to 1+ x 2 π (A) 4 tan–1 x (B) 0 (C) (D) π 2 35. If cos–1 α + cos–1 β + cos–1 γ = 3π, then α (β + γ) + β (γ + α) + γ (α + β) equals (A) 0 (B) 1 (C) 6 (D) 12 36. The number of real solutions of the equation π 1+ cos 2 x = 2 cos –1 (cos x )in , π is 2 (A) 0 (B) 1 (C) 2 (D) Infinite –1 –1 37. If cos x > sin x, then 1 1 (A) < x≤1 (B) 0≤x< 2 2 1 (C) −1≤ x < (D) x>0 2 20/04/2018 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com Download FREE NCERT Books with Solutions, Examplar from TutorialsDuniya.com 40 MATHEMATICS Fill in the blanks in each of the Exercises 38 to 48. 1 38. The principal value of cos–1 – is__________. 2 3π 39. The value of sin–1 sin is__________. 5 40. If cos (tan–1 x + cot–1 3 ) = 0, then value of x is__________. 1 41. The set of values of sec–1 is__________. 2 42. The principal value of tan–1 3 is__________. 14π 43. The value of cos–1 cos is__________. 3 44. The value of cos (sin–1 x + cos–1 x), |x| ≤ 1 is______. sin –1 x + cos –1 x 3 45. The value of expression tan ,when x = is_________. 2 2 2x 46. If y = 2 tan–1 x + sin–1 for all x, then____< y , n∈N , is valid is 5. π 4 –1 1 π 55. The principal value of sin–1 cos sin is. 2 3 20/04/2018 Download FREE CBSE Question Papers with Solutions from TutorialsDuniya.com TutorialsDuniya.com TutorialsDuniya.com is one of the most trusted academic FREE Study Material website for all School & College Students. Download FREE Study Material for CBSE and ICSE Board School Students of Class 6 to 12 from TutorialsDuniya.com NCERT Books for 2023-24 NCERT Solutions CBSE Notes and ICSE Notes NCERT Exemplar with Solutions CBSE Question Papers with Answers CBSE Practicals and Lab Manual CBSE Topper Answer Sheet CBSE Sample Papers CBSE and ICSE Syllabus CBSE and ICSE Projects Chapter-wise Revision Notes Chapter-wise Assignments FREE Scholarships for Students Reference Books with Solutions College Notes College Projects Delhi University Question Papers Govt. Exams Question Papers FREE Courses with Certificate FREE CUET Study Material Download TutorialsDuniya Android App Please Share these FREE Study Material with your Friends
