Question 1

sin^-1(-1/2)

Accepted Answer

(-π)/6. The principal value branch of sin^-1 is [-(π)/2,(π)/2], so we need the angle y in that interval with sin y=-1/2. Since sin (π)/6=1/2 and sine is an odd function, sin (-(π)/6)=- sin (π)/6=-1/2, and -(π)/6 does lie in [-(π)/2,(π)/2]. The other standard angle with sine -1/2, namely (7π)/6, is outside the branch and is rejected. sin^-1(-1/2)=-(π)/6

Question 2

cos^-1((√ 3 )/2)

Accepted Answer

(π)/6. The principal value branch of cos^-1 is [0,π], so we need y∈[0,π] with cos y=(√ 3 )/2. From the standard table, cos (π)/6=(√ 3 )/2, and (π)/6∈[0,π]. (The angle -(π)/6 has the same cosine but is not in the branch.) cos^-1((√ 3 )/2)=(π)/6

Question 3

cosec^-1(2)

Accepted Answer

(π)/6. The principal value branch of cosec^-1 is [-(π)/2,(π)/2]- 0. Let y= cosec^-1(2). Then cosec y=2⇔ sin y=1/2. The angle in [-(π)/2,(π)/2]- 0 with sin y=1/2 is y=(π)/6 (the second solution (5π)/6 lies outside the branch). cosec^-1(2)=(π)/6

Question 4

tan^-1(-√ 3 )

Accepted Answer

(-π)/3. The principal value branch of tan^-1 is the open interval (-(π)/2,(π)/2), so we need y there with tan y=-√ 3. Since tan (π)/3=√ 3 and tangent is odd, tan (-(π)/3)=- tan (π)/3=-√ 3, and -(π)/3∈(-(π)/2,(π)/2). The angle (2π)/3 also has tangent -√ 3 but is not in the branch. tan^-1(-√ 3 )=-(π)/3

Question 5

cos^-1(-1/2)

Accepted Answer

(2 π)/3. The principal value branch of cos^-1 is [0,π], so we need y∈[0,π] with cos y=-1/2. Note that cosine is negative in the second quadrant, so y lies between (π)/2 and π — writing -(π)/3 here would be the standard error. Using cos (π-θ)=- cos θ with cos (π)/3=1/2, cos (π-(π)/3)= cos (2π)/3=-1/2, and (2π)/3∈[0,π]. cos^-1(-1/2)=(2π)/3

Question 6

tan^-1(-1)

Accepted Answer

-(π)/4. The principal value branch of tan^-1 is (-(π)/2,(π)/2), so we need y there with tan y=-1. Since tan (π)/4=1 and tangent is odd, tan (-(π)/4)=-1, -(π)/4∈(-(π)/2,(π)/2). (3π)/4 also satisfies tan y=-1 but lies outside the branch. tan^-1(-1)=-(π)/4

Question 7

sec^-1(2/(√ 3 ))

Accepted Answer

(π)/6. The principal value branch of sec^-1 is [0,π]- (π)/2. Let y= sec^-1(2/(√ 3 )). Then sec y=2/(√ 3 )⇔ cos y=(√ 3 )/2. The angle in [0,π]- (π)/2 with cos y=(√ 3 )/2 is y=(π)/6. sec^-1(2/(√ 3 ))=(π)/6

Question 8

cot^-1(√ 3 )

Accepted Answer

(π)/6. The principal value branch of cot^-1 is the open interval (0,π). Let y= cot^-1(√ 3 ). Then cot y=√ 3 ⇔ tan y=1/(√ 3 ), and the angle in (0,π) with tan y=1/(√ 3 ) is y=(π)/6. cot^-1(√ 3 )=(π)/6

Question 9

cos^-1(-1/(√ 2 ))

Accepted Answer

(3 π)/4. The principal value branch of cos^-1 is [0,π], so we need y∈[0,π] with cos y=-1/(√ 2 ). A negative cosine forces y into the second quadrant, so the answer is π- (the acute angle), not a negative angle. Since cos (π)/4=1/(√ 2 ), the identity cos (π-θ)=- cos θ gives cos (π-(π)/4)= cos (3π)/4=-1/(√ 2 ), (3π)/4∈[0,π]. cos^-1(-1/(√ 2 ))=(3π)/4

Question 10

cosec^-1(-√ 2 )

Accepted Answer

(-π)/4. The principal value branch of cosec^-1 is [-(π)/2,(π)/2]- 0. Let y= cosec^-1(-√ 2 ). Then cosec y=-√ 2 ⇔ sin y=-1/(√ 2 ). Since sin (π)/4=1/(√ 2 ) and sine is odd, sin (-(π)/4)=-1/(√ 2 ), and -(π)/4 lies in the branch. cosec^-1(-√ 2 )=-(π)/4

Question 11

tan^-1(1)+ cos^-1-1/2+ sin^-1-1/2

Accepted Answer

(3 π)/4. Evaluate each inverse function separately, each in its own principal value branch, and only then add. tan^-1(1): need y∈(-(π)/2,(π)/2) with tan y=1, so tan^-1(1)=(π)/4. cos^-1(-1/2): need y∈[0,π] with cos y=-1/2; since cos (π-(π)/3)=-1/2, this is (2π)/3. sin^-1(-1/2): need y∈[-(π)/2,(π)/2] with sin y=-1/2, which is -(π)/6. Note the two negative arguments give answers of opposite sign — cos^-1 never returns a negative angle, sin^-1 does. Adding, with denominator 12, (π)/4+(2π)/3-(π)/6=(3π+8π-2π)/12=(9π)/12=(3π)/4. Value: (3π)/4.

Question 12

cos^-1 1/2+2 sin^-1 1/2

Accepted Answer

(2 π)/3. Evaluate each inverse function in its principal value branch first. cos^-11/2: the angle y∈[0,π] with cos y=1/2 is y=(π)/3. sin^-11/2: the angle y∈[-(π)/2,(π)/2] with sin y=1/2 is y=(π)/6. Therefore cos^-11/2+2 sin^-11/2=(π)/3+2·(π)/6=(π)/3+(π)/3=(2π)/3. Value: (2π)/3.

Question 13

If sin^-1 x=y, then (A) 0 ≤ y ≤ π (B) -(π)/2 ≤ y ≤ (π)/2 (C) 0<y<π (D) -(π)/2<y<(π)/2

Accepted Answer

B. By definition, sin^-1 is the inverse of the restriction of sin to the principal value branch [-(π)/2,(π)/2], on which sine is one-one and onto [-1,1]. So if sin^-1x=y, then by definition y∈[-(π)/2,(π)/2]. The interval is closed, not open: at x=1 we get y=(π)/2 and at x=-1 we get y=-(π)/2, both of which are attained. Hence (D) is wrong. Options (A) and (C) describe the range of cos^-1 and of cot^-1, not of sin^-1. Correct option: (B) -(π)/2≤ y≤(π)/2.

Question 14

tan^-1 √ 3 - sec^-1(-2) is equal to (A) π (B) -(π)/3 (C) (π)/3 (D) (2 π)/3

Accepted Answer

B. Take the two terms separately, each in its own principal value branch. tan^-1√ 3: need y∈(-(π)/2,(π)/2) with tan y=√ 3, so tan^-1√ 3 =(π)/3. sec^-1(-2): the branch is [0,π]- (π)/2, and sec y=-2⇔ cos y=-1/2. Since cos (π-(π)/3)=-1/2, we get sec^-1(-2)=(2π)/3. (Writing -(π)/3 here is the usual slip; sec^-1 never returns a negative angle.) Hence tan^-1√ 3 - sec^-1(-2)=(π)/3-(2π)/3=-(π)/3. Correct option: (B) -(π)/3.

Question 15

3 sin^-1 x= sin^-1(3 x-4 x^3), x ∈[-1/2, 1/2]

Accepted Answer

Let θ= sin^-1x, so that x= sin θ. Settle the range before the algebra. For x∈[-1/2,1/2] we get θ∈[-(π)/6,(π)/6], and therefore 3θ∈[-(π)/2,(π)/2]. By the triple-angle identity sin 3θ=3 sin θ-4 sin^3θ, sin 3θ=3x-4x^3. The principal-value branch of sin^-1 is [-(π)/2,(π)/2], and 3θ has just been shown to lie in it, so sin^-1 undoes sin here: sin^-1(3x-4x^3)=3θ=3 sin^-1x. Hence 3 sin^-1x= sin^-1(3x-4x^3) for all x∈[-1/2,1/2]. The restriction is doing real work: once x 1/2, 3θ leaves the principal branch and the equality breaks.

Question 16

3 cos^-1 x= cos^-1(4 x^3-3 x), x ∈[1/2, 1]

Accepted Answer

Let θ= cos^-1x, so that x= cos θ. Range first. cos^-1 is decreasing, so x∈[1/2,1] gives θ∈[0,(π)/3], and therefore 3θ∈[0,π]. By the triple-angle identity cos 3θ=4 cos^3θ-3 cos θ, cos 3θ=4x^3-3x. The principal-value branch of cos^-1 is [0,π], and 3θ lies in it, so cos^-1(4x^3-3x)=3θ=3 cos^-1x. Hence 3 cos^-1x= cos^-1(4x^3-3x) for x∈[1/2,1]. Without the restriction 3θ can exceed π and the identity fails.

Question 17

tan^-1 √ 1+x^2 -1 x, x ≠ 0

Accepted Answer

1/2 tan^-1 x. Substitute x= tan θ, i.e. θ= tan^-1x, with θ∈(-(π)/2,(π)/2) and θ≠ 0 (since x≠ 0 ). Then 1+x^2= sec^2θ. On (-(π)/2,(π)/2) we have cos θ 0, so sec θ 0 and the positive square root is √ 1+x^2 = sec θ. Therefore, using 1- cos θ=2 sin^2(θ)/2 and sin θ=2 sin (θ)/2 cos (θ)/2, √ 1+x^2 -1 x =( sec θ-1)/( tan θ)=(1- cos θ)/( sin θ)=(2 sin^2(θ)/2)/(2 sin (θ)/2 cos (θ)/2)= tan (θ)/2. Since θ∈(-(π)/2,(π)/2), we have (θ)/2∈(-(π)/4,(π)/4), which sits inside the principal branch of tan^-1, so tan^-1( tan (θ)/2)=(θ)/2. tan^-1 √ 1+x^2 -1 x =1/2 tan^-1x, x≠ 0.

Question 18

tan^-1(√ (1- cos x)/(1+ cos x) ), 0<x<π

Accepted Answer

x/2. Use the half-angle forms 1- cos x=2 sin^2x/2 and 1+ cos x=2 cos^2x/2: (1- cos x)/(1+ cos x)=(2 sin^2x/2)/(2 cos^2x/2)= tan^2x/2. Now take the square root, and this is the step to be careful with: √ t^2 = t. Given 0<x<π, we have x/2∈(0,(π)/2), where /2 0, so √ tan^2x/2 = /2. Since x/2∈(0,(π)/2) (-(π)/2,(π)/2), the principal branch of tan^-1 returns the angle itself: tan^-1(√ (1- cos x)/(1+ cos x) )=x/2, 0<x<π.

Question 19

tan^-1(( cos x- sin x)/( cos x+ sin x)), (-π)/4<x<(3 π)/4

Accepted Answer

(π)/4-x. Do not divide by cos x here: x=(π)/2 lies in the given interval. Write both parts as single trigonometric functions instead: cos x- sin x=√ 2 cos (x+(π)/4), cos x+ sin x=√ 2 sin (x+(π)/4). For -(π)/4<x<(3π)/4 we have x+(π)/4∈(0,π), so sin (x+(π)/4) 0 and the quotient is defined throughout. Hence ( cos x- sin x)/( cos x+ sin x)= cot (x+(π)/4)= tan ((π)/2-x-(π)/4)= tan ((π)/4-x). Check the branch: -(π)/4<x<(3π)/4 gives (π)/4-x∈(-(π)/2,(π)/2), the principal branch of tan^-1. Therefore tan^-1(( cos x- sin x)/( cos x+ sin x))=(π)/4-x, -(π)/4<x<(3π)/4.

Question 20

tan^-1 x √ a^2-x^2, x <a

Accepted Answer

sin^-1 x/a. Since x <a, necessarily a 0. Substitute x=a sin θ, i.e. θ= sin^-1x/a; as x/a <1, θ∈(-(π)/2,(π)/2). Then a^2-x^2=a^2 cos^2θ, and since cos θ 0 on that interval the positive root is √ a^2-x^2 =a cos θ. So x √ a^2-x^2 =(a sin θ)/(a cos θ)= tan θ. Because θ already lies in (-(π)/2,(π)/2), tan^-1 x √ a^2-x^2 =θ= sin^-1x/a, x <a.

Question 21

tan^-1((3 a^2 x-x^3)/(a^3-3 a x^2)), a 0; -a/(√ 3 )<x<a/(√ 3 )

Accepted Answer

3 tan^-1 x/a. Divide numerator and denominator by a^3 (legitimate, a 0 ): (3a^2x-x^3)/(a^3-3ax^2)=(3(x/a)-(x/a)^3)/(1-3(x/a)^2). Put x/a= tan θ, i.e. θ= tan^-1x/a. By the triple-angle formula for the tangent, tan 3θ=(3 tan θ- tan^3θ)/(1-3 tan^2θ), so the expression inside tan^-1 is exactly tan 3θ. Range check, which is what the given interval is for: -a/(√ 3 )<x<a/(√ 3 ) means tan θ <1/(√ 3 ), so θ <(π)/6 and hence 3θ∈(-(π)/2,(π)/2). (The same interval keeps 1-3 tan^2θ≠ 0, i.e. the denominator a(a^2-3x^2)≠ 0.) Therefore tan^-1((3a^2x-x^3)/(a^3-3ax^2))=3θ=3 tan^-1x/a.

Question 22

tan^-1[2 cos (2 sin^-1 1/2)]

Accepted Answer

(π)/4. Work from the inside out, using principal values at each stage. sin^-11/2 is the angle in [-(π)/2,(π)/2] whose sine is 1/2, so sin^-11/2=(π)/6. Then 2 sin^-11/2=(π)/3, cos (π)/3=1/2, 2 cos (2 sin^-11/2)=2·1/2=1. Finally tan^-11 is the angle in (-(π)/2,(π)/2) with tangent 1: tan^-1[2 cos (2 sin^-11/2)]= tan^-11=(π)/4.

Question 23

tan 1/2[ sin^-1 (2 x)/(1+x^2)+ cos^-1 (1-y^2)/(1+y^2)], x <1, y 0 and x y<1

Accepted Answer

(x+y)/(1-x y). Convert both inverse functions to tan^-1 using the standard identities sin^-12x/(1+x^2)=2 tan^-1x ( x ≤ 1), cos^-1(1-y^2)/(1+y^2)=2 tan^-1y (y≥ 0). The given conditions x <1 and y 0 are exactly what makes each identity applicable (for x 1 or y<0 the right-hand sides would need a correction term). Hence the bracket equals 2 tan^-1x+2 tan^-1y, and half of it is 1/2[ sin^-12x/(1+x^2)+ cos^-1(1-y^2)/(1+y^2)]= tan^-1x+ tan^-1y. Write A= tan^-1x, B= tan^-1y, so tan A=x, tan B=y, and apply the addition formula tan (A+B)=( tan A+ tan B)/(1- tan A tan B)=(x+y)/1-xy, which is finite because xy<1. tan 1/2[ sin^-12x/(1+x^2)+ cos^-1(1-y^2)/(1+y^2)]=(x+y)/1-xy.

Question 24

sin^-1( sin (2 π)/3)

Accepted Answer

(π)/3. sin^-1( sin θ)=θ only when θ lies in the principal branch [-(π)/2,(π)/2]. Here (2π)/3 (π)/2, so the answer is not (2π)/3; replace the angle by one in the branch with the same sine. Using sin (π-θ)= sin θ, sin (2π)/3= sin (π-(2π)/3)= sin (π)/3, (π)/3∈[-(π)/2,(π)/2]. Therefore sin^-1( sin (2π)/3)=(π)/3.

Question 25

tan^-1( tan (3 π)/4)

Accepted Answer

(-π)/4. tan^-1( tan θ)=θ only for θ∈(-(π)/2,(π)/2), and (3π)/4 is not in that interval, so the answer is not (3π)/4. The tangent has period π, so subtract π to bring the angle into the principal branch: tan (3π)/4= tan ((3π)/4-π)= tan (-(π)/4), -(π)/4∈(-(π)/2,(π)/2). Therefore tan^-1( tan (3π)/4)=-(π)/4.

Question 26

tan ( sin^-1 3/5+ cot^-1 3/2)

Accepted Answer

17/6. Name the two angles and find their tangents. Let A= sin^-13/5. Then A∈(0,(π)/2) with sin A=3/5, so cos A=√ 1-9/25 =4/5 (positive, as A is in the first quadrant) and tan A=3/4. Let B= cot^-13/2, so cot B=3/2 and tan B=2/3. Apply the addition formula: tan (A+B)=( tan A+ tan B)/(1- tan A tan B)=(3/4+2/3)/(1-3/4·2/3)=(17/12)/( 1 2 )= 17 6. tan ( sin^-13/5+ cot^-13/2)=17/6.

Question 27

cos^-1( cos (7 π)/6) is equal to (A) (7 π)/6 (B) (5 π)/6 (C) (π)/3 (D) (π)/6

Accepted Answer

B. cos^-1( cos θ)=θ only for θ∈[0,π], the principal branch. Here (7π)/6 π, so option (A) is the trap; convert to an angle in [0,π] with the same cosine. Using cos (2π-θ)= cos θ, cos (7π)/6= cos (2π-(7π)/6)= cos (5π)/6, (5π)/6∈[0,π]. Therefore cos^-1( cos (7π)/6)=(5π)/6. Answer: (B) (5π)/6.

Question 28

sin ((π)/3- sin^-1(-1/2)) is equal to (A) 1/2 (B) 1/3 (C) 1/4 (D) 1

Accepted Answer

D. First evaluate the inverse function as a principal value. sin^-1 takes values in [-(π)/2,(π)/2], and sin (-(π)/6)=-1/2, so sin^-1(-1/2)=-(π)/6. Watch the double negative in the subtraction: (π)/3- sin^-1(-1/2)=(π)/3+(π)/6=(π)/2. Hence the expression is sin (π)/2=1. Answer: (D) 1.

Question 29

tan^-1 √ 3 - cot^-1(-√ 3 ) is equal to (A) π (B) -(π)/2 (C) 0 (D) 2 √ 3

Accepted Answer

B. tan^-1√ 3 =(π)/3, since (π)/3∈(-(π)/2,(π)/2) and tan (π)/3=√ 3. For the second term, the principal branch of cot^-1 is (0,π), so cot^-1 of a negative number is an obtuse angle, not a negative one. Using cot^-1(-t)=π- cot^-1t, cot^-1(-√ 3 )=π- cot^-1√ 3 =π-(π)/6=(5π)/6, and indeed cot (5π)/6=-√ 3. Therefore tan^-1√ 3 - cot^-1(-√ 3 )=(π)/3-(5π)/6=(2π-5π)/6=-(π)/2. Answer: (B) -(π)/2.

Question 30

cos^-1( cos (13 π)/6)

Accepted Answer

(π)/6. The principal value branch of cos^-1 is [0,π], so the identity cos^-1( cos θ)=θ may be used only when θ∈[0,π]. Here (13π)/6 π, so the angle must be reduced first. Using the 2π -periodicity of cosine, (13π)/6=2π+(π)/6 ⇒ cos (13π)/6= cos (2π+(π)/6)= cos (π)/6. Now (π)/6 does lie in [0,π], so the identity applies: cos^-1( cos (13π)/6)= cos^-1( cos (π)/6)=(π)/6.

Question 31

tan^-1( tan (7 π)/6)

Accepted Answer

(π)/6. The principal value branch of tan^-1 is (-(π)/2,(π)/2), and (7π)/6 is not in it, so tan^-1( tan (7π)/6)≠(7π)/6. Reduce the angle using the fact that tangent has period π (not 2π ): tan (7π)/6= tan (π+(π)/6)= tan (π)/6. Since (π)/6∈(-(π)/2,(π)/2), tan^-1( tan (7π)/6)= tan^-1( tan (π)/6)=(π)/6.

Question 32

2 sin^-1 3/5= tan^-1 24/7

Accepted Answer

Let θ= sin^-13/5, so that sin θ=3/5 with θ∈(0,(π)/2). Since θ is in the first quadrant, the cosine is taken positive: cos θ=√ 1-9/25 =4/5, tan θ=3/4. Apply the double angle formula tan 2θ=(2 tan θ)/(1- tan^2θ): tan 2θ=(2·3/4)/(1-9/16)=(3/2)/(7/16)=3/2·16/7= 24 7. Before concluding 2θ= tan^-124/7, check that 2θ lies in the branch (-(π)/2,(π)/2). Because sin θ=3/5=0.6<1/(√ 2 )= sin (π)/4 and sine is increasing on (0,(π)/2), we get θ<(π)/4, hence 0<2θ<(π)/2. (This is exactly the step that fails when sin^-1 of a larger value is doubled.) Therefore 2 sin^-13/5=2θ= tan^-124/7.

Question 33

sin^-1 8/17+ sin^-1 3/5= tan^-1 77/36

Accepted Answer

Let x= sin^-18/17 and y= sin^-13/5; both lie in (0,(π)/2), so all their cosines and tangents are positive. sin x=8/17⇒ cos x=√ 1-64/289 =15/17⇒ tan x=8/15, sin y=3/5⇒ cos y=4/5⇒ tan y=3/4. By the addition formula tan (x+y)=( tan x+ tan y)/(1- tan x tan y): tan (x+y)=(8/15+3/4)/(1-8/15·3/4)=((32+45)/60)/(1-24/60)= 77 60 36 60 = 77 36. Quadrant check: tan x tan y=24/60=2/5<1 with both tangents positive, so x+y is an angle whose tangent is positive and which is less than (π)/2; that is, x+y∈(0,(π)/2), the principal branch of tan^-1. Hence sin^-18/17+ sin^-13/5= tan^-177/36.

Question 34

cos^-1 4/5+ cos^-1 12/13= cos^-1 33/65

Accepted Answer

Let x= cos^-14/5 and y= cos^-112/13. Both cosines are positive, so x,y∈(0,(π)/2) and both sines are positive: sin x=√ 1-16/25 =3/5, sin y=√ 1-144/169 =5/13. By the cosine addition formula cos (x+y)= cos x cos y- sin x sin y: cos (x+y)=4/5·12/13-3/5·5/13=48/65-15/65=33/65. The branch of cos^-1 is [0,π], and since 0<x<(π)/2 and 0<y<(π)/2 we have 0<x+y<π. So x+y is the unique angle in [0,π] with cosine 33/65: cos^-14/5+ cos^-112/13=x+y= cos^-133/65.

Question 35

cos^-1 12/13+ sin^-1 3/5= sin^-1 56/65

Accepted Answer

Let x= cos^-112/13 and y= sin^-13/5, both in (0,(π)/2). Then cos x=12/13, sin x=5/13; sin y=3/5, cos y=4/5. By the sine addition formula sin (x+y)= sin x cos y+ cos x sin y: sin (x+y)=5/13·4/5+12/13·3/5=20/65+36/65=56/65. The branch of sin^-1 is [-(π)/2,(π)/2], so we must confirm x+y lies there before inverting. Compute cos (x+y)= cos x cos y- sin x sin y=12/13·4/5-5/13·3/5=48-15/65=33/65 0, and x+y 0; a positive cosine with 0<x+y<π forces 0<x+y<(π)/2. Therefore cos^-112/13+ sin^-13/5=x+y= sin^-156/65.

Question 36

tan^-1 63/16= sin^-1 5/13+ cos^-1 3/5

Accepted Answer

Work on the right-hand side. Let x= sin^-15/13 and y= cos^-13/5; both lie in (0,(π)/2), so all ratios below are positive. sin x=5/13⇒ cos x=12/13⇒ tan x=5/12, cos y=3/5⇒ sin y=4/5⇒ tan y=4/3. By the addition formula for tangent, tan (x+y)=(5/12+4/3)/(1-5/12·4/3)=((5+16)/12)/(1-20/36)= 21 12 16 36 = 21 12 · 36 16 = 63 16. Since tan x tan y=20/36=5/9<1 with both tangents positive, x+y∈(0,(π)/2), which is inside the principal branch of tan^-1. Hence x+y= tan^-163/16, i.e. tan^-163/16= sin^-15/13+ cos^-13/5.

Question 37

tan^-1 √ x =1/2 cos^-1 1-x/(1+x), x ∈[0,1]

Accepted Answer

Substitute √ x = tan θ, i.e. θ= tan^-1√ x, which is the left-hand side. Range of the substitution: for x∈[0,1] we have √ x ∈[0,1], so θ∈[0,(π)/4] and therefore 2θ∈[0,(π)/2]. This is the step the whole proof rests on. Since x= tan^2θ, 1-x/(1+x)=(1- tan^2θ)/(1+ tan^2θ)= cos 2θ, using the standard identity cos 2θ=(1- tan^2θ)/(1+ tan^2θ). Hence cos^-11-x/(1+x)= cos^-1( cos 2θ)=2θ, the last equality being legitimate because 2θ∈[0,(π)/2] [0,π], the principal branch of cos^-1. Therefore 1/2 cos^-11-x/(1+x)=θ= tan^-1√ x, x∈[0,1].

Question 38

cot^-1((√ 1+ sin x +√ 1- sin x )/(√ 1+ sin x -√ 1- sin x ))=x/2, x ∈(0, (π)/4)

Accepted Answer

Write both surds using the half-angle identities 1+ sin x= cos^2x/2+ sin^2x/2+2 /2 /2=( /2+ /2)^2, 1- sin x=( /2- /2)^2. Taking square roots requires the sign of each bracket, and this is where the given domain is used. For x∈(0,(π)/4) we have x/2∈(0,(π)/8), so /2 /2 0 and both brackets are positive: √ 1+ sin x = /2+ /2, √ 1- sin x = /2- /2. Substituting, (√ 1+ sin x +√ 1- sin x )/(√ 1+ sin x -√ 1- sin x )=(2 /2)/(2 /2)= /2. Since x/2∈(0,(π)/8) (0,π), the principal branch of cot^-1, we may cancel the functions: cot^-1((√ 1+ sin x +√ 1- sin x )/(√ 1+ sin x -√ 1- sin x ))= cot^-1( /2)=x/2.

Question 39

tan^-1((√ 1+x -√ 1-x )/(√ 1+x +√ 1-x ))=(π)/4-1/2 cos^-1 x,-1/(√ 2 ) ≤ x ≤ 1 [Hint: Put x= cos 2 θ ]

Accepted Answer

Follow the hint: put x= cos 2θ, so that 2θ= cos^-1x and θ=1/2 cos^-1x. Fix the range of θ first. Since cos^-1 takes values in [0,π] and x∈[-1/(√ 2 ),1], 2θ= cos^-1x∈[0,(3π)/4] ⇒ θ∈[0,(3π)/8], so cos θ 0 and sin θ≥ 0. Now use 1+ cos 2θ=2 cos^2θ and 1- cos 2θ=2 sin^2θ: √ 1+x =√ 2 cos θ, √ 1-x =√ 2 sin θ, the positive roots being correct precisely because of the range just established. Therefore (√ 1+x -√ 1-x )/(√ 1+x +√ 1-x )=( cos θ- sin θ)/( cos θ+ sin θ)=(1- tan θ)/(1+ tan θ)= tan ((π)/4-θ), dividing numerator and denominator by cos θ and using tan (π)/4=1. Finally (π)/4-θ∈[(π)/4-(3π)/8,(π)/4]=[-(π)/8,(π)/4], which lies inside (-(π)/2,(π)/2), so tan^-1((√ 1+x -√ 1-x )/(√ 1+x +√ 1-x ))=(π)/4-θ=(π)/4-1/2 cos^-1x.

Question 40

2 tan^-1( cos x)= tan^-1(2 cosec x)

Accepted Answer

x=n π+(π)/4, n ∈ Z. For the equation to make sense cosec x must exist, so sin x≠ 0; consequently cos x <1. That condition is exactly what licenses the doubling formula 2 tan^-1t= tan^-12t/(1-t^2) for t <1, with t= cos x: 2 tan^-1( cos x)= tan^-1((2 cos x)/(1- cos^2x))= tan^-1((2 cos x)/( sin^2x)). The equation becomes tan^-1((2 cos x)/( sin^2x))= tan^-1(2/( sin x)). Since tan^-1 is one-one on its branch, the arguments are equal: (2 cos x)/( sin^2x)=2/( sin x) ⇒ ( cos x)/( sin^2x)=1/( sin x). Multiplying by sin^2x (allowed, sin x≠ 0 ) gives cos x= sin x ⇒ tan x=1. Every step is reversible, so the solution set is x=nπ+(π)/4, n∈ Z, the principal solution being x=(π)/4. Check at x=(π)/4: 2 tan^-11/(√ 2 )= tan^-1(2/√ 2 )/(1- 12)= tan^-1(2√ 2 )= tan^-1(2 cosec (π)/4).

Question 41

tan^-1 1-x/(1+x)=1/2 tan^-1 x,(x 0)

Accepted Answer

x=1/(√ 3 ). Use the subtraction formula tan^-1a- tan^-1b= tan^-1a-b/(1+ab), valid when ab -1. With a=1, b=x and x 0 the condition ab=x -1 holds, so tan^-11-x/(1+x)= tan^-11- tan^-1x=(π)/4- tan^-1x. The equation therefore reads (π)/4- tan^-1x=1/2 tan^-1x. Collecting the terms in tan^-1x, (π)/4=3/2 tan^-1x ⇒ tan^-1x=(π)/6. Hence x= tan (π)/6=1/(√ 3 ), which satisfies the stated restriction x 0 and is the only solution.

Question 42

sin ( tan^-1 x), x <1 is equal to (A) x √ 1-x^2 (B) 1 √ 1-x^2 (C) 1 √ 1+x^2 (D) x √ 1+x^2

Accepted Answer

D. Let θ= tan^-1x, so tan θ=x and, by the definition of the principal branch, θ∈(-(π)/2,(π)/2); in particular cos θ 0 and sec θ 0. From the identity sec^2θ=1+ tan^2θ, the positive root is taken: sec θ=√ 1+x^2 ⇒ cos θ= 1 √ 1+x^2. Therefore sin ( tan^-1x)= sin θ= tan θ cos θ= x √ 1+x^2. (The sign is automatically right: sin θ has the same sign as x.) Option (D).

Question 43

sin^-1(1-x)-2 sin^-1 x=(π)/2, then x is equal to (A) 0, 1/2 (B) 1, 1/2 (C) 0 (D) 1/2

Accepted Answer

C. Let θ= sin^-1x, so sin θ=x. The equation becomes sin^-1(1-x)=(π)/2+2θ. A necessary restriction comes from the range of sin^-1, which is [-(π)/2,(π)/2]: (π)/2+2θ≤(π)/2 ⇒ θ≤ 0 ⇒ x≤ 0. Keep this; it is what discards a spurious root below. Take the sine of both sides and use sin ((π)/2+2θ)= cos 2θ=1-2 sin^2θ: 1-x=1-2x^2. Hence 2x^2-x=0 ⇒ x(2x-1)=0 ⇒ x=0 or x=1/2. Test both in the original equation, since squaring/taking sines can create extra roots. x=1/2 violates x≤ 0; directly, sin^-11/2-2 sin^-11/2=-(π)/6≠(π)/2. Rejected. x=0: sin^-1(1-0)-2 sin^-10=(π)/2-0=(π)/2. Satisfied. So x=0, option (C).

EXERCISE 2.1 1–10 (part 1 of 6)

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Exercise 6

Exercise 7

Exercise 8

Exercise 9

Exercise 10