Inverse circular function

INVERSE CORCULAR FUNCTIONS
1.00 Inverse Function Definition
111
If1 function is one to one and onto from A to B, then function g which associates each element y
a B to
one and only one element x A, such that y = f(x), then g is called the inverse function of f, denoted by x
= g(y). Usually we denote g f –1 {Read as f inverse}

x f –1 ( y).

1.01 Inverse Trigonometric Function
111
We have seen that the trigonometric functions, sin, cos etc. are all periodic and thus, each of them
1
1
achieves the same numerical value at an infinite number of points. Thus, the equation sin x has an
2

infinite number of solutions, viz., x , – etc. If one is to answer the question : “ What is the angle
6 6
1
whose sine is ?”, there is no unique answer. The difficulty arises as the function f : R R defined by
2
f ( x) sin x is not one to one and thus, does not admit of an inverse. To achieve a unique answer to the

aforesaid question we restrict the domain of sin x so that the resulting function is invertible. Thus, the

function g : – , [–1, 1] defined by g ( x) sin x is one to one and onto and admits of an inverse
2 2

(denoted by h sin –1 and read as sin inverse or arc sin) defined as h :[–1, 1] – , where
2 2

h( y) x if y sin x . The function sin –1 is the inverse of the sin function when the sin function is viewed in

a restricted sense.

We similarly define the other inverse trigonometric functions

Important Points

1: sin –1 x is an angle and denotes the smallest numerical angle, whose sine is x.

2: If there are two angles one positive and the other negative having same numerical value. Then we
shall take the positive value.

Progression / APEX INSTITUTE FOR IIT-JEE / AIEEE / PMT, 0120-4901457, +919990495952, +919910817866 Page 1

1.02 Inverse Trigonometric Function
111
Here, sin –1 x,cos ec –1 x, tan –1 x belongs to I and IV quadrant.
1

Here, cos –1 x, sec–1 x, cot –1 x belongs to I and II quadrant.

1. I quadrant is common to all the inverse functions.
2. III quadrant is not used inverse function.
3. IV quadrant is used in the clockwise direction i.e., – / 2 y 0 .

1.03 Domain, Range And Graphs of Inverse Functions

1. If sin y = x, then y sin –1 x under certain condition.
–1 sin y 1; but sin y x

–1 x 1
Again, sin y = –1 y = – /2 and sin y = 1 y = /2

Keeping in mind numerically smallest angles or real numbers.
– /2 y /2

These restrictions on the values of x and y provide us with the domain and range for the function
y sin –1 x .

i.e., Domain : x [–1, 1]
Range : y [– / 2, / 2]

2. Let cos y = x then y cos –1 x under certain condition –1 cos y 1 .


–1 x 1
cos y –1 y
cos y 1 y 0
0 y {as cos x is a decreasing function in [0, ]; hence cos cos y cos 0}

These restrictions on the values of x and y provide us the domain and range for the function
y cos –1 x .

i.e., Domain : x [–1, 1]
Range : y [0, ]

3. If tan y = x then y tan –1 x , under certain conditions.

Here, tan y R x R
– tan y – /2 y /2
Thus, domain x R
Range y (– / 2, / 2)

4. If cot y = x, then y cot –1 x (under certain conditions)
cot y R x R;
– cot y 0 y

These conditions on x and y make the function, cot y = x one–one
and onto so that the inverse function exists.
i.e., y cot –1 x is meaningful.
i.e., Domain: x R
Range : y (0, )

5. If sec y = x, then sec–1 x, where x 1 and 0 y ,y /2

Here, Domain : x R – (–1, 1)
Range : y [0, ] – { / 2}

6. If cosec y = x then y cos ec –1 x,

where x 1 and – /2 y / 2, y 0
Here, Domain : R – (–1, 1)
Range : [– / 2, / 2] – {0}


1.04 Principal values & Domains of Inverse Trigonometric / Circular Functions

Function Domain Range

(i ) y sin –1 x where –1 x 1 – y
2 2
–1
(ii ) y cos x where –1 x 1 0 y

(iii ) y tan –1 x where x R – y
2 2

(iv) y cos ec –1 x where x –1 or x 1 – y ,y 0
2 2

(v ) y sec –1 x where x –1 or x 1 0 y ;y
2
(vi ) y cot –1 x where x R 0 y

Note :

(a) 1st quadrant is common to the range of all the inverse functions.

(b) 2nd quadrant is not used in inverse functions.

(c) 4th quadrant is used in the clockwise direction i.e. – y 0.
2

(d) No inverse function is periodic. (See the graphs on page 17)

1 1
Illustration 1 : Find the value of tan cos –1 tan –1 – .
2 3

1 1
Solution : Let y tan cos –1 tan –1 –
2 3

 tan – tan
3 6 6
1
y Ans.
3

Illustration 2 : Find the domain of sin –1 (2 x2 –1) .

Solution : Let y sin –1 (2 x 2 – 1)
For y to be defined
– 1 (2 x 2 – 1) 1
0 2x2 2 0 x2 1
x [–1, 1]


1.05 Properties of Inverse Trigonometric Functions
111
Property – 2( A)
1 –1 –1
(i ) sin(sin x) x, –1 x 1 (ii ) cos(cos x) x, –1 x 1
(iii) tan(tan –1 x) x, x R (iv) cot(cot –1 x) x, x R
–1 –1
(v ) sec(sec x) x, x –1, x 1 (vi ) cos ec(cos ec x) x, x –1, x 1

These functions are equal to identity function in their whole domain which may or may not be R.
(See the graphs on page 18)

3
Illustration 3 : Fin the values of cos ec cot cot –1 .
4

3
Solution : Let y cos ec cot cot –1 .........(i )
4
3 3
 cot(cot –1 x) x, x R cot cot –1
4 4
from equation (i), we get
3
y cos ec
4
y 2 Ans.

Property – 2( B)

(i ) sin(sin x) x; – x (ii ) cos –1 (cos x) x; 0 x
2 2

(iii ) tan –1 (tan x) x; – x (iv) cot –1 (cot x) x; 0 x
2 2

(v ) sec –1 (sec x) x; 0 x , x (vi) cos ec –1 (cos ecx) x; x 0, – x
2 2 2

These are equal to identity function for a short interval of x only.(See the graphs on page)


3
Illustration 4 : Find the value of tan –1 tan .
4

3
Solution : Let y tan –1 tan
4

Note  tan –1 (tan x) x if x – ,
2 2
3
 – ,
4 2 2
3 3 3 3
tan –1 tan  ,
4 4 4 2 2
 graphs of y = tan –1 (tan x) is as :

3
 from the graph we can see that if x ,
2 2
then y tan –1 (tan x) can be written as y x–
3 3
y tan –1 tan – y – 123
4 4 4

Illustration 5 : Find the value of sin –1 (sin 7) .

Solution : Let y sin –1 (sin 7)

Note : sin –1 (sin 7) 7as 7 – ,
2 2
5
 2 7
2
 graph of y sin –1 (sin x) is as :


5
From the graph we can see that if 2 x then
2
y sin –1 (sin x) can be written as :
y x–2
–1
sin (sin 7) 7–2
Similarly if we have to find sin –1 (sin(–5)) then
3
 –2 –5 –
2
from the graph of sin –1 (sin x), we can say that
sin –1 (sin(–5)) 2 (–5)
2 –5
Property – 2(C )
(i) sin –1 (– x) – sin –1 x; –1 x 1 (ii) tan –1 (– x) – tan –1 x, x R
(iii) cos –1 (– x) – cos –1 x, –1 x 1 (iv) cot –1 (– x) – cot –1 x, x R

The function sin –1 x, tan –1 x and cos ec–1 x are odd functions and rest are neither even nor odd.

Illustration 6 : Find the value of cos –1 sin(–5) .

Solution : Let y cos –1 sin(–5)
cos –1 (– sin 5)  cos –1 (– x) – cos –1 x, x 1

– cos –1 cos –5 .........(i)
2

 –2 –5 –
2
 graph of cos –1 (cos x) is as :


 from the graph we can see that if –2 x –
–1
then y=cos (cos x) can be wriiten as y x 2
5 5
from the graph cos –1 cos –5 –5 2 –5
2 2 2
5 3
y – –5 y 5– Ans.
2 2

Property – 2( D)
1 1
(i ) cos ec –1 x sin –1 ; x –1, x 1 (ii) sec –1 x cos –1 ;x –1, x 1
x x
1
tan –1 ; x 0
x
(iii ) cot –1 x
1
tan –1 ; x 0
x

–2
Illustration 7 : Find the value of tan cot –1 .
3
–2
Solution : Let y = tan cot –1 .........(i )
3
 cot –1 (– x) – cot –1 x, x R
equation (i) can be written as
–2
y tan – cot –1
3

2 1
y – tan cot –1  cot –1 x tan –1 ifx 0
3 x
3 3
y – tan tan –1 y –
2 2

Property – 2( E )

(i ) sin –1 x cos –1 x , –1 x 1 (ii ) tan –1 x cot –1 x , x R
2 2

(iii ) cos ec –1 x sec –1 , x 1
2


1
Illustration 8 : Find the value of sin(2cos –1 x sin –1 x) when x .
5

Solution :

Let y sin[2 cos –1 x sin –1 x]

 sin –1 x cos –1 x ,x 1
2

y sin 2 cos –1 x – cos –1 x
2

sin cos –1 x
2
1
cos(cos –1 x)  x
5
1
y cos cos –1 .........(i )
5
 cos(cos –1 x) x if x [–1, 1]
1 1 1
 [–1, 1] cos cos –1
5 5 5
1
from equation (i), we get y
Property – 2( F ) 5
1
(i) sin(cos –1 x) cos(sin –1 x) 1 – x 2 , –1 x 1 (ii ) tan(cot –1 x) cot(tan –1 x) , x R, x 0
x
x
(iii ) cos ec(sec –1 x) sec(cos ec –1 x) , x 1
x –12


3
Illustration 9 : Find the value of sin tan –1 .
4

3
Solutions : Let y sin tan –1 .......(i )
4
Note : To find y we use sin(sin –1 x) x, –1 x 1
For this we convert tan –1 x in sin –1 x
3 3
Let tan –1 tan and 0,
4 4 2
3
sin
5
3
sin –1 (sin ) sin –1 ........(ii )
5

 0, sin –1 (sin )
2
equation (ii) can be written as :
3 3 3 3
sin –1  tan –1 tan –1 sin –1
5 4 4 5
3
from equation (i), we get y sin sin –1
5
3
y
5
1 5
Illustration 10 : Find the value of tan cos –1 .
2 3

Solution : 1 5
Let y tan cos –1 ...........(i )
2 3
5 5
Let cos –1 0 0, and cos
3 2 3
equation (i) becomes

y tan .........(ii )
2
5
1–
1 – cos 3 3– 5 (3 – 5) 2
 tan 2
2 1 cos 5 3 5 4
1
3
3– 5
tan .........(iii)
2 2

 0, 0,
2 2 4

tan 0
2


3– 5
from equation (iii), we get tan
2 2
3– 5
from equation (ii), we get y
2

1
Illustration 11 : Find the value of cos(2cos–1 x sin –1 x) when x .
5

Solution : Lety cos[2 cos –1 x sin –1 x]

 sin –1 x cos –1 x ,x 1
2

y cos 2 cos –1 x – cos –1 x cos cos –1 x
2 2
1
– sin(cos –1 x)  x
5
1
y – sin cos –1 ........(i )
5
 sin(cos –1 x) 1 – x2 , x 1
1 1 24
sin cos –1 1–
5 25 5
24
from equation (i), we get y –
5
1 1
Aliter : Let cos –1 cos and 0,
5 5 2
24
sin
5
24
sin –1 (sin ) sin –1 .........(ii )
5

 0, sin –1 (sin )
2
equation (ii) can be written as
24 1
sin –1  cos –1
5 5
1 24
cos –1 sin –1
5 5
Now equation (i) can be written as

24
y – sin sin –1 ........(iii )
5

24 24 24
 [–1, 1] sin sin –1
5 5 5
from equation (iii), we get
24
y –
5

1.06 Identities of Addition and Subtraction
A.

(i ) sin –1 x sin –1 y sin –1 x 1 – y 2 y 1 – x2 , x 0, y 0 & ( x2 y2 ) 1

– sin –1 x 1 – y 2 y 1 – x2 , x 0, y 0 & x2 y2 1

Note that : x 2 y2 1 0 sin –1 x sin –1 y
2

x2 y2 1 sin –1 x sin –1 y
2
(ii ) cos –1 x cos –1 y cos –1 xy – 1 – x 2 1 – y 2 , x 0, y 0
x y
(iii ) tan –1 x tan –1 y tan –1 ,x 0, y 0 & xy 1
1 – xy
x y
tan –1 ,x 0, y 0 & xy 1
1 – xy

,x 0, y 0 & xy 1
2

Note that : xy 1 0 tan –1 x tan –1 y ; xy 1 tan –1 x tan –1 y
2 2
B.

(i ) sin –1 x – sin –1 y sin –1 x 1 – y 2 – y 1 – x 2 , x 0, y 0

(ii ) cos –1 x – cos –1 y cos –1 xy – 1 – x 2 1 – y 2 , x 0, y 0, x y
x– y
(iii ) tan –1 x – tan –1 y tan –1 ,x 0, y 0
1 xy
Note : For x 0 and y 0 those identities can be used with the help of preperties 2(C )
i.e. change x and y to – x and – y which are positive.

3 15 84
Illustration 12 : Show that sin –1 sin –1 – sin –1 .
5 17 85
2 2
3 15 3 15 8226
Solution :  0, 0 and 1
5 17 5 17 7225
3 15 3 225 15 9
sin –1 sin –1 – sin –1 1– 1–
5 17 5 289 17 25
3 8 15 4
– sin –1 . .
5 17 17 5
84
– sin –1
85


12 4 63
Illustration 13 : Evaluate cos –1 sin –1 – tan –1 .
13 5 16

12 4 63
Solution : Let z cos –1 sin –1 – tan –1
13 5 16
4 4
 sin –1 – cos –1
5 2 5
12 4 63
z cos –1 – cos –1 – tan –1
13 2 5 16
4 12 63
z – cos –1 – cos –1 – tan –1 .......(i )
2 5 13 16
4 12 4 12
 – 0, 0 and
5 13 5 13
4 12 4 12 16 144 63
cos –1 – cos –1 cos –1 1– 1– cos –1
5 13 5 13 25 169 65
equation (i) can be written as
63 63
z – cos –1 – tan –1
2 65 16
63 63
z sin –1 – tan –1 .........(ii)
65 16
63 63
 sin –1 tan –1
65 16
from equation (ii), we get
63 63
z tan –1 – tan –1 z 0 Ans.
65 65

5
Illustration 14 : Evaluate tan –1 9 tan –1 .
4
5 5
Solution :  9 0, 0 and 9 1
4 4
5
9
–15 –1 –1 4
tan 9 tan tan
4 5
1 – 9.
4

tan –1 (–1) –
4
5 3
tan –1 9 tan –1 .
4 4


C.
1
2sin –1 x if x
2
1
(i ) sin –1 2 x 1 – x 2 – 2sin –1 x if x
2
1
–( 2sin –1 x) if x –
2
2cos –1 x if 0 x 1
(ii) cos –1 (2x 2 –1) =
2 – 2cos –1 x if –1 x 0
2 tan –1 x if x 1
2x
(iii) tan –1 = 2 tan –1 x if x 1
1 – x2
–( 2 tan –1 x) if x –1

1 – x2 2 tan –1 x if x 0
(iv) cos –1 `
1 x2 –2 tan –1 x if x 0

Illustration 15 : Define y cos –1 (4 x3 – 3x) in terms of cos –1 x and also draw its graph.

Solution : Let y cos –1 (4 x 3 – 3 x)
Note  Domain : [–1,1] and range : [0, ]
Let cos –1 x [0, ] and x cos
y cos –1 (4cos 3 – 3cos )
y cos –1 (cos 3 ) .........(i)


 [0, ] 3 [0,3 ]
to define y cos –1 (cos 3 ), we consider the graph of cos –1 (cos x) in the interval [0,3 ].
Now from the above graphs we can see that
(i ) if 0 3 cos –1 (cos 3 ) 3
y 3 if 3

y 3 if 0
3
1
y 3cos –1 x if x 1
2
(ii ) if 3 2 cos –1 (cos 3 ) 2 –3
y 2 –3 if 3 2
2
y 2 –3 if
3 3
1 1
y 2 – 3cos –1 x if – x
2 2
(iii ) 2 3 3 cos –1 (cos 3 ) –2 3
y –2 3 if 2 3 3
2
y –2 3 if
3
1
y –2 3cos –1 x if –1 x –
2
from (i), (ii) & (iii), we get
1
3cos –1 x ; x 1
2
1 1
y cos –1 (4 x 3 – 3 x) 2 – 3cos –1 x ; – x
2 2
1
–2 3cos –1 x ; –1 x –
2


Graph :
For y cos –1 (4 x3 – 3 x)
domain :[–1,1]
range :[0, ]
1
(i ) if x 1, y 3cos –1 x
2
dy –3
–3(1 – x 2 ) –1/ 2 ..........(i )
dx 1 – x2
dy 1
0 if x ,1
dx 2
1
decreasing if x ,1
2
again if we differentiate equation (i) w.r.t. ' x ', we get
d2y 3x
–
dx 2 (1 – x 2 )3/ 2

d2y 1 1
0 if x ,1 concavity downwards if x ,1
dx 2 2 2
1 1
(ii ) if – x , y 2 – 3cos –1 x.
2 2
dy 3 dy 1 1
0 if x – ,
dx 1– x 2 dx 2 2
1 1 d2y 3x
increasing if x – , and
2 2 dx 2 (1 – x 2 )3/ 2
1 d2y
(a) if x – , 0 then 0
2 dx 2
1
concavity downwards if x – ,0
2
1 d2y
(b) if x 0, then 0
2 dx 2
1
concavity downwards if x 0,
2
1 dy d2y
(iii ) Similarly if –1 x – then 0 and 0.
2 dx dx 2
the graph of y cos –1 (4 x 3 – 3 x) is as


D.
x y z – xyz
If tan –1 x tan –1 y tan –1 z tan –1 if , x 0, y 0, z 0 & ( xy yz zx) 1
1 – xy – yz – zx
NOTE :
(i ) If tan –1 x tan –1 y tan –1 z then x y z xyz

(ii) If tan –1 x tan –1 y tan –1 z then xy yz zx 1
2
(iii) If tan –11 tan –1 2 tan –1 3
1 1
(iv) tan –1 1 tan –1 tan –1
2 3 2


Inverse Trigonometric Functions

Some Useful Graphs


1.07 General Definitions
111
1 1. sin –1 x, cos –1 x, tan –1 x etc. denote angles or real numbers whose sine is x , whose cosine is x and

whose tangent is x, provided that the answers given are numerically smallest available. These are
also written as arc sin x, arc cos x etc.

If there are two angles one positive & the other negative having same numerical value, then
positive angle should be taken.

EXERCISE-3

Part : (A)

1. If cos –1 cos –1 cos –1 v 3 then v v is equal to .

(a) –3 (b) 0 (c) 3 (d) –1

2. Range of f ( x) sin –1 x tan –1 x sec –1 x is.

3 3 3
(a) , (b) , (c) , (d) none of these
4 4 4 4 4 4

3
3. The solution of the equation sin –1 tan – sin –1 – 0 is.
4 x 6

(a) x = 2 (b) x = –4 (c) x = 4 (d) none of these

4. The value of sin –1[cos{cos –1 (cos x) sin –1 (sin x)}], where x , is
2

(a) (b) (c) – (d) –
2 4 4 2

5. The set of values of k for which x 2 – kx + sin –1 (sin 4) > 0 for all real x is

(a) {0} (b) (2, 2) (c) R (d) none of these

6. sin –1 (cos(sin –1 x)) cos –1 (sin(cos –1 x)) is equal to

3
(a) 0 (b) 4 (c) (d)
2 4

1 2 x2 x
7. cos –1 x 1– x 2 . 1– cos – – cos –1 x holds for
2 4 2

(a) x 1 (b) x R
(c) 0 x 1 (d) –1 x 0
8. tan –1 a tan –1 b, where a 0, b 0, ab 1, is equal to
a b a b
(a) tan –1 (b) tan –1 –
1 – ab 1 – ab

a b a b
(c) tan –1 (d) – tan –1
1 – ab 1 – ab
–1
9. The set of values of „x‟ for which the formula 2sin x sin –1 2 x 1– x2 is true, is.

(a) (–1, 0) (b) [0, 1]

3 3 1 1
(c) – , (d) – ,
2 2 2 2
2
10. The set of values of „a‟ for which x ax sin –1 ( x 2 – 4 x 5) cos –1 ( x 2 – 4 x 5) 0 has at
least one solution is
– , – 2 2 , – , – 2 2 ,
(a) (b)

(c) R (d) none of these
–1 3
11. All possible values of p and q for which cos p cos –1 1– p cos –1 1– q holds, is
4
1 1 1
(a) p 1, q (b) q 1, p (c) 0 p 1, q (d) none of these
2 2 2
[cot –1 x] [cos –1 x] 0
12. If , where [.] denotes the greatest integer function, then complete set of
values of „x‟ is

(a) cos1, 1 (b) (cot 1, cos 1) (c) cot1, 1 (d) none of these

–1 2 –1
13. The complete solution set of the inequality [cot x] – 6[cot x] 9 0 , where [.] denotes

greatest integer function, is

(a) – , cot 3 (b) [cot 3, cot 2] (c) cot 3, (d) none of
these

1 1
14. tan cos –1 x tan – cos –1 x , x 0 is equal to
4 2 4 2


2 x
(a) x (b) 2x (c) x (d) 2

1 –1 3sin 2
15. If sin , then tan is equal to .
2 5 4 cos 2 4

(a) 1/3 (b) 3 (c) 1 (d) –1

u
16. If u cot –1 tan – tan –1 tan , then tan – is equal to .
4 2

(a) tan (b) cot (c) tan (d) cot

–1 1 – sin x 1 sin x
17. The value of cot 1 – sin x – 1 sin x
,
2
x , is :

x x x x
– 2 –
(a) 2 (b) 2 2 (c) 2 (d) 2

–1 –1 –1
18. The number of solutions of the equation, sin x cos (1– x) sin (– x), is/are .

(a) 0 (b) 1 (c) 2 (d) more than 2

–1 1 1 2
19. The number of solutions of the equation tan 2 x 1 tan –1 tan –1 is .
4x 1 x2

(a) 0 (b) 1 (c) 2 (d) 3

1 1 1 1
If tan –1 tan –1 tan –1 ........ tan –1 tan –1 , then is equal to.
20. 1 2 1 2.3 1 3.4 1 n(n 1)

n n n 1 1
(a) n 2 (b) n 1 (c) n (d) n

n
21. If cot –1 ,n N , then the maximum value of 'n' is :
6

(a) 1 (b) 5 (c) 9 (d) none of these
( C )9
–1
22. The number of real solutions of (x, y) where, (C )3 sin x, y cos (cos x), – 2
y x 2 , is :

(a) 2 (b) 1 (c) 3 (d) 4

–1 1 1
23. The value of cos 2 cos 8 is equal to


(a) 3/4 (b) –3/4 (c) 1/16 (d) ¼

Part : (B)

1 1
24. , and are three angles given by 2 tan –1 ( 2 –1), 3sin –1 sin –1 – and
2 2

1
cos –1 . Then
3

(a) (b) (c) (d)

cos–1 x tan –1 x then
25.

5 –1 5 1
x2 x2
(a) 2 (b) 2

5 –1 5 –1
sin(cos –1 x) tan(cos –1 x)
(c) 2 (d) 2

2x tan(2 tan –1 a) 2 tan(tan –1 a tan –1 a3 )
26. For the equation , which of the following is invalid?

a2 x 2a x a2 2ax 1 0 a 0 a –1, 1
(a) (b) (c) (d)

4n
The sum of tan –1 is equal to
27. n 1 n4 – 2n2 2

tan –1 2 tan –1 3 4tan –1 1 /2 sec –1 – 2
(a) (b) (c) (d)

tan(cos –1 (4 / 5)) is a/b then.
28. If the numerical value of

(a) a + b = 23 (b) a – b = 11 (c) 3b = a + 1 (d) 2a = 3b

x 2 – x – 2 0,
29. If satisfies the in equation then a value exists for

sin –1 cos–1 sec–1 cosec–1
(a) (b) (c) (d)

x 1
If f ( x) cos –1 x cos –1 3 – 3x2 then:
30. 2 2


2 2 2
f f 2cos –1 –
(a) 3 3 (b) 3 3 3

1 1 1
f f 2cos –1 – m
(c) 3 3 (d) 3 3 3

1.01 SINE RULE
111
In1a triangle ABC, the sides are proporticnal to the sines of the angles opposite to them i.e.
a b c
sin A sin B sin C

Illustration 1 : In any ABC, prove that

EXERCISE-3

Q1: In a ABC, prove that a cot A b cot B c cot C 2( R r ) .

s s s r
Q2: In a ABC, prove that 4 –1 –1 –1 .
a b c R

Q3: If , , are the distances of the vertices of a triangle from the corresponding points of contact with
y
the incircle, then prove that r2 .
y

Q4: In a ABC prove that, r1r2 r2 r3 r3 r1 s2 .

Q5: In a ABC prove that rr1 rr2 rr3 ab bc ca – s 2 .


Inverse circular function

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Inverse circular function