Class 12-commerce RD SHARMA Solutions Maths Chapter 4: Inverse Trigonometric Functions
Inverse Trigonometric Functions Exercise Ex. 4.1
Solution 1(i)
Solution 1(ii)
Solution 1(iii)
Solution 1(iv)
Solution 1(v)
Solution 1(vi)
Solution 2(i)
Solution 2(ii)
Solution 3(i)
Solution 3(ii)
Solution 3(iii)
Solution 3(iv)
Solution 4
Solution 5
Inverse Trigonometric Functions Exercise Ex. 4.2
Solution 1
Solution 2
Solution 3
Solution 4(i)
Solution 4(ii)
Solution 4(iii)
Solution 4(iv)
Solution 5(i)
Solution 5(ii)
Solution 5(iii)
?
Solution 5(iv)
Inverse Trigonometric Functions Exercise Ex. 4.3
Solution 1(i)
Solution 1(ii)
Solution 1(iii)
Solution 1(iv)
Solution 2(i)
Solution 2(ii)
Solution 3(i)
Solution 3(ii)
Solution 3(iii)
Inverse Trigonometric Functions Exercise Ex. 4.4
Solution 1(i)
Solution 1(ii)
Solution 1(iii)
Solution 1(iv)
Solution 2(i)
Solution 2(ii)
Solution 3(i)
Solution 3(ii)
Inverse Trigonometric Functions Exercise Ex. 4.5
Solution 1(i)
Solution 1(ii)
Solution 1(iii)
Solution 1(iv)
Solution 2
Solution 3(i)
Solution 3(ii)
Solution 3(iii)
Solution 3(iv)
Inverse Trigonometric Functions Exercise Ex. 4.6
Solution 1(i)
Solution 1(ii)
Solution 1(iii)
Solution 1(iv)
Solution 2
Solution 3(i)
Solution 3(ii)
Solution 3(iii)
Solution 3(iv)
Inverse Trigonometric Functions Exercise Ex. 4.7
Solution 1(i)
We know that,
Hence, 
Solution 2(ii)
We know that,
Hence, 
Solution 2(iv)
We know that,
Hence, 
Solution 1(ii)
Solution 1(iii)
Solution 1(iv)
Solution 1(v)
Solution 1(vi)
Solution 1(vii)
Solution 1(viii)
Solution 1(ix)
Solution 1(x)
Solution 2(i)
Solution 2(iii)
Solution 2(v)
Solution 2(vi)
Solution 2(vii)
Solution 2(viii)
Solution 3(i)
Solution 3(ii)
Solution 3(iii)
Solution 3(iv)
Solution 3(v)
Solution 3(vi)
Solution 3(vii)
Solution 3(viii)
Solution 4(i)
Solution 4(ii)
Solution 4(iii)
Solution 4(iv)
Solution 4(v)
Solution 4(vi)
Solution 4(vii)
Solution 4(viii)
Solution 5(i)
Solution 5(ii)
Solution 5(iii)
Solution 5(iv)
Solution 5(v)
Solution 5(vi)
Solution 6(i)
Solution 6(ii)
Solution 6(iii)
Solution 6(iv)
Solution 6(v)
Solution 6(vi)
Solution 7(i)
Solution 7(ii)
Solution 7(iii)
Solution 7(iv)
Solution 7(v)
Solution 7(vi)
Solution 7(vii)
Solution 7(viii)
Solution 7(ix)
Solution 7(x)
Inverse Trigonometric Functions Exercise Ex. 4.8
Solution 1(i)
Solution 1(ii)
Solution 1(iii)
Solution 1(iv)
Solution 1(v)
Solution 1(vi)
Solution 1(vii)
Solution 1(viii)
Solution 1(ix)
Solution 2(i)
Solution 2(ii)
Solution 2(iii)
Solution 2(iv)
Solution 3
Solution 4
Inverse Trigonometric Functions Exercise Ex. 4.9
Solution 1(iii)
Hence, 
Solution 1(i)
Solution 1(ii)
Solution 2(i)
Solution 2(ii)
Solution 2(iii)
Solution 3
Inverse Trigonometric Functions Exercise Ex. 4.10
Solution 1(i)
Solution 1(ii)
Solution 1(iii)
Solution 1(iv)
Solution 1(v)
Solution 2
[ ∏/2 - sin-1 x ] + [ ∏/2 - sin-1y ] = ∏/4
sin-1x + sin-1y = ∏ - ∏/4
sin-1x + sin-1y = 3∏/4
Solution 3
On adding both the equation
Π/2 + sin-1y - cos-1y = Π/2
[ Π/2- cos-1y ] - cos-1y = 0
cos-1y = Π/4
y = 1/√2
on putting y=1/√2 in 2nd equation
cos-1x - Π/4 = Π/6
cos-1x = Π/4 + Π/6
x = cos(Π/4 + Π/6)
x = cos(Π/4)cos(Π/6)-sin(Π/4)sin(Π/6)
x = (√3-1)/2√2
Solution 4
cot(z) = 0 means z = Π/2, 3Π/2, 5Π/2 ………..
cos-1(3/5) + sin-1x = nΠ + Π/2
sin-1x = nΠ + Π/2 - cos-1(3/5)
sin-1x = nΠ + sin-1(3/5)
x = sin(nΠ + sin-1(3/5)) = (-1)n sin (sin-1(3/5))
x = (-1)n 3/5
Solution 5
[ Π/2 - cos-1x ]2 + (cos-1x)2 =17Π2/36
Π2/4 - Πcos-1x + 2(cos-1x)2 =17Π2/36
Let, cos-1x=u
2u2 - Πu + Π2/4 - 17Π2/36 = 0
2u2 - Πu - 2Π2/9 = 0
18u2 - 9Πu -2Π2 = 0
On factorizing
18u2 - 12Πu + 3Πu -2Π2 = 0
6u( 3u -2Π ) + Π( 3u -2Π ) = 0
( 3u -2Π )(6u + Π) = 0
u = -Π/6, 2Π/3
i.e. cos-1x = -Π/6, 2Π/3
but range of cos-1x is [0, π]
x = cos(Π/2 + Π/6)
x = -1/2
Solution 6
sin-1(1/5) + [ Π/2 - sin-1x ] = sin-11
sin-1(1/5) + Π/2 - sin-1x = Π/2
sin-1(1/5) - sin-1x = 0
x = 1/5
Solution 7
Π/2 - cos-1x = Π/6 + cos-1x
Π/3 = 2cos-1x
cos-1x = Π/6
x = √3/2
Solution 8
4sin-1x+cos-1x=Π
3sin-1x+sin-1x+cos-1x=Π
3sin-1x=Π/2 [sin-1x+cos-1x=Π/2]
sin-1x=Π/6
x = sinΠ/6=0.5
Solution 9
tan-1x+cot-1x=Π/2 so the above equation reduces to
cot-1x =2Π/3-Π/2 =Π/6
x= cotΠ/6 =√3
Solution 10
Given: 
Hence,
Inverse Trigonometric Functions Exercise Ex. 4.11
Solution 1(i)
Solution 1(ii)
Solution 1(iii)
Solution 2
Solution 3(i)
Solution 3(ii)
Solution 3(iii)
Solution 3(iv)
Solution 3(v)
Solution 3(vi)
Solution 3(vii)
Solution 3(viii)
Solution 3(ix)
Solution 3(x)
Given: 
As 
Solution 3(xi)
Given: 
As
But 
does not satisfy the given equation.
Therefore, 
Solution 4
Inverse Trigonometric Functions Exercise Ex. 4.12
Solution 1
Solution 2(i)
Solution 2(ii)
Solution 2(iii)
Solution 3(i)
Solution 3(ii)
Solution 3(iii)
Given: 
Inverse Trigonometric Functions Exercise Ex. 4.13
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Let 
Let 
We know that, sin(x + y) = sinx cosy + cosx siny
Inverse Trigonometric Functions Exercise Ex. 4.14
Solution 1(i)
Solution 1(ii)
Solution 1(iii)
Solution 1(iv)
Solution 2(i)
Solution 2(ii)
Solution 2(iii)
Solution 2(iv)
Solution 2(v)
Solution 2(vi)
Solution 2(vii)
Solution 2(viii)
Solution 2(ix)
Solution 2(x)
Solution 3
Solution 4(i)
Solution 4(ii)
Solution 4(iii)
Let 
Hence, 
Solution 5
Solution 6
Solution 7(i)
Solution 7(ii)
Solution 8(i)
Solution 8(ii)
Solution 8(iii)
Solution 8(iv)
Thus, the solution is 
Solution 8(v)
Given: 
Solution 8(vi)
Given: 
Using the formula
Solution 9
Solution 10
Solution 11
Inverse Trigonometric Functions Exercise MCQ
Solution 1
Correct option: (a)
Solution 2
Correct option: (d)
Solution 3
Correct option: (c)
Solution 4
Correct option: (a)
Solution 5
Correct option: (a)
Solution 6
Correct option: (b)
Solution 7
Correct option: (a)
Solution 8
Correct option: (a)
Solution 9
Correct option: (a)
Solution 10
Correct option: (c)
Solution 11
Correct option: (b)
Solution 12
Correct option: (a)
Solution 13
Correct option: (c)
Solution 14
Correct option: (a)
Solution 15
Correct option: (b)
Solution 16
Correct option: (d)
Solution 17
Correct option: (c)
Solution 18
Correct option: (b)
Solution 19
Correct option: (b)
Solution 20
Correct option: (d)
Solution 21
Correct option: (d)
Solution 22
Correct option: (a)
Solution 23
Correct option: (a)
Solution 24
Correct option: (c)
Solution 25
Correct option: (d)
Solution 26
Correct option: (a)
Solution 27
Correct option: (b)
Solution 28
Correct option: (c)
Solution 29
Correct option:(a)
Solution 30
Correct option: (c)
Solution 31
Correct option: (d)
Solution 32
Correct option: (c)
Solution 33
Correct option: (a)
Solution 34
Correct option: (c)
Solution 35
Correct option: (a)
Solution 36
Given: 
… (i)
We know that 
As we know that 
Comparing the above inequality with (i), we get
Solution 37
Let 
We know that, 
Solution 38
Let 
We know that, 
Solution 39
Given: 
We know that 
Solution 40
One
branch of cos-1 other than the principal
value branch corresponds to 
.
Solution 41
The
principal value branch of sec-1is 
.
Solution 42
The
principal value branch of tan-1is 
.
Solution 43
The
principal value branch of cosec-1is 
.
Solution 44
Let
We
know that 
Solution 45
Given: 3tan-1x + cot-1x = π
Solution 46
Hence,
Solution 47
We know that,
when xy < 1
Therefore,
is valid when
Hence,
the given expression is valid for 
Solution 48
We know that,
, which is the principal value branch of tan-1x
Now,
As
Inverse Trigonometric Functions Exercise Ex. 4VSAQ
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Solution 21
Solution 22
Solution 23
Solution 24
Solution 25
Solution 26
Solution 27
Solution 28
Hence,
Solution 29
Solution 30
Solution 31
Solution 32
Solution 33
Solution 34
Solution 35
Solution 36
Solution 37
Solution 38
Solution 39
Let
Solution 40
Solution 41
Solution 42
Solution 43
Solution 44
Solution 45
Solution 46
We know that,
Hence,
Solution 47
Let
The
range of 
is R - (-1,
1).
So,
does not
belong to this range.
Thus,
is undefined.
Hence,
Solution 48
Hence,
Solution 49
Using
the property that, when 
we have
Solution 50
We
know that when
As
we have
Solution 51
Hence,
Solution 52
Let
Now, range of cosec x is R - (-1, 1)
As
does not
belong to this range, so cosec x also does not belong to this range.
Thus, the value of x will be undefined.
Hence,
Solution 53
For x < 0,
Hence,
Solution 54
for all 
Solution 55
We
know that,
Solution 56
Given:
Solution 57
Hence,
Solution 58
Given:
Hence,
the value of x is 
Solution 59
Hence,
Solution 60
Hence,
Solution 61
As
Hence,
Solution 62
As
Hence,
Solution 63
Let
Now,
Let
be the third
angle.
Hence,
the third angle is 