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  Exercise 8.1 1. Evaluate the following without using the table: Solution: a. Sin -1  1 = π 2 b. Sin -1   = ( − 1 2 ) = –sin -1 ( 1 2 ) = ...

 

Exercise 8.1

1. Evaluate the following without using the table:

Solution:

a. Sin^-1 1 = π2

b. Sin^-1 = (−12)
= –sin^-1(12) = −π6

c. Cos^-1(−√31)= π – cos^-1(√32)= π –
π6 = 5π6.

d. Tan^-1(1) = π4

e. Arc cot(-1) = cos^-1(-1) = 3π4

f. Arc tan(−1√3) = tan^-1(−1√3) = -tan^-1(1√3) = −π6.

2. Express each of the following in terms of x:

Solution:

a. Cos tan^-1x

Let tan^-1x = θ
               
[ tan θ = x.]

Now, costan^-1x = cosθ = 1√1+x2 [x = tanθ]

b. sin tan^-1x

Let, cot^-1x =
θ               
[cotθ = x]

Now, sin.cot^-1x = sinθ = 1√1+x2    [cotθ = x]

c. Tan(Arc cotx) = tancot^-1x = tantan^-11x
= 1x.

d. Cos.sin^-1x = cos.cos^-1x
√1−x2 = √1−x2

e. Tan(2tan^-1x) = tantan^-1(2x1−x2) = 2x1−x2

f. Sin(2tan^-1x) = sin.sin^-1(2x1+x2) = 2x1+x2

g. Cos(2cot^-1x)

Let, cos(2cot^-1x) = cos2θ = cot2θ−1cot2θ+1 = x2−1x2+1

h. Cot(2 Arc cotx) = cot(2cot^-1x) = cot
cot^-1(x2−12x) =
x2−12x.

3. Evaluate each of the following using the table if
necessary:

a. Sin.cos^-1(35)
= sin.sin^-1√1−(35)2 =
sin.sin^-145 = 45.

b. cos(Arccos23) =
cos.cos^-123 = 23

c. Arc tan (tanπ6) = tan^-1.tan π6 = π6.

d. Sin (tan−134)
= sin.sin^-135 = 35.

e. sin(2cos−112)

Let cos^-1 12= θ

So, cosθ = 12.

Now, sin(2cos−112)=
sin2θ = 2sinθ.cosθ.

= 2. 12

= 1

f. Sin^-1(2cosπ3) = sin^-1(2.12) = sin^-1(1)
= π2.

4.Prove each of the following:

a. 2cos^-1x = cos^-1(2x² –
1)

Solution:

Let, cos^-1x = θ.

So, cosθ = x

Now, cos2θ = 2cos²θ – 1

Or, cos2θ = 2x² – 1   [cosθ = x]

Or, 2θ = cos^-1(2x² – 1)

So, 2cos^-1x = cos^-1(2x² –
1).

b. 3cos^-1x = cos^-1(4x³ –
3x)

Solution:

Let, cos^-1x = θ.

So, cosθ = x

Now, cos3θ = 4cos³θ – 3cosθ

Or, cos3θ = 4x³ – 3x   [cosθ = x]

Or, 3θ = cos^-1(4x³ – 3x)

So, 3cos^-1x = cos^-1(4x³ –
3x).

c. 3tan^-1x = tan^-13x−x31−3x2

Solution:

Let tan^-1x = θ

So, tanθ = x

Now, tan3θ = 3tanθ−tan3θ1−3tan2θ

Or, tan3θ = 3x−x31−3x2

Or, 3θ = tan^-13x−x31−3x2

So, 3tan^-1x = tan^-13x−x31−3x2.

d. sin(Arc.cost)= cos(Arc.sint)

Solution:

L.H.S. = sin(Arc.cost) = sin(cos^-1) = sin.sin^-1√1−t2 = √1−t2.

R.H.S = cos(Arc.sint) = cos.sin^-1t = cos.cos^-1√1−t2 = √1−t2.

So, L.H.S. = R.H.S.

e. Cos(2 Arc.cost) =2t² – 1

Solution:

Cos(2 Arc.cost) = cos(2cos^-1t) = cos.cos^-1(2t² –
1)

So, cos(2 Arc.cost) = 2t² – 1

f. sin(2sin^-1x) = 2x.√1−x2

Solution:

Let, sin^-1x = θ.

So. sinθ = x.

Now, sin(2sin^-1x) = sin2θ = 2sinθ.cosθ

So, sin(2sin^-1x) = 2x.√1−x2  [sinθ = x]

g. cos(sin^-1u + cos^-1v) = v√1−u2 – u√1−v2

Solution:

Let, sin^-1u = A → sinA = u and cosA = √1−u2

And cos^-1v = b → cosB = v and sinB = √1−v2

We have,

Cos(A + B) = cosA.cosB – sinA.sinB

So, cos(sin^-1u + cos^-1v) = v√1−u2 – u√1−v2

h. tan^-115+ tan^-117= tan^-1(617)                                                                                   

Solution:

L.H.S. = tan^-115 + tan^-117
= tan^-1(15) + tan^-1(17) = tan^-115+171−15.17

= tan^-1(7+535−1) = tan^-1(1234)

So, tan^-115+ tan^-117= tan^-1(617)

(i) tan^-1 a– tan^-1c= tan^-1a−b1+ab + tan^-1b−c1+bc

Solution:

Here, R.H.S. = tan^-1a−b1+ab + tan^-1b−c1+bc

= tan^-1a – tan^-1b + tan^-1b –
tan^-1c = tan^-1 a– tan^-1c = L.H.S.

(j) tan(Arc.tanu – Arc.tanv) = u−v1+uv

Solution:

L.H.S. = tan(Arc.tanu – Arc.tanv) = tan(tan^-1u –
tan^-1v)

= tan.tan^-1u−v1+uv = u−v1+uv = R.H.S.

(k) tan(2tan^-1x) =2
tan (tan^-1x + tan^-1x³)

Solution:

(l) tan^-1x + tan^-1y + tan^-1z
= tan^-1x+y+z−xyz1−xy−zx−yz

Solution:

L.H.S. = tan^-1x + tan^-1y + tan^-1z.

= tan^-1(x+y1−xy) + tan^-1z = tan^-1(x+y1−xy+Z)1−x+y1−xy.z

= tan^-1x+y+z−xyz1−xy−zx−yz.

(m) sin^-145 + sin^-1513
+ sin^-11665= π2

Solution:

L.H.S. = sin^-1{45√1−(513)2+513√1−(45)2} + sin^-11665.

= sin^-1(45.1213+513.35) + sin^-11665= sin^-1(6365) + sin^-11665.

= sin^-1{6365√1−(1665)2+1665√1−(6365)2}

= sin^-1(6365.6365+1665.1665) = sin^-13969+2564225.

= sin^-1(42254225) =
sin^-11 = π2 = R.H.S.

(n) Prove that: tan^-1√x =
12 cos^-11−x1+x =
12sin^-1(2√x1+x)

Solution:

Or, 12cos^-11−x1+x = 12cos^-11−tan2θ1+tan2θ       [x
= tan²θ say]

= 12cos^-1 cos2θ =
12.2θ = θ = tan^-1√x.

And, 12sin^-12√x1+x = 12 sin^-12tanθ1+tan2θ       [x = tan²θ
say]

= 12 sin^-1 sin2θ.

= 12.2θ = θ tan^-1√x [x = tan²θ say]

Hence, tan^-1√x = 12 cos^-11−x1+x = 12sin^-1(2√x1+x).

5. Find the value of each of the following:

a. cos(sin−145+tan−1512)

Solution:

Let sin^-1 = 45 = A.

SO, sinA = 45

And, cosA = 35

And tan^-1512 = B

So, tanB = 512.

CosB = 1213, sinB = 513.

Now, cos(sin−145+tan−1512)

= cos(A + B) = cosA.cosB – sinA.sinB

= 35.1213 –
45.513 = 36−2065 =
1665.

b. sin(cos112+sin−135)

Solution:

Let, cos^-112  = A

So, cos A = 12, sinA = √32

And sin^-135 = B

So, sinB = 35, cosB = 45.

Now, sin(cos112+sin−135)

= sin(A + B) = sinA.cosB + cosA.sinB

= √32.45 +
12.35 = 4√3+310.

c. tan(cos−145−sin−11213)

Solution:

Let, cos^-145 = A

So, cos A = 45

So, tanA = 34

Or, sin^-11213 = B

So, sinB = 1213

So, tanB = 125.

Now, tan(cos−145−sin−11213) = tan(A – B)

= tanA−tanB1+tanA.tanB = 34−1251+34.125 = 15−4820+36 = −3356.

d. tan(tan^-1x – tan^-12y)

Solution:

tan(tan^-1x – tan^-12y)

= tan^-1(x−2y1+x.2y)[Formula of tan^-1x – tan^-1y]

= x−2y1+2xy

e. tan^-13 + tan^-113

Solution:

tan^-13 + tan^-113

= tan^-1(3+13)1−3.13

= tan^-1(9+13−3)
= tan^-1(100)= tan^-1 ∞ =
π2.

f. tan^-13 + tan^-113

Solution:

Arc sint – Arc cos(-t)

= sin^-1t – cos^-1(-t) = sin^-1t
– {π – cos^-1t}

= sin^-1t – π + cos^-1t

= π2 - π  [sin^-1x + cos^-1x
= π2]

= −π2.

6. Solve each of the following equations:

a. Cos^-1x - sin^-1x= 0

Solution:

Cos^-1x = sin^-1x

Or, cos^-1x = cos^-1 √1−x2

Or, x = √1−x2

Or, x² = 1 – x² [on
squaring]

Or. 2x² = 1

So, x =±1√2.

Since x = −1√2 does not satisfy the
given equation

Hence, x = 1√2.

b. Cos^-1x = sin^-1x

Solution:

Here, given equation is:

Tan^-1x – cot^-1x = 0

Or, tan^-1x = tan^-11x.

Or, x = 1x

Or, x² = 1

So, x = 1.

c. Sin^-1x2 = cos^-1x

Solution:

Sin^-1x2 = cos^-1x

Or, sin^-1x2 = sin^-1√1−x2

Or, x2 = √1−x2

Squaring , we have, x24 = 1 – x².

Or, x24 + x² = 1

Or, 5x24 = 1

Or, x² = 45

So, x = ±2√5

Since x = −2√5 does not satisfy the
given equation.

So, x = 2√5.

d.Cos^-1x = cos^-112x

Solution:

Her, given equation is:

Cos^-1x = cos^-112x.

Or. X = 12x

Or, x² = 12

So, x = ±1√2.

e. tan^-12x=2tan^-1x

Solution:

tan^-12x = tan^-12x1−x2

Or, 2x = 2x1−x2

Or, 2x – 2x³ = 2x

Or, -2x³ = 0

So, x = 0

f. cos^-1x + cos^-12x=12 π  

Solution:

Here the given equation is, cos^-1x + cos^-12x
= π2.

Or, cos^-12x = π2 - cos^-1x.

Or, cos^-12x = sin^-1x

Or,cos^-12x = cos^-1+ √1−x2

Or, 2x = √1−x2

Or, 4x² = 1 – x² [on
squaring]

Or, 5x² = 1

So, x = ±1√5^.

g. Sin^-1x + cos^-1(1 – x) =
π2

Solution:

Sin^-1x + cos^-1(1 – x) =
π2

Or, cos^-1(1 – x) = π2 - sin^-1x

Or, cos^-1(1 – x) = cos^-1 x

Or, 1 – x = x

Or, 1 = 2x

So, x = 12.

h. Sin^-1 2a1+a2- cos^-11−b21+b2 = 2tan^-1x

Solution:

The given equation is:

Sin^-1 2a1+a2- cos^-11−b21+b2 = 2tan^-1x.

Or, 2tan^-1a – 2tan^-1b = 2tan^-1x 
[2tan^-1x = sin^-12x1+x2.]

Or, tan^-1a – tan^-1b = tan^-1x

Or, tan^-1(a−b1+ab) = tan^-1x

Or, x = a−b1+ab

So, x = a−b1+ab.

i. tan^-1x−1x−2 + tan^-1x+1x+2 = tan^-11

Solution:

Given equation is tan^-1x−1x−2 + tan^-1x+1x+2=
tan^-11.

Or, tan^-1{x−1x−2+x+1x+21−x−1x−2.x+1x+2} =
tan^-1 1.

Or, (x+1)(x+2)+(x−2)(x+1)(x−2)(x+1)−(x−1)(x+1)=1

Or, x2+2x−x−2+x2+x−2x−2x2−4−(x2−1) = 1

0r, 2x² – 4 = -3

Or, 2x² = 1

Or, x² = 12

So, x = ±1√2.

j.tan^-12x + tan^-13x =
π4

Solution:

The given equation is:

Tan^-12x + tan^-13x = π4.

Or, tan^‑1(2x+3x1−2x.3x) ⁼ π4^.

Or, 5x1−6x2 =
tanπ4 = 1

Or, 5x – 1 – 6x² = 0

Or, 6x² + 5x – 1 = 0

Or, 6x² + 6x – x – 1 = 0

Or, 6x(x + 1) – 1 (x + 1) = 0

Or, (x + 1)(6x – 1) = 0

So, x = -1, 16.

k. 3tan^-112+√3 - tan^-11x
= tan^-113

Solution:  The
given equation is:

3tan^-112+√3 - tan^-11x
= tan^-113.

Or, 3tan^-1 (2 – √3) – tan^-11x
= tan^-113.               
[3tan^-1x = tan^-1(3x−x31−3x2)]

Or, tan^-1{(3(2−√3)−(2−√3)2)1−3(2−√3)2} – tan^-113= tan^-11x

Or, tan^-1{12√3−2012√3−20} - tan^-113 = tan^-11x.

Or, tan^-1(1−131+1.13) = tan^-11x

Or, tan^-1(23.34) = tan^-11x.

Or, tan^-112 = tan^-11x.

Or, 12 = 1x

So, x = 2.

7. Prove that:

a. x+y+z=xyz, if tan^-1x+ tan^-1y+tan^-1z
= π

Solution:

Tan^-1 + tan^-1y + tan^-1z
= π

Or, tan^-1(x+y1−xy) = π – tan^-1z

Or, x+y1−xy = tan (π
– tan^-1z)

Or, x+y1−xy =
-z   [tan(π – tan^-1) = –tan.tan^-1z = -z]

Or, x + y = -z + xuz

So, x + y + z = xyz.

b. xy + yz + zx =1, if Tan^-1 + tan^-1y
+ tan^-1z = π2.

Solution:

Tan^-1 + tan^-1y + tan^-1z
= π2

Or, tan^-1(x+y1−xy) = π2 – tan^-1z

Or, x+y1−xy = tan
(π2 – tan^-1z) = cot.tan^-1z.

Or, x+y1−xy =
-z   [tan(π – tan^-1) = -tan.tan^-1z = -z]

Or, x+y1−xy= cot.cot^-11z
= 1z

Or, zx + yz = 1 – xy

So, xy + yz + zx = 1.

8. If cos^-1x + cos^-1y + cos^-1z
= π, show that x² + y² + z² +
2xyz = 1.

Solution:

Here, cos^-1x + cos^-1y + cos^-1z
= π

Or, cos^-1 { xy - √(1−x2)(1−y2) } = π – cos^-1z

Or, xy - √(1−x2)(1−y2) = cos (π – cos^-1z)

Or, xy - √(1−x2)(1−y2) = -cos.cos^-1z = -z

Or, xy + z = √(1−x2)(1−y2)

Squaring, we have,

X²y² + 2xyz + z² =
1 – y² – x² + x²y².

So, x² + y² + z² +
2xyz = 1.

9. Prove that:

(i) tan^-135 + sin^-135
= tan^-12711

Solution:

Let, sin^-135 = θ →sin θ =
35.

So, p = 3 , b = 4, h = 5.

So,tanθ = 34.

Or, θ = tan^-134.

So, sin^-135 = tan^-134.

L.H.S. = tan^-135 + sin^-135
= tan^-135 + tan^-134.

= tan^-135+341−35.34 = tan^-112+1520−9 = tan^-12711
= R.H.S.

(ii) tan^-113+ tan^-115
+ tan^-117 + tan^-118=π4

Solution:

L.H.S. = tan^-113+ tan^-115
+ tan^-117 + tan^-118.

= tan^-113+151−13.15 + tan^-117+181−17.18

= tan^-1(5+314) +
tan^-1(8+755) = tan^-147
+ tan^-1311.

= tan^-147+3111−47.311 = tan^-144+2177−12
= tan^-1(6565) = tan^-1 1

= π4 = R.H.S.

(iii) 2tan^-113 + tan^-117
= π4

L.H.S. = 2tan^-113 + tan^-117

= tan^-1(2.131−19) + tan^-117  
[2tan^-1x = tan^-1(2x1−x2)]

= tan^-1(23.98)
+ tan^-117 = tan^-134 + tan^-117.

= tan^-1(34+171−34.17) = tan^-1(21+428−3) = tan^-1(2525)

= tan^-1 1 = π4 = R.H.S.

(iv) 4(cot^-13 + cot^-1 2) = π

Solution:

Cosec^-1√5= θ.

So, cosec θ = √5,  (p = 1, h = √5, b =
2)

So, cot θ = 2

So, θ = cot^-1 2

So, cosec^-1√5 = cos^-12 …(i)

L.H.S. = 4(cot^-1 3 + cosec^-1√5)

= 4(cot^-13 + cot^-1 2)   
[from(i)]

= 4 (tan−113+tan−112) = 4tan^-1(13+121−13.12) = 4tan^-1(55)

= 4tan^-1 1 = 4. π4 = π
= R.H.S.

(v) tan^-11 + tan^-12 + tan^-13
= π

Solution:

L.H.S. = tan^-11 + tan^-12 + tan^-13

= tan^-1 + tan^-1(2+31−2.3) = tan^-1.tan^-1(-1) =
π4 + 3π4 = π.

R.H.S. = 2(tan−11+tan−112+tan−113)

= 2{tan−11+tan−1(12+131−12.13)}

= 2{tan−11+tan−1(56.64)}

= 2(tan^-11 + tan^-11) = 2(π4+π4) = π.

10. If sin^-1x + Sin^-1y + sin^-1z
= π, prove that x√1−x2 + y√1−y2 +
z√1−z2.

Solution:

Let, sin^-1x = A → sinA = x and cosA = √1−x2

Sin^-1y = B → sinB = y and cosB = √1−y2

And sin^-1z = C → sinC = z and cosC = √1−z2

And sin^-1x + sin^-1 y + sin^-1 z
= π.

Or, A + B + C = π.

So, A + B = π – c

Or, sin(A + B) = sin(π – C) = sinC.

Cos(A + B) = cos(π – C) = - cosC

Now,

L.H.S. = x√1−x2 + y√1−y2 + z√1−z2

= sinA.cosA + sinB.cosB + sinC.cosC.

= 12 (2sinA.cosA +2 sinB.cosB) + sinC.cosC

= 12 (sin2A + sin2B) + sinC.cosC

= 12.2 sin2A+2B2
.cos2A−2B2 + sinC.cosC.

= sin(A + B). cos(A – B) + sinC.cosC.

= sinC.cos(A – B) + sinC.cosC 
                
[A + B = π – C]

= sinC {cos(A – B) + cosC}

= sinC {cos(A – B) – cos(A +
B)}           
   [A + B = π – C]

= sinC.2sinA.sinB = 2sinA.sinB.sinC = 2xy = R.H.S.

Hence, x√1−x2 + y√1−y2 + z√1−z2

11. If sin^-1x + Sin^-1y + sin^-1z
= $\frac{π}{2}$, prove
that x² + y² + z² + 2xyz = 1.

Solution:

Proof:

Let sin^-1x → x = sinA

Sin^-1y = B → y sinB

And sin^-1z = C → z = sinC.

Then the given question changed to:

If  A + B + C = π2, prove that,

Sin²A + sin²B + sin²C = 1 –
2sinA.sinB.sinC.

From the given part, A + B = π2 – C.

Sin(A + B) = cosC

And cos (A + B) = sinC.

Now, L.H.S.= 12(2sin²A + 2sin²B)
+ sin²C.

= 12 [(1 – cos2A) + (1 – cos2B)] + sin²C.

= 12 [2 – cos2A – cos2B] + sin²c.

= 1 – 12 (cos2A + cos2B) + sin²C.

= 1 – 12 2cos(A + B).cos(A – B) + sin²C.

= 1 – sinC [cos(A – B) – cos(A + B)]

= 1 – sinC [2sinA−B+A+B2.sinA+B−A+B2]

= 1 – sinC [2sinA.sinB]

= 1 – 2sinA.sinB.sinC = R.H.S>

Hence, x² + y² + z² +
2xyz = 1