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Page 1
JEE Solved Example on Inverse Trigonometric Function
JEE Mains
Q?? : ?????? ?????? -?? ?? is equal to
(A)
v ?? -?? ?? ??
(B)
?? v ?? +?? ??
(C)
v ?? +?? ?? ??
(D) ?? v?? + ?? ??
Ans: ( ?? )= tan cos
-1
?? =
v1-?? 2
??
Q?? : |?????? -?? ?? |
?? + |?????? -?? ?? |
?? + ?? |?????? -?? ?? ||?????? -?? ?? | = ?? ?? then ?? ?? + ?? ?? is equal to
(A) 1
(B) ?? /??
(C) 2
(D) ½
Ans: (C) ( |sin
-1
?? | + |sin
-1
?? |
2
)= ?? 2
? ( |sin
-1
?? | + |sin
-1
?? |)= ?? ? |sin
-1
?? | =
?? 2
= |sin
-1
?? |
? ?? = ±?? = ±1
? ?? 2
+ ?? 2
= 2
Q3: Number of solution(s) of the equation ?????? -?? v?? ?? + ???? + ?? + ?????? -?? v ?? ?? - ???? - ?? =
?? ?? is
(A) 2
(B) 1
(C) 4
(D) Infinite
Ans: (A)
cot
-1
v( ?? - 1) ( ?? - 2)+ cos
-1
v
( ?? -
3
4
)
2
+
3
4
=
?? 2
Domain for cot
-1
v( ?? - 1) ( ?? - 2) is
?? ? ( -8, 1] ? [2, 8)
Page 2
JEE Solved Example on Inverse Trigonometric Function
JEE Mains
Q?? : ?????? ?????? -?? ?? is equal to
(A)
v ?? -?? ?? ??
(B)
?? v ?? +?? ??
(C)
v ?? +?? ?? ??
(D) ?? v?? + ?? ??
Ans: ( ?? )= tan cos
-1
?? =
v1-?? 2
??
Q?? : |?????? -?? ?? |
?? + |?????? -?? ?? |
?? + ?? |?????? -?? ?? ||?????? -?? ?? | = ?? ?? then ?? ?? + ?? ?? is equal to
(A) 1
(B) ?? /??
(C) 2
(D) ½
Ans: (C) ( |sin
-1
?? | + |sin
-1
?? |
2
)= ?? 2
? ( |sin
-1
?? | + |sin
-1
?? |)= ?? ? |sin
-1
?? | =
?? 2
= |sin
-1
?? |
? ?? = ±?? = ±1
? ?? 2
+ ?? 2
= 2
Q3: Number of solution(s) of the equation ?????? -?? v?? ?? + ???? + ?? + ?????? -?? v ?? ?? - ???? - ?? =
?? ?? is
(A) 2
(B) 1
(C) 4
(D) Infinite
Ans: (A)
cot
-1
v( ?? - 1) ( ?? - 2)+ cos
-1
v
( ?? -
3
4
)
2
+
3
4
=
?? 2
Domain for cot
-1
v( ?? - 1) ( ?? - 2) is
?? ? ( -8, 1] ? [2, 8)
while cos
-1
v
( ?? -
3
2
)
2
+
3
4
is defined for ?? = [1,2]
At ?? = 1
? cot
-1
( 0)+ cos
-1
( 1)=
?? 2
At ?? = 2
? cot
-1
( 0)+ cot
-1
( 1)=
?? 2
Hence two solutions.
Q4: If ?????? ( ?????? -?? ?? )= ?? , then ?? ?? is equal to
(A)
v ?? +?? ??
(B)
v ?? -?? ??
(C)
v ?? +?? ??
(D)
v ?? -?? ??
Ans: (D) cos ( tan
-1
?? )=
1
v 1 + ?? 2
= ?? 1
1 + ?? 2
= ?? 2
? ?? 4
+ ?? 2
- 1 = 0
?? 2
= ?? ? ?? 2
+ ?? - 1 = 0
? ( ?? +
1
2
)
2
=
5
4
?? 2
= -
1
2
±
v 5
2
= positive
?? 2
=
v 5 - 1
2
Page 3
JEE Solved Example on Inverse Trigonometric Function
JEE Mains
Q?? : ?????? ?????? -?? ?? is equal to
(A)
v ?? -?? ?? ??
(B)
?? v ?? +?? ??
(C)
v ?? +?? ?? ??
(D) ?? v?? + ?? ??
Ans: ( ?? )= tan cos
-1
?? =
v1-?? 2
??
Q?? : |?????? -?? ?? |
?? + |?????? -?? ?? |
?? + ?? |?????? -?? ?? ||?????? -?? ?? | = ?? ?? then ?? ?? + ?? ?? is equal to
(A) 1
(B) ?? /??
(C) 2
(D) ½
Ans: (C) ( |sin
-1
?? | + |sin
-1
?? |
2
)= ?? 2
? ( |sin
-1
?? | + |sin
-1
?? |)= ?? ? |sin
-1
?? | =
?? 2
= |sin
-1
?? |
? ?? = ±?? = ±1
? ?? 2
+ ?? 2
= 2
Q3: Number of solution(s) of the equation ?????? -?? v?? ?? + ???? + ?? + ?????? -?? v ?? ?? - ???? - ?? =
?? ?? is
(A) 2
(B) 1
(C) 4
(D) Infinite
Ans: (A)
cot
-1
v( ?? - 1) ( ?? - 2)+ cos
-1
v
( ?? -
3
4
)
2
+
3
4
=
?? 2
Domain for cot
-1
v( ?? - 1) ( ?? - 2) is
?? ? ( -8, 1] ? [2, 8)
while cos
-1
v
( ?? -
3
2
)
2
+
3
4
is defined for ?? = [1,2]
At ?? = 1
? cot
-1
( 0)+ cos
-1
( 1)=
?? 2
At ?? = 2
? cot
-1
( 0)+ cot
-1
( 1)=
?? 2
Hence two solutions.
Q4: If ?????? ( ?????? -?? ?? )= ?? , then ?? ?? is equal to
(A)
v ?? +?? ??
(B)
v ?? -?? ??
(C)
v ?? +?? ??
(D)
v ?? -?? ??
Ans: (D) cos ( tan
-1
?? )=
1
v 1 + ?? 2
= ?? 1
1 + ?? 2
= ?? 2
? ?? 4
+ ?? 2
- 1 = 0
?? 2
= ?? ? ?? 2
+ ?? - 1 = 0
? ( ?? +
1
2
)
2
=
5
4
?? 2
= -
1
2
±
v 5
2
= positive
?? 2
=
v 5 - 1
2
Q5: The value of a for which ?? ?? + ???? + ?????? -?? ( ?? ?? - ?? ?? + ?? )+ ?????? -?? ( ?? ?? - ?? ?? + ?? )= ?? , is
(A)
?? ?? + ??
(B)
?? ?? + ??
(C) - |
?? ?? + ?? |
(D) - (
?? ?? + ?? )
Ans: (D) ?? 2
- 4?? + 5 = ( ?? - 2)
2
+ 1
?? = 2, to define sin
-1
( ?? 2
- 4?? + 5)
So 4 + 2?? +
?? 2
+ 0 = 0 ? ?? = -
?? 4
- 2
Q6: Domain of the function ?? ( ?? )= v?????? -?? ( ?????? ?? )+ v?????? -?? ( ?????? ?? ) is
(A) |?? ???? , ?? ???? +
?? ?? | , ?? ? ??
(B) [( ?? ?? + ?? ) ?? , ( ?? ?? + ?? ) ?? ], ?? ? ??
(C) [?? ???? , ( ?? ?? + ?? ) ?? ], ?? ? ??
(D) |?? ???? +
?? ?? , ?? ???? +
?? ?? ?? | , ?? ? ??
Ans: (C) ?? ( ?? )= v sin
-1
sin ?? + vcos
-1
cos ??
sin ?? must not be negative to define ?? ( ?? ) . So the domain is ?? ? [2???? , ( 2?? + 1) ?? ], ?? ? ??
Q7: If ?? - ?????? -?? ?? + ?????? -?? ?? - ?????? -?? ?? , ?? ? [?? , ?? ] Then the interval in which ?? lies is given by
(A) [?? ,
?? ?? ]
(B) [
?? ?? ,
?? ?? ]
(C) [?? ,
?? ?? ]
(D) [
?? ?? ,
?? ?? ?? ]
Ans: (B) 0 = sin
-1
?? + cos
-1
?? - tan
-1
?? =
?? 2
- tan
-1
?? ?? ? [0,1] ?
?? 4
? 0 =
?? 2
Q8: The function
?? ( ?? )= ?????? -?? v( ?? + ?? ) ?? + ?????? -?? v?? ?? + ?? ?? + ??
is defined on the set ?? where ?? =
(A) {?? , ?? }
(B) ( ?? , ?? )
Page 4
JEE Solved Example on Inverse Trigonometric Function
JEE Mains
Q?? : ?????? ?????? -?? ?? is equal to
(A)
v ?? -?? ?? ??
(B)
?? v ?? +?? ??
(C)
v ?? +?? ?? ??
(D) ?? v?? + ?? ??
Ans: ( ?? )= tan cos
-1
?? =
v1-?? 2
??
Q?? : |?????? -?? ?? |
?? + |?????? -?? ?? |
?? + ?? |?????? -?? ?? ||?????? -?? ?? | = ?? ?? then ?? ?? + ?? ?? is equal to
(A) 1
(B) ?? /??
(C) 2
(D) ½
Ans: (C) ( |sin
-1
?? | + |sin
-1
?? |
2
)= ?? 2
? ( |sin
-1
?? | + |sin
-1
?? |)= ?? ? |sin
-1
?? | =
?? 2
= |sin
-1
?? |
? ?? = ±?? = ±1
? ?? 2
+ ?? 2
= 2
Q3: Number of solution(s) of the equation ?????? -?? v?? ?? + ???? + ?? + ?????? -?? v ?? ?? - ???? - ?? =
?? ?? is
(A) 2
(B) 1
(C) 4
(D) Infinite
Ans: (A)
cot
-1
v( ?? - 1) ( ?? - 2)+ cos
-1
v
( ?? -
3
4
)
2
+
3
4
=
?? 2
Domain for cot
-1
v( ?? - 1) ( ?? - 2) is
?? ? ( -8, 1] ? [2, 8)
while cos
-1
v
( ?? -
3
2
)
2
+
3
4
is defined for ?? = [1,2]
At ?? = 1
? cot
-1
( 0)+ cos
-1
( 1)=
?? 2
At ?? = 2
? cot
-1
( 0)+ cot
-1
( 1)=
?? 2
Hence two solutions.
Q4: If ?????? ( ?????? -?? ?? )= ?? , then ?? ?? is equal to
(A)
v ?? +?? ??
(B)
v ?? -?? ??
(C)
v ?? +?? ??
(D)
v ?? -?? ??
Ans: (D) cos ( tan
-1
?? )=
1
v 1 + ?? 2
= ?? 1
1 + ?? 2
= ?? 2
? ?? 4
+ ?? 2
- 1 = 0
?? 2
= ?? ? ?? 2
+ ?? - 1 = 0
? ( ?? +
1
2
)
2
=
5
4
?? 2
= -
1
2
±
v 5
2
= positive
?? 2
=
v 5 - 1
2
Q5: The value of a for which ?? ?? + ???? + ?????? -?? ( ?? ?? - ?? ?? + ?? )+ ?????? -?? ( ?? ?? - ?? ?? + ?? )= ?? , is
(A)
?? ?? + ??
(B)
?? ?? + ??
(C) - |
?? ?? + ?? |
(D) - (
?? ?? + ?? )
Ans: (D) ?? 2
- 4?? + 5 = ( ?? - 2)
2
+ 1
?? = 2, to define sin
-1
( ?? 2
- 4?? + 5)
So 4 + 2?? +
?? 2
+ 0 = 0 ? ?? = -
?? 4
- 2
Q6: Domain of the function ?? ( ?? )= v?????? -?? ( ?????? ?? )+ v?????? -?? ( ?????? ?? ) is
(A) |?? ???? , ?? ???? +
?? ?? | , ?? ? ??
(B) [( ?? ?? + ?? ) ?? , ( ?? ?? + ?? ) ?? ], ?? ? ??
(C) [?? ???? , ( ?? ?? + ?? ) ?? ], ?? ? ??
(D) |?? ???? +
?? ?? , ?? ???? +
?? ?? ?? | , ?? ? ??
Ans: (C) ?? ( ?? )= v sin
-1
sin ?? + vcos
-1
cos ??
sin ?? must not be negative to define ?? ( ?? ) . So the domain is ?? ? [2???? , ( 2?? + 1) ?? ], ?? ? ??
Q7: If ?? - ?????? -?? ?? + ?????? -?? ?? - ?????? -?? ?? , ?? ? [?? , ?? ] Then the interval in which ?? lies is given by
(A) [?? ,
?? ?? ]
(B) [
?? ?? ,
?? ?? ]
(C) [?? ,
?? ?? ]
(D) [
?? ?? ,
?? ?? ?? ]
Ans: (B) 0 = sin
-1
?? + cos
-1
?? - tan
-1
?? =
?? 2
- tan
-1
?? ?? ? [0,1] ?
?? 4
? 0 =
?? 2
Q8: The function
?? ( ?? )= ?????? -?? v( ?? + ?? ) ?? + ?????? -?? v?? ?? + ?? ?? + ??
is defined on the set ?? where ?? =
(A) {?? , ?? }
(B) ( ?? , ?? )
(C) {?? , -?? }
(D) ( -?? , ?? )
Ans: (C) ?? ( ?? )= cot
-1
v( ?? + 3) ?? + cos
-1
v?? 2
+ 3?? + 1
= cot
-1
v?? ( ?? + 3)+ cos
-1
v
( ?? +
3
2
)
2
-
5
4
?? ( ?? + 3)= 0 ? ?? ? ( -8, -3] ? [0, 8)
( ?? +
3
2
)
2
-
5
4
? 0
? ?? ? (-8,
-3
2
-
v 5
2
] ? [
-3
2
+
v 5
2
, 8)
( ?? +
3
2
)
2
-
5
4
= 1 ? ?? ? [-3,0]
So the answer ?? ? {0, -3}
Q9: ?? = ?????? -?? ( ?????? ( ?????? -?? ?? ) ) and ?? = ?????? -?? ( ?????? ( ?????? -?? ?? ) ) then
(A) ?????? ?? - ?????? ??
(B) ?????? ?? - -?????? ??
(C) ?????? ?? = ?????? ??
(D) ?????? ?? = -?????? ??
Ans: (A) ?? = sin
-1
cos sin
-1
x
?? = cos
-1
sin cos
-1
?? tan ?? = tan sin
-1
cos sin
-1
?? = tan sin
-1
v1 - ?? 2
=
v 1 - ?? 2
?? tan ?? = tan cos
-1
sin cos
-1
?? = tan cos
-1
v1 - ?? 2
=
?? v 1 - ?? 2
cot ?? = tan ??
Q10: If ?? = ?? ?????? -?? (
?? ?? )+ ?????? -?? (
?? ?? )+ ?????? -?? ( v ?? ) and ?? = ?????? (
?? ?? ?????? -?? ( ??????
?? ?? ) ) then which of
the following statements hold good?
(A) ?? = ??????
?? ?? ????
(B) ?? = ??????
?? ?? ????
Page 5
JEE Solved Example on Inverse Trigonometric Function
JEE Mains
Q?? : ?????? ?????? -?? ?? is equal to
(A)
v ?? -?? ?? ??
(B)
?? v ?? +?? ??
(C)
v ?? +?? ?? ??
(D) ?? v?? + ?? ??
Ans: ( ?? )= tan cos
-1
?? =
v1-?? 2
??
Q?? : |?????? -?? ?? |
?? + |?????? -?? ?? |
?? + ?? |?????? -?? ?? ||?????? -?? ?? | = ?? ?? then ?? ?? + ?? ?? is equal to
(A) 1
(B) ?? /??
(C) 2
(D) ½
Ans: (C) ( |sin
-1
?? | + |sin
-1
?? |
2
)= ?? 2
? ( |sin
-1
?? | + |sin
-1
?? |)= ?? ? |sin
-1
?? | =
?? 2
= |sin
-1
?? |
? ?? = ±?? = ±1
? ?? 2
+ ?? 2
= 2
Q3: Number of solution(s) of the equation ?????? -?? v?? ?? + ???? + ?? + ?????? -?? v ?? ?? - ???? - ?? =
?? ?? is
(A) 2
(B) 1
(C) 4
(D) Infinite
Ans: (A)
cot
-1
v( ?? - 1) ( ?? - 2)+ cos
-1
v
( ?? -
3
4
)
2
+
3
4
=
?? 2
Domain for cot
-1
v( ?? - 1) ( ?? - 2) is
?? ? ( -8, 1] ? [2, 8)
while cos
-1
v
( ?? -
3
2
)
2
+
3
4
is defined for ?? = [1,2]
At ?? = 1
? cot
-1
( 0)+ cos
-1
( 1)=
?? 2
At ?? = 2
? cot
-1
( 0)+ cot
-1
( 1)=
?? 2
Hence two solutions.
Q4: If ?????? ( ?????? -?? ?? )= ?? , then ?? ?? is equal to
(A)
v ?? +?? ??
(B)
v ?? -?? ??
(C)
v ?? +?? ??
(D)
v ?? -?? ??
Ans: (D) cos ( tan
-1
?? )=
1
v 1 + ?? 2
= ?? 1
1 + ?? 2
= ?? 2
? ?? 4
+ ?? 2
- 1 = 0
?? 2
= ?? ? ?? 2
+ ?? - 1 = 0
? ( ?? +
1
2
)
2
=
5
4
?? 2
= -
1
2
±
v 5
2
= positive
?? 2
=
v 5 - 1
2
Q5: The value of a for which ?? ?? + ???? + ?????? -?? ( ?? ?? - ?? ?? + ?? )+ ?????? -?? ( ?? ?? - ?? ?? + ?? )= ?? , is
(A)
?? ?? + ??
(B)
?? ?? + ??
(C) - |
?? ?? + ?? |
(D) - (
?? ?? + ?? )
Ans: (D) ?? 2
- 4?? + 5 = ( ?? - 2)
2
+ 1
?? = 2, to define sin
-1
( ?? 2
- 4?? + 5)
So 4 + 2?? +
?? 2
+ 0 = 0 ? ?? = -
?? 4
- 2
Q6: Domain of the function ?? ( ?? )= v?????? -?? ( ?????? ?? )+ v?????? -?? ( ?????? ?? ) is
(A) |?? ???? , ?? ???? +
?? ?? | , ?? ? ??
(B) [( ?? ?? + ?? ) ?? , ( ?? ?? + ?? ) ?? ], ?? ? ??
(C) [?? ???? , ( ?? ?? + ?? ) ?? ], ?? ? ??
(D) |?? ???? +
?? ?? , ?? ???? +
?? ?? ?? | , ?? ? ??
Ans: (C) ?? ( ?? )= v sin
-1
sin ?? + vcos
-1
cos ??
sin ?? must not be negative to define ?? ( ?? ) . So the domain is ?? ? [2???? , ( 2?? + 1) ?? ], ?? ? ??
Q7: If ?? - ?????? -?? ?? + ?????? -?? ?? - ?????? -?? ?? , ?? ? [?? , ?? ] Then the interval in which ?? lies is given by
(A) [?? ,
?? ?? ]
(B) [
?? ?? ,
?? ?? ]
(C) [?? ,
?? ?? ]
(D) [
?? ?? ,
?? ?? ?? ]
Ans: (B) 0 = sin
-1
?? + cos
-1
?? - tan
-1
?? =
?? 2
- tan
-1
?? ?? ? [0,1] ?
?? 4
? 0 =
?? 2
Q8: The function
?? ( ?? )= ?????? -?? v( ?? + ?? ) ?? + ?????? -?? v?? ?? + ?? ?? + ??
is defined on the set ?? where ?? =
(A) {?? , ?? }
(B) ( ?? , ?? )
(C) {?? , -?? }
(D) ( -?? , ?? )
Ans: (C) ?? ( ?? )= cot
-1
v( ?? + 3) ?? + cos
-1
v?? 2
+ 3?? + 1
= cot
-1
v?? ( ?? + 3)+ cos
-1
v
( ?? +
3
2
)
2
-
5
4
?? ( ?? + 3)= 0 ? ?? ? ( -8, -3] ? [0, 8)
( ?? +
3
2
)
2
-
5
4
? 0
? ?? ? (-8,
-3
2
-
v 5
2
] ? [
-3
2
+
v 5
2
, 8)
( ?? +
3
2
)
2
-
5
4
= 1 ? ?? ? [-3,0]
So the answer ?? ? {0, -3}
Q9: ?? = ?????? -?? ( ?????? ( ?????? -?? ?? ) ) and ?? = ?????? -?? ( ?????? ( ?????? -?? ?? ) ) then
(A) ?????? ?? - ?????? ??
(B) ?????? ?? - -?????? ??
(C) ?????? ?? = ?????? ??
(D) ?????? ?? = -?????? ??
Ans: (A) ?? = sin
-1
cos sin
-1
x
?? = cos
-1
sin cos
-1
?? tan ?? = tan sin
-1
cos sin
-1
?? = tan sin
-1
v1 - ?? 2
=
v 1 - ?? 2
?? tan ?? = tan cos
-1
sin cos
-1
?? = tan cos
-1
v1 - ?? 2
=
?? v 1 - ?? 2
cot ?? = tan ??
Q10: If ?? = ?? ?????? -?? (
?? ?? )+ ?????? -?? (
?? ?? )+ ?????? -?? ( v ?? ) and ?? = ?????? (
?? ?? ?????? -?? ( ??????
?? ?? ) ) then which of
the following statements hold good?
(A) ?? = ??????
?? ?? ????
(B) ?? = ??????
?? ?? ????
(C) ?? - ?? ?????? -?? ??
(D) None of these
Ans: ( ?? ) ?? = 2cos
-1
(
1
2
)+ sin
-1
(
-1
2
)+ tan
-1
v 3
?? =
2?? 3
+
?? 4
+
?? 3
; ?? =
5?? 4
?? = cos (
1
2
sin
-1
( sin
?? 2
) ) = cos
1
2
sin
-1
sin
5?? 8
= cos
1
2
[?? -
5?? 8
] = cos
3?? 16
Q11: The set values of ?? satisfying the equation ?????? ?? ( ?????? -?? ?? )> ?? is
(A) [-?? , ?? ]
(B) [-
v ?? ?? ,
v ?? ?? ]
(C) ( -?? , ?? )- [-
v ?? ?? ,
v ?? ?? ]
(D [-?? , ?? ] - ( -
v ?? ?? ,
v ?? ?? )
Ans: ( ?? ) [tan ( sin
-1
?? ) ]
2
= [
?? v 1 - ?? 2
]
2
> 1
?
?? 2
1 - ?? 2
> 1
?
-1 + 2?? 2
1 - ?? 2
> 0
?
( v 2?? - 1) ( v 2?? + 1)
( ?? - 1) ( ?? + 1)
> 0
?? ? ( -1,1)- [-
v 2
2
,
v 2
2
]
Q12: The equation ?????? -?? ?? - ?? ?????? -?? a has a solution for
(A) All real values of ??
(B) ?? < -??
(C) ?? > ??
(D)
-?? v ?? = ?? =
?? v ??
Ans: (D) sin
-1
?? = 2sin
-1
??
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FAQs on Solved Example: Inverse Trigonometric Function - Mathematics (Maths) for JEE Main & Advanced
| 1. How to find the inverse trigonometric function of a given value? |  |
Ans. To find the inverse trigonometric function of a given value, you can use the inverse trigonometric functions such as arcsin, arccos, arctan, etc. Simply input the value into the corresponding inverse trigonometric function to find the angle.
| 2. What are the properties of inverse trigonometric functions? | |
Ans. Some properties of inverse trigonometric functions include limited output ranges, periodicity, and their relationships with trigonometric functions. It is important to understand these properties when working with inverse trigonometric functions.
| 3. How are inverse trigonometric functions used in solving equations? | |
Ans. Inverse trigonometric functions are commonly used to solve trigonometric equations. By applying the appropriate inverse trigonometric function to both sides of the equation, you can isolate the variable and solve for its value.
| 4. Can inverse trigonometric functions be used to find angles in a triangle? | |
Ans. Yes, inverse trigonometric functions can be used to find angles in a triangle. By using the trigonometric ratios and their inverses, you can determine the measures of angles in a triangle based on the given side lengths.
| 5. How can inverse trigonometric functions be applied in real-life situations? | |
Ans. Inverse trigonometric functions have various applications in real-life situations such as engineering, physics, and navigation. They are used to calculate angles, distances, and heights in a wide range of practical scenarios.
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Nov 24, 2024 Last updated |
Document Description: Solved Example: Inverse Trigonometric Function for JEE 2024 is part of Mathematics (Maths) for JEE Main & Advanced preparation.
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The content of Solved Example: Inverse Trigonometric Function is prepared as per the latest JEE syllabus.
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