Chapter 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) –ππ y = sin–1x [–1,1] ,22 y = cos–1x [–1,1] [0,π] y = cosec–1x R– (–1,1) –ππ ,22 – {0} π y = sec–1x R– (–1,1) [0,π] – 2 –ππ y = tan–1x R ,22 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. (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 1. sin–1 (sin x) = x : cos–1(cos x) = x : tan–1(tan x) = x : cot–1(cot x) = x : sec–1(sec x) = x : cosec–1(cosec x) = x : 2. sin (sin–1 x) = x : cos (cos–1 x) = x : tan (tan–1 x) = x : cot (cot–1 x) = x : sec (sec–1 x) = x : cosec (cosec–1 x) = x : 1–1 –1sin cosec x3. : x–11 –1 cos sec x : x– x ,22 x [0, ] ⎛ –ππ⎞ x ,∈⎜ ⎟⎝ 22 ⎠ x ∈(0, π) x [0, π]– –ππ x ,22 x ∈[–1,1] x ∈[–1,1] x ∈R x ∈R π 2 – {0} x ∈R – (–1,1) x ∈R – (–1,1) x ∈R – (–1,1) x ∈R – (–1,1) –1tan 1 x –1 cot x : x > 0 = – π + cot–1x : x < 0 4. sin–1 (–x) = –sin–1x cos–1 (–x) = π−cos–1x tan–1 (–x) = –tan–1x cot–1 (–x) = π–cot–1x sec–1 (–x) = π–sec–1x cosec–1 (–x) = –cosec–1x : : : : : : x ∈[–1,1] x ∈[–1,1] x ∈R x ∈R x ∈R –(–1,1) x ∈R –(–1,1) π5. sin–1x + cos–1x = : x ∈[–1,1]2 π tan–1x + cot–1x = :2 x ∈R π sec–1x + cosec–1x = :2 xy x ∈R–[–1,1] 6. tan–1x + tan–1y = tan–1 1– xy : xy < 1 ⎛ xy−⎞ ; xy > –1 tan–1x – tan–1y = tan–1 ⎜1+ xy ⎟⎝⎠ 2x7. 2tan–1x = sin–1 2 :–1 ≤ x ≤ 11 x1– x2 2tan–1x = cos–1 : x ≥ 01 x2 2x2tan–1x = tan–1 2 : –1 < x < 11– x2.2 Solved Examples Short Answer (S.A.) 3Example 1 Find the principal value of cos–1x, for x = .2 3 3Solution If cos–1 2 = θ , then cos θ = .2 3Since we are considering principal branch, θ∈ [0, π]. Also, since > 0, θ being in23 π the first quadrant, hence cos–1 = 26. –π Example 2 Evaluate tan–1 sin 2 . –π ⎛ π ⎞ π− ⎛⎞ Solution tan–1 sin = tan–1 ⎜sin ⎜⎟⎟= tan–1(–1) = −.⎝⎠2 ⎝ 2 ⎠ 4 13π cosExample 3 Find the value of cos–1 .6 13π ⎛π –1 ⎛ π⎞Solution cos–1 cos = cos–1 ⎜cos(2 π+)⎞⎟= cos ⎜cos ⎟6 ⎝ 6 ⎠⎝ 6 ⎠π = 6. 9π Example 4 Find the value of tan–1 tan .8 9π ⎛π Solution tan–1 tan = tan–1 tan ⎜π+ ⎟⎞ 8⎝ 8 ⎠⎛ π⎞ π⎛⎞tan–1 tan = ⎜ ⎜⎟⎟= ⎝⎠⎝ 8 ⎠8 Example 5 Evaluate tan (tan–1(– 4)). Solution Since tan (tan–1x) = x, ∀x ∈ R, tan (tan–1(– 4) = – 4. Example 6 Evaluate: tan–1 3 – sec–1 (–2) . Solution tan–1 3– sec–1 (– 2) = tan–1 3– [π – sec–12]π –11 2ππ π⎛⎞−π+cos =− +=−= ⎜⎟.3 2 333⎝⎠ –1 –13sin cos sin Example 7 Evaluate: .2 3 –1 π –1 1 π–1 –1sin cos sin sin cos sinSolution = 6.2 32 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 π –1 π or, tan – θ = x ⇒ tan x = – θ22 –1 ⎛π ⎞⎛π –1 ⎞ –1So tan(cot x) =tanθ=cot – θ=cot −cot x =cot(tan x)⎜⎟⎜ ⎟⎝2 ⎠⎝2 ⎠ The equality is valid for all values of x since tan–1x and cot–1x are true for x ∈ R. ⎛ y ⎞Example 9 Find the value of sec ⎜tan–1 ⎟. ⎝ 2 ⎠ y ⎛ππ⎞ ytan–1 =θθ∈−,Solution Let , where ⎜⎟. So, tanθ = ,2 ⎝22 ⎠2 4 y2 which gives secθ=.2 ⎛ –1 y ⎞ 4 +y2 Therefore, sec ⎜tan ⎟=secθ=. ⎝ 2 ⎠ 2 cos –1 8 Example 10 Find value of tan (cos–1x) and hence evaluate tan .17 Solution Let cos–1x = θ, then cos θ = x, where θ∈ [0,π] 1– cos 2 θ x2 Therefore, tan(cos–1x) = tan θ = =1– . cosθ x ⎛ 8 ⎞2 1– ⎜⎟⎛ 8 ⎞⎝17 ⎠ 15 Hence –1 .tan cos = =⎜⎟⎝ 17 ⎠ 88 17 –1 –5 Example 11 Find the value of sin 2cot 12 ⎛ –5 ⎞−5Solution Let cot–1 ⎜⎟ = y . Then cot y = .⎝ 12 ⎠12 –1 –5 Now sin 2cot 12 = sin 2y 12 –5 ⎡⎛ π ⎞⎤ = 2siny cosy = 2 ⎢since cot y <0, so y∈⎜ , π⎟⎥13 13 ⎣⎝ 2 ⎠⎦ –120 169 –11 –14 Example 12 Evaluate cos sin 4 sec 3 –11 –14 ⎡ –11 –1 3⎤Solution cos sin sec = cos ⎢sin +cos ⎥43⎣ 44⎦ –11 –13 –11 –13 = cos sin cos cos – sin sin sin cos 44 44 3 121 32 1– –1– = 4 44 4 315 17 315– 7 – = 4444 16 . Long Answer (L.A.) 3 17 πExample 13 Prove that 2sin–1 – tan–1 = 5314 33 ⎡−π π⎤ ,Solution Let sin–1 = θ, then sinθ = 5 , where θ∈ ⎢⎥5 ⎣ 22 ⎦ 33Thus tan θ = 4 , which gives θ = tan–1 4 . 3 17Therefore, 2sin–1 – tan–1 531 17 317= 2θ – tan–1 = 2 tan–1 – tan–1 31431 ⎛ 3 ⎞2.⎜⎟ 17–14 –1 tan ⎜⎟ – tan 24 –1 17 = 9 31 = tan–1 −tan ⎜1– ⎟ 7 31 ⎝ 16 ⎠⎛ 24 17 ⎞−⎜⎟–1 731 πtan ⎜⎟= 2417 = ⎜1+ .⎟ 4 ⎝ 731 ⎠Example 14 Prove that cot–17 + cot–18 + cot–118 = cot–13 Solution We have cot–17 + cot–18 + cot–118 111 1 = tan–1 + tan–1 + tan–1 (since cot–1 x = tan–1 , if x > 0)7818x ⎛ 11 ⎞+ –1 ⎜ 78 ⎟ –1 1 11tan ⎜ ⎟+ tan = 11 18 (since x. y = . < 1)⎜1−×⎟ 78⎝ 78 ⎠⎛31 ⎞+⎜⎟–1 1118–13 –1 1 tan ⎜⎟= tan +tan = 31 (since xy < 1)11 18 ⎜1−× ⎟ ⎝ 11 18 ⎠–1 65 –11 = tan = tan = cot–1 3195 3Example 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 ⎛−ππ⎞ ππ⎜ ,⎟, since 1 > ⇒ tan 1 > tan . This givesY tan x tan 1 > 1 π⇒ tan 1 > 1 > ⇒ tan 1 > 1 > tan–1 (1). Example 16 Find the value of 4 ⎛ –1 2 ⎞–1sin 2tan +cos(tan 3) .⎜⎟⎝ 3 ⎠ 2 ⎝22 ⎠44 O –/21 1/4 1/2 2Solution Let tan–1 = x and tan–1 3 = y so that tan x = and tan y = 3.33⎛ –1 2 ⎞–1Therefore, sin ⎜2tan ⎟+cos(tan ⎝ 3 ⎠ = sin (2x) + cos y 22.2tan x 1 31++= 22 =4 21tan x+ 1tan y 1+ 1+(3)+9 12 137 = += .13 226 Example 17 Solve for x –1 ⎛1−x⎞1 –1 tan = tan xx>0,⎜⎟1 x 2⎝+⎠ –1 ⎛1−x⎞ –1 Solution From given equation, we have 2tan ⎜ ⎟=tan x 1 x⎝+⎠ –1–1 –1 ⇒ 2 ⎡tan 1−tan x⎤=tan x⎣⎦ ππ⎛⎞2 =3tan –1 x =tan –1 x⇒ ⎜⎟⇒46⎝⎠ 1 x=⇒ 3 Example 18 Find the values of xwhich 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) +−(1 x)1−x =1−x 2 22(1 x1) −−= 0 ⇒x= 0 or 2x– x2 = 1 – x2 1 ⇒x= 0 or x= 2. Example 19 Solve the equation sin–16x+ sin–1 6 3 x= −π 2 πSolution From the given equation, we have sin–1 6x= −−sin–16 3 x 2 ⎛π ⇒ sin (sin–1 6x) = sin ⎜−−sin–16 3 x ⎞ ⎟⎝ 2 ⎠ ⇒ 6x = – cos (sin–1 63 x) ⇒ . Squaring, we get 1⇒ 144x2 = 1 ⇒ x = ± 12 11Note that x = – is the only root of the equation as x = does not satisfy it.1212Example 20 Show that ⎧ α ⎛π β⎞⎫ sin αcos β 2 tan–1 ⎨tan .tan ⎜ − ⎟⎬=tan –1 ⎩ 2 ⎝42 ⎠⎭ cos α+sin β α ⎛π β⎞2tan .tan −⎜⎟ –1 2 ⎝42 ⎠⎛ –1 –1 2x ⎞ tan since 2tan x = tan Solution L.H.S. = ⎜ 2 ⎟ 2 α 2 ⎛π β⎞ 1−x⎝⎠1tan − tan −⎜⎟2 ⎝42 ⎠ β1tan −α 22tan 2 β1tan + –1 2 = tan ⎛ β⎞2 1tan ⎜− ⎟2 α 21tan ⎜ β⎟− 2 ⎜1tan +⎟ ⎝ 2 ⎠ α⎛ 2 β⎞2tan .1 tan ⎜− ⎟ –1 2 ⎝ 2 ⎠= tan ⎛ β⎞22 α⎛ β⎞2 1+tan − tan 1−tan ⎜⎟ ⎜⎟⎝ 2 ⎠ 2 ⎝ 2 ⎠ α⎛ 2 β⎞2tan 1 tan ⎜− ⎟ –1 2 ⎝ 2 ⎠= tan ⎛ 2 β⎞⎛ 2 α⎞ β⎛ 2 α⎞1+tan 1−tan +2 tan 1+tan ⎜⎟⎜ ⎟ ⎜⎟⎝ 2 ⎠⎝ 2 ⎠ 2 ⎝ 2 ⎠ α 2 β2 tan 1−tan 22 2 α 2 β1tan + 1tan + –12 2= tan αβ1tan 2 2tan − 22+ 2 α 2 β1tan + 1tan +22 –1 ⎛sin αcos β⎞ tan = ⎜⎟ = R.H.S.cos sin α+ 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? ⎛ππ⎞ ⎡ππ⎤(A) ⎜− ,⎟ (B) ⎢− ,⎥⎝ 22 ⎠⎣22 ⎦ ⎛ ππ⎞(C) ⎜− ,⎟ – {0} (D) (0, π)⎝ 22 ⎠Solution (A) is the correct answer. Example 22 The principal value branch of sec–1 is ⎡ππ⎤ ⎧⎫− 0 π−(A) ⎢ , ⎥−{} (B) []0, ⎨⎬π⎣22 2⎦ ⎩⎭ ⎛ ππ⎞(C) (0, π) (D) ⎜− ,⎟⎝ 22 ⎠ Solution (B) is the correct answer. Example 23 One branch of cos–1 other than the principal value branch corresponds to ⎡ππ3 π⎤ ⎧⎫ ] 3,2 ⎨⎬(A) ⎢ ,⎥ (B) [ππ−⎣22 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−π π7 π(A) (B) (C) (D) – 5 510 10 –1 ⎛ 40π+π 3 ⎞ –1 ⎛ 3π⎞Solution (D) is the correct answer. sin ⎜cos ⎟=sin cos ⎜8π+ ⎟⎝ 5 ⎠⎝ 5 ⎠ –1 ⎛ 3π⎞ –1 ⎛ ⎛π 3π⎞⎞sin cos =sin sin −= ⎜⎟ ⎜⎜⎟⎟⎝ 5 ⎠ ⎝⎝25 ⎠⎠ ⎛ ⎛−π⎞⎞ πsin–1 sin =−= ⎜⎜ ⎟⎟ .⎝⎝ 10 ⎠⎠ 10 Example 25 The principal value of the expression cos–1 [cos (– 680°)] is 2π −π ππ2 34(A) (B) (C) (D)9 999 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 2 x1+x(A) (B) 21+x 1 1−x2 (C) (D) . x x Solution (D) is the correct answer. Let sin–1 x = θ, then sinθ = x 1 1 ⇒ cosec θ = ⇒ cosec2θ = 2x x 11−x2 ⇒ 1 + cot2 θ = 2 ⇒ cotθ = . x x πExample 27 If tan–1x = for some x ∈ R, then the value of cot–1x is10 π 2π 3π 4π(A) (B) (C) (D)5 555 πSolution (B) is the correct answer. We know tan–1x + cot–1x = . Therefore2 ππ cot–1x = – 2 10 ππ2π⇒ cot–1x = – = .2 10 5 Example 28 The domain of sin–1 2x is (A) [0, 1] (B) [– 1, 1] ⎡11 ⎤(C) ⎢−,⎥ (D) [–2, 2]⎣22 ⎦ Solution (C) is the correct answer. Let sin–12x = θ so that 2x = sin θ. 11−≤≤.Now – 1 ≤ sin θ≤ 1, i.e.,– 1 ≤ 2x ≤ 1 which gives x22 ⎛−3 ⎞ Example 29 The principal value of sin–1 ⎜⎟ is2⎝⎠2ππ 4π 5π(A) − (B) − (C) (D) .333 3 Solution (B) is the correct answer. –1 ⎛− 3 ⎞ –1 ⎛ π⎞ –1 ⎛π⎞ πsin =sin –sin =–sin sin = –⎜⎟⎜⎟ ⎜⎟ . ⎝ 2 ⎠⎝ 3 ⎠⎝ 3 ⎠ 3 Example 30 The greatest and least values of (sin–1x)2 + (cos–1x)2 are respectively 5π2 π2 π −π(A) and (B) and 48 22 22 2π−π π(C) and (D) and0 .44 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 –1 ⎛π –1 ⎞−2sin x −sin x= ⎜⎟4 ⎝ 2 ⎠ π22–1 –1= −πsin x +2 sin x 4 () 2⎡ –1 2 π –1 π⎤ = ⎢( x − sin x2sin )+ ⎥28⎣⎦ ⎡⎛ π⎞2 π2 ⎤ 2sin⎢ –1 x − +⎥= ⎜⎟ .⎝ 4 ⎠ 16⎢⎣ ⎥⎦ 2⎛π2 ⎞π2 ⎡⎛−π π⎞2 π⎤ 2 −+Thus, the least value is 2⎜⎟i.e. and the Greatest value is ⎢⎜ ⎟⎥ ,16 8 ⎢⎝ 24 ⎠ 16 ⎥⎝⎠⎣⎦ 5π2 i.e. .4 Example 31 Let θ = sin–1 (sin (– 600°), then value of θ is ππ 2π −π2(A) (B) (C) (D) .323 3 Solution (A) is the correct answer. –1 ⎛ π⎞ –1 ⎛− π 10 ⎞sin sin −600 ×=sin sin ⎜ ⎟⎜⎟⎝ 180 ⎠⎝ 3 ⎠–1 ⎡⎛ 2π⎞⎤ –1 ⎛ 2π⎞sin −sin ⎜4π− sin sin = ⎢ ⎟⎥ = ⎜⎟⎣⎝ 3 ⎠⎦⎝ 3 ⎠–1 ⎛ ⎛ π⎞⎞ –1 ⎛ π⎞πsin 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.. −≤ ≤ 1 xie 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 (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. πExample 36 If sin–1 x + sin–1 y = , then value of cos–1 x + cos–1 y is2 π 2π(A) (B) π (C) 0 (D)2 3 πSolution (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 ⎛ –13 –11 ⎞Example 37 The value of tan ⎜cos +tan ⎟is⎝ 54 ⎠19 8193(A) (B) (C) (D)8 1912 4 ⎛ –13 –11 ⎞⎛ –14 –1 1 ⎞Solution (A) is the correct answer. tan ⎜cos +tan ⎟= tan ⎜tan +tan ⎟⎝ 54 ⎠⎝ 34 ⎠ = tan tan –1 –1 41 19 1934 tan tan 41 8 81 34 ⎛ ⎞+⎜ ⎟ ⎛ ⎞ = =⎜ ⎟ ⎜ ⎟⎝ ⎠⎜ ⎟−× ⎝ ⎠ . Example 38 The value of the expression sin [cot–1 (cos (tan–1 1))] is (A) 0 (B) 1 (C) 1 3 (D) 2 3 . Solution (D) is the correct answer. sin [cot–1 (cos 4 π )] = sin [cot–1 1 2 ]= –1 2 2sin sin 3 3 ⎡ ⎤ =⎢ ⎥ ⎣ ⎦ Example 39 The equation tan–1x – cot–1x = tan–1 1 3 ⎛ ⎞ ⎜ ⎟⎝ ⎠ has (A) no solution (B) unique solution (C) infinite number of solutions (D) two solutions Solution (B) is the correct answer. We have tan–1x – cot–1x = 6 π and tan–1x + cot–1x = Adding them, we get 2tan–1x = 2 3 π ⇒ tan–1x = 3 π i.e., x = Example 40 If α≤2 sin–1x + cos–1x ≤β, then (A) ,2 2 −π πα= β= (B) (C) 3 ,2 2 −π πα= β= (D) 2 π α=0, β=π 0, 2α= β= π −π πSolution (B) is the correct answer. We have ≤ sin–1 x ≤22 −ππ πππ⇒ + ≤ sin–1x + ≤ +22 222 ⇒ 0 ≤ sin–1x + (sin–1x + cos–1x) ≤ π ⇒ 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.) –1 ⎛ 5π⎞ –1 ⎛ 13π⎞tan tan +cos cos 1. Find the value of ⎜⎟⎜ ⎟⎠.⎝ 6 ⎠⎝ 6 2. Evaluate cos cos –1 .3. Prove that cot –2cot –13 7.26 4 –11 –1 1 –1– 4. Find the value of tan – cot tan sin .33 2 ⎛ 2π⎞tan 5. Find the value of tan–1 ⎜⎟⎝ 3 ⎠ . – ⎛ –4 ⎞tan 6. Show that 2tan–1 (–3) = + –1 ⎜⎟ .2⎝ 3 ⎠ 7. Find the real solutions of the equation –1 2 πtan –1 ( ) ++sin xx 1xx 1 ++= .2 ⎛ –1 1 ⎞–18. Find the value of the expression sin ⎜2 tan ⎟+cos(tan 22 ).⎝ 3 ⎠ π 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 ⎞ cos 2tan =sin 4tan 10. Show that ⎜ ⎟⎜⎟.⎝ 7 ⎠⎝ 3 ⎠–1 ⎛ –1 3 ⎞11. Solve the following equation cos(tan x)=sin ⎜cot .⎟⎝ 4 ⎠ Long Answer (L.A.) –1 1 x2 1– x2 1 –1 2tan cos x12. Prove that 2 242 1 x– 1– x –134 –3 cos x13. Find the simplified form of cos sin x, where x∈ ,.55 44 –1 8 –13 –1 77 14. Prove that sin sin sin .17 5 85 –1 5 –13 –163 15. Show that sin cos tan13 516 . –11 –12 −11tan +tan =sin 16. Prove that 49 11–1 –117. Find the value of 4tan – tan .5 239 1 34–7 4 +718. Show that tan sin –1 and justify why the other value2 433 is ignored? 19. If a1, a2, a3,...,an is an arithmetic progression with common difference d, then evaluate the following expression. ⎡⎛⎞⎛⎞⎛⎞ ⎛ ⎞⎤–1 d –1 d –1 d –1 dtan tan +tan +tan ++ tan .⎢⎜⎟⎜ ⎟⎜ ⎟... ⎜ ⎟⎥1+aa 1+aa 1+aa 1+aa⎣⎝ 12 ⎠⎝ 23 ⎠⎝ 34 ⎠⎝ n–1 n ⎠⎦ 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? ⎡–ππ⎤(A) ⎢ ,⎥ (B) (0, π)⎣22 ⎦ π⎧⎫(C) [0, π] (D) (0, π) – ⎨⎬2⎩⎭ 21. Which of the following is the principal value branch of cosec–1x? ⎛–ππ π⎞ ⎧⎫(A) ⎜ ,⎟ (B) [0, π] – ⎨⎬22 2⎝ ⎠ ⎩⎭ ⎡–ππ⎤ ⎡–ππ⎤(C) ⎢ ,⎥ (D) ⎢ ,⎥– {0}⎣22 ⎦ ⎣22 ⎦22. If 3tan–1 x + cot–1 x = π, then x equals 1 (A) 0 (B)1 (C)–1 (D) 2. 33 cos 23. The value of sin–1 is53π –7ππ –π(A) (B) (C) (D)5 510 10 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 (A) [1, 2] (B) [–1, 1] (C) [0, 1] (D) none of these 26. If cos –1 –12sin cos 05 x ⎛ ⎞+ =⎜ ⎟⎝ ⎠ , then x is equal to (A) 1 5 (B) 2 5 (C) 0 (D) 1 27. The value of sin (2 tan–1 (.75)) is equal to (A) .75 (B) 1.5 (C) .96 (D) sin 1.5 28. The value of –1cos 3 cos 2 is equal to (A) 2 π (B) 3 2 π (C) 5 2 π (D) 7 2 π 29. The value of the expression 2 sec–1 2 + sin–1 1 2 is π5π7π (A) (B) (C) (D) 16 66 4π 30. If tan–1 x + tan–1y = , then cot–1 x + cot–1 y equals5 π2π3 (A) 5 (B) 5 (C) 5 (D) π 31. If sin–1 2 2 1 a a –1cos 2 2 1– 1 a a –1tan 2 2 1– x x , where a, x ∈ ]0, 1, then the value of x is (A) 0 (B) 2 a (C) a (D) 2 2 1– a a cos –1 7 32. The value of cot is2525 2524 7 (A) (B) (C) (D)24 7 2524 1 –12 33. The value of the expression tan cos is2 5(A) 25 (B) 5–2 52(C) (D) 522 ⎡ θ ⎢Hint :tan = 2⎣ 2x 34. If | x | ≤ 1, then 2 tan–1 x + sin–1 2 is equal to1 x(A) 4 tan–1 x (B) 0 (C) (D) π2 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 –1 ⎡π ⎤(cos x)in ,⎢ π⎥is⎣2 ⎦(A) 0 (B) 1 (C) 2 (D) Infinite 37. If cos–1x > sin–1x, then 1 1<≤ xx1 (B) 0 ≤<(A) 2 2 −≤< (C) 1 x1 (D) x > 02 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______ . –1 –1⎛sin x +cos x ⎞ 45. The value of expression tan ⎜⎟,when x = 3 is_________.2⎝⎠22x 46. If y = 2 tan–1 x + sin–1 2 for all x, then____< y <____.1 x⎛x −y ⎞ 47. The result tan–1x – tan–1y = tan–1 ⎜⎟is true when value of xy is _____⎝1+xy ⎠48. The value of cot–1 (–x) for all x ∈ R in terms of cot–1x is _______. State True or False for the statement in each of the Exercises 49 to 55. 49. All trigonometric functions have inverse over their respective domains. 50. The value of the expression (cos–1 x)2 is equal to sec2 x. 51. The domain of trigonometric functions can be restricted to any one of their branch (not necessarily principal value) in order to obtain their inverse functions. 52. The least numerical value, either positive or negative of angle θ is called principal value of the inverse trigonometric function. 53. The graph of inverse trigonometric function can be obtained from the graph of their corresponding trigonometric function by interchanging x and y axes. n π54. The minimum value of n for which tan–1 > ,n∈N , is valid is 5.π 4 ⎡⎛ –1 1 ⎞⎤ π55. The principal value of sin–1 ⎢cos ⎜sin ⎟⎥ is .⎣⎝ 2 ⎠⎦3

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