Chapter 2 – Inverse Trigonometric Functions

Find the principal value of cos–1x, for x = . INVERSE TRIGONOMETRIC FUNCTIONS 21  3 Solution If cos  2  = θ , then cos θ = –1 . Since we are considering principal branch, θ ∈ [0, π]. Also, since > 0, θ being in  3 π the first quadrant, hence cos–1  2  = .   6   –π  

Evaluate tan–1  sin    .  2    –π     π  π Solution tan–1  sin    = tan–1  − sin    = tan–1(–1) = − .  2     2 4  13π 

Find the value of cos–1  cos . 6   13π   π  –1  π Solution cos–1  cos  = cos–1  cos (2π + )  = cos  cos 

Find the value of tan–1  tan  .  9π   π Solution tan–1  tan  = tan–1 tan  π + 

Evaluate tan (tan (– 4)). Solution Since tan (tan–1x) = x, ∀ x ∈ R, tan (tan–1(– 4) = – 4.

Evaluate: tan–1 3 – sec–1 (–2) . 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 3  

Evaluate: sin cos  sin 2   . –1       –1 3   –1   π  –1  1  π Solution sin cos  sin  = sin cos    = sin   = . –1    2     3  2 6

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  π or, tan  – θ = x ⇒ tan x = – θ –1 2 2 π  π –1  So tan(cot x) = tan θ = cot  – θ  = 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 

Find the value of sec  tan .  2 y  π π y Solution Let tan –1 =θ , where θ ∈  − ,  . So, tanθ = , 2  2 2 2 4 + y2 which gives secθ= .  –1 y  4 + y2 Therefore, sec  tan  = secθ = .  2 2  –1 8 

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] INVERSE TRIGONOMETRIC FUNCTIONS 23 1 – cos 2 θ 1 – x2 Therefore, tan(cos–1x) = tan θ = = . cosθ x  8 1–   Hence  8   17  15 . tan  cos –1  = =  17  8 8  –1  –5  

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    π  = 2siny cosy = 2     since cot y < 0, so y ∈ 2 , π   13 13    –120 =  –1 1 4

Evaluate cos sin + sec –1   4 3  –1 1 4  1 3 Solution cos sin + sec –1  = cos sin –1 + cos –1   4 3  4 4  –1 1   –1 3   –1 1   –1 3  = cos  sin  cos  cos  – sin  sin  sin  cos   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 . 24 MATHEMATICS

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 

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   7+8  –1 1  + tan 1 1 –1 tan  . < 1) = 1 1 (since x . y =  1− ×  18 7 8  7 8 INVERSE TRIGONOMETRIC FUNCTIONS 25  3 1   +  –1 3 –1 1 tan –1  11 18  = tan + tan = (since xy < 1)

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 tan 1 > 1 π ⇒ tan 1 > 1 > ⇒ tan 1 > 1 > tan–1 (1).

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 tan x 1 3 + 1 + ( ) = 1 + tan x 2 = 4 1+ tan 2 y 1+ 1+ 3

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 ⇒ x=

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 – x − 1− x ) = 0 2 2 ⇒x = 0 or 2x – x2 = 1 – x2 ⇒x = 0 or x= . π

Solve the equation sin–16x + sin–1 6 3 x = − π Solution From the given equation, we have sin–1 6x = − − sin 6 3 x –1 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 ⇒ 144x2 = 1 ⇒ x= ± 1 1 Note that x = – is the only root of the equation as x = does not satisfy it. 12 12

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  2  2 = tan –1 2 2  β 2 α  β 1 + tan  − tan 1 − tan   2 2  2 28 MATHEMATICS α  2 β 1 − tan  2 tan 2  2 = tan –1  2 β  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 

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.

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 INVERSE TRIGONOMETRIC FUNCTIONS 29 Solution (B) is the correct answer.

One branch of cos–1 other than the principal value branch corresponds to  π 3π  3π 

• (A)2, 2 
• (B)[ π , 2π] −     2
• (C)(0, π)
• (D)[2π, 3π] Solution (D) is the correct answer. –1   43π  

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

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° = .

The value of cot (sin–1x) is 1+ x 2 x

• (A)
• (B)x 1+ x 2

If tan–1x = for some x ∈ R, then the value of cot–1x is 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 π π cot–1x = – 2 10 π π 2π ⇒ cot–1x = – = . 2 10 5

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

The principal value of sin–1  2  is   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

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 ( ) = − π sin –1 x + 2 sin –1 x  π –1 π2  ( ) =  2 sin –1 x − sin x +   2 8   –1 π  π2  =  2 sin x −  +   4  16  .  π2  π2  −π π  2 π2  Thus, the least value is   2 i.e. and the Greatest value is  2 − 4  + 16  ,  16  8    5π2 i.e. .

Let θ = sin–1 (sin (– 600°), then value of θ is

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

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 

The domain of the function defined by f (x) = sin–1x + cosx is 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].

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

If sin–1 x + sin–1 y = , then value of cos–1 x + cos–1 y is 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 = . π –1  π –1  π Therefore,  – cos x  +  – cos y  = 2  2  2 π ⇒ cos–1x + cos–1y = .  –1 3 1

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  4 1 Solution (A) is the correct answer. tan  cos + tan –1  = tan  tan –1 + tan –1   5 4   3 4

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 

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 = π ⇒ tan–1x = i.e., x = 3 .

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 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

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.