Some Identities:
1) sin⁻¹x + cos⁻¹x= π/2
2) tan⁻¹x+ cot⁻¹x=π/2
3) sec⁻¹x+ cosec⁻¹x =π/2
EXERCISE - A
Write down the following in inverse notation:
1) sinx =3/4. x= sin⁻¹(3/4)
2) cosx = 1. x= cos⁻¹1
3) tanx =-4. x= tan⁻¹(-4)
EXERCISE- B
Verify each of the following:
Type-1
1) sin sin⁻¹(-1/2) = -1/2
2) sin cos⁻¹(4/5) = 3/5.
3) sin⁻¹ tan(3π/4) = -π/2
4) sin cos⁻¹(- √3/2) = 1/2.
5) sin⁻¹(sin 2π/3) = π/3
Type -2
1) cos sin⁻¹(- 1/√2) = 1/√2.
2) cos sin⁻¹(- 12/13) = 5/13.
3) cos⁻¹sin 220= 130°.
4) cos cos⁻¹(1/√2) = 1/√2.
5) cos⁻¹tan(- 5π/4) = π.
6) Cos[π/3 + cos⁻¹(-1/2)]= 1
Type-3
1) tan tan⁻¹(- 1) = - 1.
2) tan sin⁻¹0 = 0
3) tan⁻¹ cot(230) = 40°
EXERCISE - C
Find the principal values of:
Type-1
1) sin⁻¹(1/2). π/6
2) sin⁻¹(√3/2). π/3
3) Sin⁻¹0. 0
4) sin⁻¹(- √3/2). - π/3
Type -2
1) cosec⁻¹(-1). -π/2
2) cosec⁻¹(-√2). -π/4
Type -3
1) cos⁻¹(-√3/2). 5π/6
2) cos⁻¹(- 1/2). 2π/3
3) Cos⁻¹(-1) π
4) cos⁻¹(0). π/2
Type -4
1) sec⁻¹(-√2). 3π/4
2) Sec⁻¹(1). 0
3) sec⁻¹(2). π/3
4) sec⁻¹(-2). 2π/3
Type-5
1) Tan⁻¹(-√3). -π/3
2) Tan⁻¹(0). 0
Type-6
1) cot⁻¹(0). π/2
2) cot⁻¹(-1). -π/4
3) Cot⁻¹1. π/4
4) cot⁻¹(√3). π/6
EXERCISE - D
*** Find the general values of:
Type -1
1) sin⁻¹(0). nπ
2) sin⁻¹(1/√2). nπ+ (-1)ⁿ π/4
Type-2
1) cosec⁻¹(-2). 2nπ±2π/3
2) cosec⁻¹(√2). nπ+ (-1)ⁿ π/4
Type -3
1) cos⁻¹(√3/2). 2nπ± π/6
Type -4
1) tan⁻¹(-1). nπ-π/4
2) Tan⁻¹(-√3). nπ - π/3
Type -5
1) sec⁻¹(- 1). (2n+1)π
Type -6
1) cot⁻¹(-√3). nπ - π/6
EXERCISE- E
**Find the value of:
Type -1
1) sin⁻¹(2 sin 150). π/2
2) sin{π/2 - sin⁻¹(3/5)}. 4/5
3) sin⁻¹tan(3π/4). -π/2
4) Sin cos⁻¹(-√3/2). 1/2
5) sin⁻¹cos(-2π/3). -π/6
6) sin(2 tan⁻¹3/4). 24/25
7) sin(sin⁻¹π/4+ cos⁻¹π/4). 1
8) sin(sin⁻¹1/3 + sec⁻¹3) + cos(tan⁻¹1/2 + tan⁻¹2). 1
9) sin(sin⁻¹1/2 + cos⁻¹1/2). 1
10) If sin⁻¹x=k, then cosec⁻¹{1/√(1- x²)} is. π/2 - k
Type -2
1) 1/2 cos⁻¹(1/2). 1/2
2) cos sin⁻¹(-√3/2). 1/2
3) cos sin⁻¹(3/5). 4/5
4) If cos⁻¹(√5)=k, then cosec⁻¹(√5) is. π/2 - k
Type-3
1) sec tan⁻¹(- 5/12). 13/12
Type -4
1) Tan sin⁻¹(1/√2). π/4
2) tan⁻¹(1/√2 sec π/4). π/4
3) tan⁻¹(cot 330°). -π/3
4) tan sin⁻¹(√3/2). √3
5) tan sec⁻¹(-13/12). -5/12
6) tan 1/2(tan⁻¹x + cot⁻¹x). 1
7) If k= tan⁻¹a, find
A) sin 2a. 2a/(1+ a²)
B) cos 2a. (1- a²)/(1+ a²)
C) tan 2a. 2a/(1- a²)
8) If 2 tan⁻¹{2x/(1- x²)}=π, then find x. 1
Type -5
1) cosec sec⁻¹(√5). √5/2
Type -6
1) cot cos⁻¹(-1/√2). -1
2) cot sin⁻¹(-4/5). - 3/4
Formula Used:
1) sin⁻¹x + sin⁻¹y= sin⁻¹{x √(1- y²)+ y√(1- x²)}
2) sin⁻¹x - sin⁻¹y= sin⁻¹{x √(1- y²) - y√(1- x²)}
3) 2 sin⁻¹x = sin⁻¹{2x √(1- x²)} = cos⁻¹(1- 2x²).
4) 3sin⁻¹x = sin⁻¹(3x - 4x³)
EXERCISE -F
*** Prove:
Type -1(a)
1) a) sin⁻¹a = tan⁻¹a/√(1- a²)
b) sin⁻¹(1/√5)+ sin⁻¹(2/√5)= π/2
2) a) sin⁻¹3/5 +sin⁻¹7/25 =π/2.
b) sin⁻¹4/5 +sin⁻¹3/5 =π/2.
3) sin⁻¹(77/85)- sin⁻¹(3/5)= cos⁻¹(15/17)
4) sin⁻¹4/5 + sin⁻¹5/13 + sin⁻¹16/15 =π/2.
5) 2sin⁻¹x = sin⁻¹{2x √(1- x²)}
Type -1(b)
1) 2sin⁻¹(3/5) + sin⁻¹(7/25) =π/2.
Type -1(c)
1) 3sin⁻¹x = sin⁻¹(3x - 4x³).
Formula Used:
1) cos⁻¹x + cos⁻¹y= cos⁻¹[xy - √{(1- y²)(1- x²)}] .
2) cos⁻¹x - cos⁻¹y= cos⁻¹[xy + √{(1- y²)(1- x²)}] .
3) 2 cos⁻¹x = cos⁻¹(2x² -1) .
4) 3 cos⁻¹x = cos⁻¹(4x³ - 3x).
Type-2(a)
** Prove:
1) cos⁻¹1/√5 + cos⁻¹2/√5 =π/2
2) Cos⁻¹4/5 -Cos⁻¹5/13 = Cos⁻¹56/65.
3) cos⁻¹3/5 + cos⁻¹12/13 + cos⁻¹63/65 = π/2.
Type-2(b)
1) 2cos⁻¹x= cos⁻¹(2x²- 1).
Type -2(c)
1) 3 cos⁻¹x = cos⁻¹(4x³- 3x).
Formula Used:
1) tan⁻¹x + tan⁻¹y= tan⁻¹{(x+ y)/(1- xy)}
2) tan⁻¹x - tan⁻¹y= tan⁻¹{(x- y)/(1+ xy)}
3) tan⁻¹x + tan⁻¹y + tan⁻¹z = + tan⁻¹{(x+ y + z - xyz)/(1- xy - yz - zx)}.
4) 2tan⁻¹x = tan⁻¹{2x/(1- x²)}
= Sin⁻¹{2x/(1+ x²)}
= cos⁻¹{(1- x²)/(1+ x²)}
5) 3 tan⁻¹x = tan⁻¹{(3x - x³)/(1- 3x²)}
Type-3(a)
**Prove:
1) tan⁻¹1/2 + tan⁻¹1/3=π/4.
2) tan⁻¹1/4 + tan⁻¹3/5=π/4.
3) tan⁻¹3/5 + tan⁻¹1/4=π/4.
4) tan⁻¹x + tan⁻¹{2x/(1- x²)} =tan⁻¹{(3x - x³)/(1- 3x²)}
5) Tan⁻¹{(2a - b)/b√3} + tan⁻¹{(2b - a)/a√3= π/3.
6) tan⁻¹1/(p+q)+ tan⁻¹{q/(p²+ pq+1)= tan⁻¹ 1/p.
7) tan⁻¹a+ tan⁻¹c = tan⁻¹ {(a- b)/(1+ ab)} + tan⁻¹{(b - c)/(1+ bc).
8) tan⁻¹1/4 + tan⁻¹1/2=tan⁻¹6/7
9) tan⁻¹4 - tan⁻¹3 = cot⁻¹13.
10) tan⁻¹1/4 + tan⁻¹2/9 =1/2 cos⁻¹3/5.
11) tan⁻¹1/2 + tan⁻¹2/11 = cos⁻¹4/5
12) tan⁻¹4/3 + tan⁻¹12/5 =π - tan⁻¹56/33.
13) tan⁻¹1/7 + tan⁻¹1/8 + tan⁻¹1/18= tan⁻¹1/8
14) tan⁻¹1/4 + tan⁻¹1/5 + tan⁻¹2/9 + tan⁻¹1/8 = π/4
15) tan⁻¹1/3 + tan⁻¹1/5+ tan⁻¹1/7 + tan⁻¹1/8 = π/4.
16) tan⁻¹1 + tan⁻¹2 + tan⁻¹3 =π = 2(tan⁻¹1/2 + tan⁻¹1/3 + tan⁻¹1)
17) tan⁻¹1 + cot⁻¹1/2 + tan⁻¹3 =π = 2(tan⁻¹1 + tan⁻¹1/2 + cot⁻¹3).
18) tan(tan⁻¹x + tan⁻¹y + tan⁻¹z) = cot (cot⁻¹x + cot⁻¹y + cot⁻¹z).
Type-3(b)
1) 2tan⁻¹2/3 = tan⁻¹12/5.
2) 2tan⁻¹1/3+ tan⁻¹1/7 =π/4
3) 2tan⁻¹1/5 +tan⁻¹1/4=tan⁻¹32/43
4) 2(tan⁻¹1/4 + tan⁻¹2/9) = cos⁻¹3/5
5) 2tan⁻¹1/4 + tan⁻¹1/5 +tan⁻¹6/61= π/4
Type -3(c)
1) 3 tan⁻¹x = tan⁻¹{(3x - x³)/(1- 3x²)}.
2) 3 tan⁻¹1/(2+√3) - tan⁻¹1/2= tan⁻¹1/3
Type -3(d)
1) 4tan⁻¹1/5 - tan⁻¹1/239 = π/4
2) 4 tan⁻¹1/5 - tan⁻¹1/70 + tan⁻¹1/99 =π/4.
3) 4(tan⁻¹1 + tan⁻¹1/2 + tan⁻¹1/3) =π.
Type-3(e)
1) tan⁻¹1+ tan⁻¹2+ tan⁻¹3 = π= tan⁻¹1/2+ tan⁻¹1/3 +tan⁻¹1
2) tan⁻¹(x -y)/(1+ xy) + tan⁻¹(y - z)/(1+ yz) +tan⁻¹(z - x)/(1+ zx)= 0
3) tan(2tan⁻¹p)= 2 tan(tan⁻¹p +tan⁻¹p³)
** Prove
Type-4(a)
1) Cot⁻¹1/2 - 1/2Cot⁻¹4/3=π/4
2) cot⁻¹(1+xy)/(x - y) + cot⁻¹(1+yz)/(y-z) + cot⁻¹(1+zx)/(z - x)= 0
3) cot⁻¹(xy+1)/(x - y) + cot⁻¹(yz +1)/(y-z) + cot⁻¹(zx +1)/(z - x)= 0
Type -4(b)
1) 2cot⁻¹5 + cot⁻¹7+ 2 cot⁻¹8 =π/4.
EXERCISE - G
**Prove
Type -1
1) sin⁻¹1/√17 + cos⁻¹9/√85 = tan⁻¹1/2
2) sin⁻¹1/√5 + Cot⁻¹3 = π/4
3) sin⁻¹1/√10 + cos⁻¹2/√5 = tan⁻¹1.
4) sin⁻¹12/13 + cos⁻¹4/5+ cos⁻¹63/16 =π.
5) sin⁻¹3/5 +Cos⁻¹15/17 = sin⁻¹77/85
6) sin⁻¹4/5 + 2 tan⁻¹1/3 =π/2.
7) sin⁻¹√3/2 + 2 tan⁻¹1/√3 =2π/3.
8) sin⁻¹4/5 + 2tan⁻¹1/3 = π/2.
9) sin⁻¹(4/5)+ tan⁻¹(3/4)= π/2.
10) sin(sin⁻¹1/2 - cos⁻¹1/3= (1- 2√6)/6.
11) Sin⁻¹√{(x -q)/(p - q)} = cos⁻¹{(p - x)/(p - q)} = cot⁻¹√{(p- x)/(x - p)}.
Type -2
1) cos⁻¹4/5 + cot⁻¹5/3 = tan⁻¹27/11.
2) Cos⁻¹x = 2 sin⁻¹√{(1- x)/2}
= 2 cos⁻¹√{(1+ x)/2}
= 2 tan⁻¹{√(1- x²)/(1+x)}
3) cos(tan⁻¹15/8 - sin⁻¹7/25)= 297/425
4) {Cos(sin⁻¹x)}² = {sin(cos⁻¹x)}²
Type-3(a)
1) tan⁻¹1/4 + cos⁻¹3 /5 =tan⁻¹19/8
2) tan⁻¹x+Cot⁻¹(x+1) = tan⁻¹(x²+x+1)
3) tan⁻¹1/2 + 1/2 cos⁻¹4/5 =π/4
4) tan⁻¹x +Cot⁻¹y=(tan⁻¹{(xy+ 1)/(y - x)}.
5) tan⁻¹√x = 1/2 cos⁻¹{(1- x)/(1+ x)}.
6) tan(2 sin⁻¹4/5+ cos⁻¹12/13)= -253/204
Type-3 (b)
1) 2tan⁻¹1/3 +Cot⁻¹7 = π/4
2) 2 tan⁻¹1/5 + cos⁻¹ 63/65 = tan⁻¹3/4.
3) 2tan⁻¹1/5 +cos⁻¹63/65 = tan⁻¹3/4.
4) 2 tan⁻¹√x + 2 tan⁻¹y= sin⁻¹[{2(x+ y)(1- xy)}/{(1+ x²)(1+ y²)}.
Type-3 (d)
1) 4(tan⁻¹1/3 + cos⁻¹2/√5) =π.
2)
Type -4
1) 4(cot⁻¹3+ cosec⁻¹√5) =π.
2) cot⁻¹1/2 - 1/2 tan⁻¹4/3 =π/4
3) Cot⁻¹(43/32)- tan⁻¹(1/4)=cos⁻¹(12/13).
4) Cot⁻¹(tan2x)+Cot⁻¹(-tan3x)=x
5) Cot⁻¹3 +Cosec⁻¹√5 =π/4.
EXERCISE -H
Prove the Following:
Type -1
*1) sin⁻¹{x/√(1+ x²)} + cos⁻¹{(x+1)/√(x²+ 2x+2)} = tan⁻¹(x²+ x +1)
2) sin(sin⁻¹1/2 + cos⁻¹3/5) = (3+ 4√3)/10.
3) sin⁻¹{2a/(1 + a²)} - cos⁻¹{(1- b²)/(1+ b²)}= 2tan⁻¹{(a -b)/(1+ ab)}
4) sin⁻¹√{(x-b)/(a-b)} = cos⁻¹{(a-x)/(a-b)}= tan⁻¹{(x-b)/(a-x)}.
Type -2
1) cos⁻¹{(cosx + cosy)/(1+ cosx cos y)} = 2 tan⁻¹(tan x/2 tan y/2)
2) Cos⁻¹{(b+ a cosx)/(a+ b cosx)} = 2 tan⁻¹[√{(a- b)/(a+ b)} . tan(x/2)]
3) cos⁻¹x =2 sin⁻¹√{(1- x)/2}= 2 cos⁻¹√{(1+x)/2}=2 tan⁻¹√{(1- x)/(1+x)}
Type -3
1) tan⁻¹x +cot⁻¹(x+1)=tan⁻¹(x²+ x +1)
2) 1/2 tan⁻¹x= cos⁻¹√[{1+ √(1+x²)}/2√(1+ x²)]
3) tan(2tan⁻¹a)= 2 tan(tan⁻¹ a + tan⁻¹a³).
4) tan⁻¹{(x - y)/(1+ xy)} + tan⁻¹ {(y - z)/(1+ yz)} + tan⁻¹{(z - x)/(1+ zx)} = 0.
5) tan⁻¹x + tan⁻¹y = sin⁻¹[(x+y)/√{(1+ x²)(1+ y²)}].
6) tan[1/2 sin⁻¹{2x/(1+ x²)}] + 1/2 cos⁻¹{(1- x²)/(1+ x²)} = 2x/(1 - x²).
*7) tan⁻¹(1/√3 tan x/2) = 1/2 cos⁻¹{(1+ 2 cosx)/(2+ cosx)
*8) tan(π/4 + 1/2 cos⁻¹ a/b) + tan⁻¹(π/4 - 1/2 cos⁻¹ a/b)= 2b/a
*9) tan⁻¹√x = 1/2 cos⁻¹{(1- x)/(1+ x)}.
10) tan[2 tan⁻¹{(1+ cosx)/(1- cosx)}] + tan x = 0.
*11) tan⁻¹x = 2 tan⁻¹(cosec tan⁻¹x + tan cot⁻¹x).
*12) 1/2 tan⁻¹x = cos⁻¹[{(1+ √(1+ x²)}/{2√(1+ x²)}]¹⁾²
13) tan⁻¹[{√(1+ x²)+ √(1- x²)}/{√(1+ x²)- √(1- x²)}]= π/4 + 1/2 cos⁻¹x²
14) tan⁻¹{(√2+1) cotx} - tan⁻¹{(√2-1)cotx= tan⁻¹(sin 2x).
15) tan⁻¹x + tan⁻¹y= sin⁻¹[(x+y)/√{(1+x²)(1+y²)}]
16) tan(sin⁻¹x+ sin⁻¹y) + tan(Cos⁻¹x + Cos⁻¹y) simplify.
17) Tan⁻¹(cotx)+Cot⁻¹(tanx)=π-2x
18) tan⁻¹(1/2 tan 2A) + tan⁻¹(cot A)+ tan⁻¹(cot³A)= 0
Type -4
1) cot(cos⁻¹a + cos⁻¹b) + cot(sin⁻¹a + sin⁻¹b) = 0.
2) cot(cot⁻¹x +cot⁻¹y +cot⁻¹z)=tan (tan⁻¹x + tan⁻¹y + tan⁻¹z)=0.
3) cot⁻¹ x= 1/2 sin⁻¹{2x/(1+ x²)}
EXERCISE - I
Prove the Following:
Type -1
1) sin cosec⁻¹cot tan⁻¹x = x
2) sin cot⁻¹tan cos⁻¹x = x.
3) sin cot⁻¹tan cos⁻¹x = x.
4) sin cos⁻¹ tan sec⁻¹x = √(2- x²)
5) sin cot⁻¹cos tan⁻¹x = √{(x²+1)/(x²+2)}
6) sin⁻¹cos sin⁻¹x + cos⁻¹sincos⁻¹x = π/2
7) sin⁻¹√{(x - b)/(a - b)} = cos⁻¹√{(a- x)/(a - b)} = tan⁻¹√{(x - b)/(a - x)}.
8) 3 sin⁻¹{2x/(1+ x²)} - 4 cos⁻¹{(1- x²)/(1+ x²)} + 2 tan⁻¹ {2x/(1+ x²) = π.
Type -2
1) cosec⁻¹cot tan⁻¹x = x
Type -3
1) cos⁻¹cos sin⁻¹x + cos⁻¹ sin cos⁻¹x =π/2.
2) 1/2 cos⁻¹{(5 cosx +3)/(5+ 3 cosx)} = tan⁻¹(1/2 tan x/2).
3) cos⁻¹{(2+ 3 cosx)/(3 + 2 cosx)} = 2 tan⁻¹(1/√5 tan x/2).
4) cos⁻¹{(cosx+ cos y)/(1 + cosx cos y)} = 2 tan⁻¹(tan x/2 tan y/2).
Rype-4
1) sec²(tan⁻¹3) + cosec²(cot⁻¹5) = 36.
2) sec²(tan⁻¹2)+ cosec²(Cot⁻¹3) =15
3) sec²(Cot⁻¹2) + cosec²(tan⁻¹3) = 85/36
4) sec²cot⁻¹(1/√3)+ tan²cosec⁻¹(√2)= 5
) sec²cot⁻¹(1/2)+ cosec²tan⁻¹(1/3)=15
Type -5
1) tan⁻¹(1/2 tan 2A)+ tan⁻¹(cot A) + tan⁻¹(cot³A) = 0.
2) tan⁻¹(cot x) + cot⁻¹(tanx) = π - 2x.
3) tan⁻¹(cot 2x) + tan⁻¹(- cot 3x) = x.
4) tan(π/4 + 1/2 cos⁻¹2/3) + tan (π/4 - 1/2 cos⁻¹2/3) = 3.
5) tan[2 tan⁻¹√{(1+ cosx)/(1- cosx)} + tan x = 0.
6) tan⁻¹x = 2 tan⁻¹(cosec tan⁻¹x - tan cot⁻¹x).
7) tan(π/4 + 1/2 cos⁻¹a/b) + tan (π/4 - 1/2 cos⁻¹a/b) = 2b/a.
8) tan(π/6 + 1/2 cos⁻¹x) + tan (π/3 - 1/2 cos⁻¹x) = 2/x.
9) tan⁻¹[√{(a - b)/(a+b)} tan x/2] = 1/2 cos⁻¹{a cosx +b)/(a+ b cosx)}.
10) tan(1/√3 + tan x/2) = 1/2 cos⁻¹{(1+ 2cosx)/(2+ cosx-)}.
11) tan⁻¹{(√2 +1) cot x} - tan⁻¹ {(√2 - 1) cot x)= tan⁻¹(sin 2x).
12) 1/2 tan⁻¹x = cos⁻¹√[{1+√(1+ x²)}/2√(1+ x²)]
13) tan[1/2 sin⁻¹{2x/(1 -x²)} + 1/2 cos⁻¹{(1- y²)/(1+y²)} = (x+ y)/(1- xy).
EXERCISE - J
SOLVE:
Type-1(a)
1) sin⁻¹x + sin⁻¹(1 - x)= cos⁻¹x. 0, 1/2
2) sin⁻¹ cos sin⁻¹x = π/3. ±1/2
3) sin⁻¹x + sin⁻¹2x =π/3. ±√21/14
4) sin⁻¹5/x + sin⁻¹12/x = π/2. 13
5) sin⁻¹6x + sin⁻¹(6√3 x)= -π/2. -1/12
6) sin{2 cos⁻¹ cot(2 tan⁻¹x)} = 0.
7) sin⁻¹(√3/2)+ 2 tan⁻¹ 1/√3 = 2π/3.
7) sin⁻¹{2a/(1 +a²)} + sin⁻¹{2b/(1+ b²)} = 2 tan⁻¹x.
9) sin⁻¹{2a/(1 +a²)} - cos⁻¹{(1- b²)/(1+ b²)} = tan⁻¹{2x/(1 - x²)}. (a- b)/(1+ab)
10) sin⁻¹{x/√(1+ x²)} - sin⁻¹{1/√(1 + x²)}= sin⁻¹{(1+x)/(1+ x²)}. 2
11) a) sin⁻¹(1/x) = Cos⁻¹√(3)/4
b) sin⁻¹(1/x) = Cos⁻¹√3/x. ±2
12) sin⁻¹(6x) + sin⁻¹(6√3x) = - π/2
13) sin⁻¹x + sin⁻¹2x = π/3
14) Sin⁻¹x + sin⁻¹(1 -x) = Cos⁻¹x
15 ) sin⁻¹√(3)x + sin⁻¹x = π/2
16) sin cosec⁻¹cot tan⁻¹x=1. 1
17) sin⁻¹5/x + sin⁻¹12/x=π/2. ±13
18) sin⁻¹x - sin⁻¹y =π/3 ; cos⁻¹x + cos⁻¹y = 2π/3. √3/2, 0
Type-1(b)
1) 2sin⁻¹x + sin⁻¹(1 - x)= π/2. 0, 1/2
2) 2sin⁻¹x = cos⁻¹x. -1, 1/2
Type -1(c)
1) 3sin⁻¹{2x/(1+x²)} - 4Cos⁻¹{(1-x²)/(1+x²)} + 2 tan⁻¹{2x/(1- x²)} = π/3. 1/√3
2) sin⁻¹x + sin⁻¹y= π; tan⁻¹x + tan⁻¹y=π/2. 1,1
Type -2
1) cos⁻¹(x √3)+ cos⁻¹x=π/2. 1/2
2) 1/2 cos⁻¹x= tan⁻¹1/2. 3/5
3) cos⁻¹x- cos⁻¹π/3 ; sin⁻¹x + sin⁻¹ y= 2π/3. 1/2, 1
Type -3
1) sec⁻¹x/2 + sec⁻¹x/b = sec⁻¹a + sec⁻¹b. ab
Type-4(a)
1)a) tan⁻¹x + 2 cot⁻¹x = 2π/3. √3
b) tan⁻¹x = cot⁻¹x. ±1
2) tan⁻¹(2x) + tan⁻¹(3x) = π/4. -1, 1/6
3) tan⁻¹(2/x) + tan⁻¹(x/4) = tan⁻¹3. 2,4
4) tan(2 tan⁻¹x)= √3. 1/√3, -√3
5) tancos⁻¹x = sin cot⁻¹1/2. ±√5/3
6) tan(Cos⁻¹x) = sin (tan⁻¹2)
7) tan⁻¹(cotx) + cot⁻¹(tanx) =π/4. 3π/8
8) tan(cos⁻¹x) = sin(cot⁻¹1/2)
9) tan⁻¹(2+x) + tan⁻¹(2- x) = tan⁻¹ 2/3. ±3
10) tan⁻¹1/(1 + 2x)+ tan⁻¹1/(1 + 4x)= tan⁻¹ 2/x²
11) tan⁻¹(1 +x)+ cot⁻¹(x -1)= sin⁻¹4/5 + tan⁻¹3/4.
12) tan⁻¹(x - 1) + tan⁻¹x + tan⁻¹(x +1)= tan⁻¹3x. 0, ±1/2
13) tan⁻¹{(1 - x)/(1+x)} = 1/2 sin⁻¹{x/√(1+ x²)}. 1/√3
14) tan⁻¹(1/2 secx) + cot⁻¹(2 cos x)= π/3. π/6
15) tan⁻¹(x +1) + tan⁻¹(x - 1)= tan⁻¹6/17. 1/3
16) tan⁻¹(x -1) + tan⁻¹(x+ 1)= tan⁻¹3x.
17) tan⁻¹x +tan⁻¹2x +tan⁻¹3x = π. 0,±1
18) tan⁻¹{2x/(1 -x²)} = sin⁻¹{2a/1+a²)} + sin⁻¹{2b/(1+b²)}
19) tan⁻¹{2x/(1 -x²)} = sin⁻¹{2a/1+a²)} - cos⁻¹{(1-b²)/(1+b²)}. (a- b)/(1+ ab)
20) tan⁻¹3x = tan⁻¹x + tan⁻¹(x+1) + tan⁻¹(x-1)
21) tan⁻¹1/4 + 2tan⁻¹1/5+ tan⁻¹1/6 + tan⁻¹1/x = π/4
22) Tan⁻¹(x-1)+Tan⁻¹x+Tan⁻¹(x+1)= tan⁻¹x
Type-4(b)
1) 2 tan⁻¹(cosx) = tan⁻¹(2 cosecx). π/4
2) 2 tan⁻¹{2x/(1- x²)} =π/3. 2-√3
3) 2Tan⁻¹(cosx)= Tan⁻¹(2cosecx).
4) 2 tan⁻¹{(1- x)/(1+ x)} =tan⁻¹x. 1/√3
Type -4(c)
1) 3 tan⁻¹{1/(2+ √3)} - tan⁻¹1/x = tan⁻¹1/3. 2
Type -5(a)
1) cot⁻¹x + cot⁻¹(a² - x +1) = cot⁻¹(a - 1). a, a²- a+ 1
Type -5(b)
1) 2 cot⁻¹{(1+x)/(1- x)} = tan⁻¹x. ±1/√3
EXERCISE -K
Type-1
1) If sin⁻¹ x + sin⁻¹y = 2π/3; and cos⁻¹x + cos⁻¹y π/3, then find the value of x and y.
2) If sin⁻¹ x + sin⁻¹y= π and tan⁻¹x + tan⁻¹ y= π/2, then find the value of x and y.
3) If sin⁻¹ x = tan⁻¹y, then show that 1/x² - 1/y² = 1.
4) If sin⁻¹ x + sin⁻¹y + sin⁻¹ z= π, then show that x√(1- x²) + y √(1- y²) + z √(1- z²) = 2xyz.
5) If sin(π cosx) = cos(π sinx), show that x= ± 1/2 sin⁻¹3/4
Type -2
1) If cos⁻¹ x + cos⁻¹y + cos⁻¹ z = π, then prove x²+ y² + z² +2xyz= 1.
2) If cos⁻¹ x/a + cos⁻¹y/b = m, then prove x²/a²- (2xy cos m)/ab +y²/b² = sin²m
3) If cos⁻¹ x + cos⁻¹y = m then show that x² - 2xy cos m + y²= sin² m.
4) If 4 cos 4A + 6 cos 2A - 5= 0, then show that A= 1/2 cos⁻¹ 3/4.
5) If x+ y = cos⁻¹m and x - y = cos⁻¹n, show that cos²x + cos²y = 1+ mn.
Type -3
1) If sec⁻¹ x = cosec⁻¹y, then show that cos⁻¹ 1/x + cos⁻¹ 1/y = π/2.
2) If sec⁻¹ x = cosec⁻¹y, then show that 1/x² + 1/y² = 1.
3) If sec A - cosec A = 4/3, show that A= 1/2 sin⁻¹ 3/4.
Type -4
1) If x² + y² + z²= r², prove tan⁻¹ yz/xr + tan⁻¹zx/yr + tan⁻¹ xy/zr = π/2.
2) If tan⁻¹ x + tan⁻¹y + tan⁻¹ z = π, then prove x+ y + z = xyz
3) if xy = 1+ a², show that tan⁻¹ {1/(a+ x)} + tan⁻¹{1/(a+y)= , tan⁻¹ 1/a, (x + y + 2a)≠ 0.
4) If tan⁻¹ x + tan⁻¹y + tan⁻¹z = π/2. Then prove xy + yz + zx = 1.
5) If tan⁻¹ x + tan⁻¹y + tan⁻¹ z = π/2, then prove (y + x)/(1- xy) = 1/z
6) tan⁻¹(yz/xr) + tan⁻¹(zx/yr) + tan⁻¹(xy/zr) =π/2, when x²+ y²+ z²= r²
7) If m= tan⁻¹{3x/(2a - x)} and n= tan⁻¹{(2x -a)/√3 a show that m - n = π/6.
8) If m=tan⁻¹{1/√(cos 2a)} - tan⁻¹√(cos2a) then show that sec m = ±1/2 (√sec2a + √cos 2a).
Type -5
1) If cot(sin⁻¹ 13/17) = cos(cot⁻¹x), then show that x = 2/3.
2) If m= cot⁻¹√cos 2x - tan⁻¹√cos 2x, then show sin m = tan²x
EXERCISE - L
Find the value of:
Type -1
1) sin(2tan⁻¹ 1/3) + cos(tan⁻¹2√2). 14/15
Type -2
1) tan[1/2 sin⁻¹{(2x/(1+ x²)} + 1/2 cos⁻¹(1- y²)/(2+ y²)]; xy≠ 1. (x+y)/(1- xy)
2) tan(tan⁻¹1/2 + tan⁻¹1/5 + tan⁻¹1/8. 1
Mg. A- R.1
Prove:
1) 2 tan⁻¹1/3 + tan⁻¹1/7 =π/4.
2) 2 tan⁻¹2 + tan⁻¹3 =π + tan⁻¹1/3
3) 2 cos⁻¹2/√5 + cos⁻¹3/5 = sin⁻¹24/25.
4) 4 tan⁻¹1/5 + tan⁻¹1/239 =π/4.
5) sin(π/2 - sin⁻¹3/5)= 4/5.
6) tan(cos⁻¹3/5 + tan⁻¹2/3) = 18.
7) sec²(tan⁻¹3) + cosec²(cot⁻¹5)= 36.
8) sin⁻¹cos(sin⁻¹x) + cos⁻¹ sin(cos⁻¹x) =π/2.
9) tan[1/2 sin⁻¹{2x/(1 +x²)} + 1/2 cos⁻¹{(1 - y²)/(1+ y²)}] = (x+y)/(1- xy).
10) tan⁻¹1/2 + 1/2 cos⁻¹4/5 =π/4.
Mg. A- R-2
1) 1) tan⁻¹1 + tan⁻¹2 + tan⁻¹3 =π.
2) cos⁻¹8/17 + cos⁻¹3/5 + cos⁻¹36/85 =π.
3) tan(1/2 cos⁻¹12/13)= 1/5
4) sin[cos⁻¹{(1- x²)/(1+ x²)} + cot⁻¹{2x/(1- x²)} = 1
5) cos⁻¹a - sin⁻¹b = cos⁻¹[b √(1- a²) + a √(1- b²)].
6) cos(cos⁻¹1/2 + 2 sin⁻¹1/2= 1/2
7) tan{π/4 + 1/2 cos⁻¹(a/b)}+ tan{π/4 - 1/2 cos⁻¹(a/b) =2b/a
8) tan⁻¹4/3 + tan⁻¹12/5 + tan⁻¹56/33 = π
9) tan⁻¹{x cos k/(1- x sin k)} - cot⁻¹{cos k/(x - sin k)} is independent of x and find it.
10) tan(tan⁻¹a + tan⁻¹b+ tan⁻¹c)= cot(cot⁻¹a+ cot⁻¹b + cot⁻¹c)
Mg. A- R.3
1) cos tan⁻¹ sin cot⁻¹x = √{(x²+1)/(x² +2)}
2) If sin⁻¹x + sin⁻¹y =π/2, show that 2(x²- xy + y²) = 1+ x⁴ + y⁴.
3) cos⁻¹{(cosx+ cosy)/(1+ cosx cosy) = 2 tan⁻¹(tan x/2 tan y/2)
4) tan⁻¹x= 2 tan⁻¹(cosec tan⁻¹x - tan cot⁻¹x)
5) tan⁻¹(1/2 tan 2A) + tan⁻¹(cot A) + tan⁻¹(cot³A) = 0
6) tan⁻¹p/q - tan⁻¹{(p - q)/(p+ q) =π/4.
7) 8 cosec²(1/2 tan⁻¹2/3) + 27 sec²(1/2 tan⁻¹3/2) =130
8) tan⁻¹x + tan⁻¹y + tan⁻¹{(1- x - y - xy)/(1+ x + y - xy)} =π/4.
9) x= tan⁻¹{(tanx)/4} + tan⁻¹{(3 sin 2x)/(5+ 3 cos 2x).
10) If tan⁻¹y + 4 tan⁻¹x, express y as algebraic function. y=(4x- 4x³)/(1- 6x²+ x⁴)
Mg. A- R.4
1) If cos⁻¹n= 2 sin⁻¹n then value of n is
a) 1 b)1/3 c) 1/2 d) 0 e) none
2) value of cos(cos⁻¹x + 2 sin⁻¹x) at x =3/5.
a) 3/5 b) -3/5 c) 5/3 d) -5/3 e) 1
3) If tan⁻¹x + tan⁻¹2x + tan⁻¹3x=π/4. Then x is
a) 0 b) 1 c) 1/2 d) √3 e) 1/√3
4) If cot⁻¹x + sin⁻¹1/√5 =π/4, then x is
a) 0 b) 1 c) 2 c) 3 d) 1/2 e) √3
5) If sec⁻¹x= cosec⁻¹y, then the value cos⁻¹1/x + cos⁻¹1/y is
a) π b) π/2 c) π/3 d) 2π/3 e) none
6) if tan⁻¹x{(x -1)/(x-2)} + tan⁻¹{(x+1)/(x+2)} =π/4 then x is
a) 1/√2 b) -1/√2 c) both of a and b d) one of them e) none
7) If cos⁻¹8/x + cos⁻¹15/x =π/2. Then x is
a) 0 b) 1/2 c) either a or b d) neither a nor b e) both a and b
8) If a+ b= cos⁻¹m and a - b = cos⁻¹n, then show cos²a + cos²b = 1 + mn.
9) If cos⁻¹x + cos⁻¹y + cos⁻¹z =π, then show x²+ y²+ z²+ 2xyz = 1
10) prove: tan⁻¹{(2a - b)/b√3} + tan⁻¹{(2b - a)/a√3)}= π/3.
Mg. A- R.5
1) If K= cos⁻¹x/a + cos⁻¹y/b then show that sin²K= x²/a² - 2xy/ab cos K + y²/b².
2) If K= cot⁻¹√cos2x - tan⁻¹√cos 2x then prove sin K = tan²x
3) If a= tan⁻¹{x√3/(2k - x)} and b= tan⁻¹{(2x - k)/k√3}, then show that one of the values of a - b is π/6.
4) Prove: cos{1/2 cos⁻¹(-1/9)}= 2/3
5) Show that: a cosx + b sin x =√(a²+ b²) cos(x - tan⁻¹b/a)= √(a²+ b²) sin(x + tan⁻¹a/b).
6) Prove: tan⁻¹{(ap - q)/(aq +p)}+ tan⁻¹{b - a)/(ab +1)}+ tan⁻¹{c - b)/(bc+1)}+ tan⁻¹ 1/c = tan⁻¹p/q.
7) Prove: 2 tan⁻¹a + 2 tan⁻¹b= sin⁻¹[{2(a+b)(1- ab)}/{(1+ a²)(1+ b²)}].
8) If tan⁻¹{(x+1)/(x-1)} + tan⁻¹{(x-1)/x}= tan⁻¹(-7), (x≠0,1)
a) 0 b) 1 c) 1/2 d) no solution
9) If tan⁻¹(yz/xr) +tan⁻¹(zx/yr)+ tan⁻¹(xy/zr)=π/2, prove x²+ y²+ z²= r².
10) If K= sin⁻¹(sin x + siny) - sin⁻¹(sinx - sin y), and sin²y + sin²x = 1/2 (y< x), then show that cos K = cos 2y - cos 2x.
Mg. A- R.6
1) value of tan⁻¹ sin cos⁻¹√(2/3) is
a) π b) π/2 c) π/3 d) π/6
2) Principal value of sin⁻¹sin(sin 5π/6)ᶜ
a) 1/2 b) 1 c) 1/2 radian d) 1 radian
3) Value of cos⁻¹x+ cos⁻¹(-x), when 0 < x < 1
a) π b) π/2 c) 2π d) π/3
4) Value of tan(cos⁻¹4/5 + tan⁻¹2/3).
a) 1 b) 1/2 c) 17 d) 17/6
5) Value of tan {1/2(tan⁻¹x+ tan⁻¹1/x)}.
a) 1 b) 2 c) 3 d) π/2
6) Value of sin{tan⁻¹(tan7π/6)+ cos⁻¹(cos 7π/3)}.
a) 0 b) 2 c) 1 d) 3
7) General value of cos⁻¹[1/2(-1)ⁿ].
a) nπ b) nπ± π/2 c) nπ+ 2π d) nπ ±π/3
8) If tan⁻¹1/3 = 18°26', then value of tan⁻¹1/2
a) 26° b) 26°34' c) 34° d) 34°26'
9) If sin⁻¹x+ sin⁻¹y= 2π/3, find the value of cos⁻¹x+ cos⁻¹y.
a) 0 b) 1 c) 1/2 d) none
10) Prove: tan(1/2 cos⁻¹a)= √{(1- a)/(1+ a)}.
Mg. A- R.7
1) Prove: cos⁻¹√(3/5)= 1/2 cos⁻¹1/5.
2) Prove: 2 cos⁻¹3/√13 + cot⁻¹16/63 + 1/2 cos⁻¹7/25 =π.
3) If sec⁻¹x/a + sec⁻¹x/b = sec⁻¹a + sec⁻¹b then x is
a) a b) b c) ab d) -a
4) If sin⁻¹x - cos⁻¹x = sin⁻¹(3x-2), then x is
a) 1 b) 1/2 c) 1,1/2 d) none
5) If cos⁻¹{(x²-1)/(x²+1)} + tan⁻¹{2x/(x²-1)}= 2π/3 then x is
a) √3, 2 -√3 b) 0,1 c) ±√3 d) 1, √3
6) If sin[2 cos⁻¹cot(2tan⁻¹x)]= 0, then x is
a) ±1 b) 1±√2 c) -1±√2 d) all
7) If tan⁻¹{(x-1)/(x+1)} + tan⁻¹{(2x-1)/(2x+1)}= tan⁻¹7/6 then x is
a) 0 b) 1 c) 2 d) 3
8) show:cos⁻¹√(2/3) - cos⁻¹{(√6+1)/2√3} =π/6
9) show: 2tan⁻¹[√50 - √18 - 1/√(3-2√2) =π/4
10) tan⁻¹(1/2 tan 2A) + tan⁻¹(cot A)+ tan⁻¹(cot³A)=π, where 0≤ A≤π/4.
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