Inverse Trigonometric Functions
Properties and Formulae
ANIL SHARMA
K.V. HIRA NAGAR
we have studied that the inverse of a function
f, denoted by f –1, exists if f is one-one and onto. There are
many functions which are not one-one, onto or both and
hence we can not talk of their inversesThe concepts of inverse
trigonometric functions is also used in science and engineering we have
studied that the inverse of a function
f, denoted by f –1, exists if f is one-one and onto. There are
many functions which are not one-one, onto or both and
hence we can not talk of their inverses.
Pretesting Questions:Q 1) Can you suggest the restricted Domains of each
T ratios separately at which the functions are
one and onto ?
Q2) When a function is said to be invertible. ?
Q3 Evaluate the followings
a) Sin-1(1/2)
b) Sin-1(-1/2)
c) Tan-1(1)
d) Sin-1(2sin∏/6)
Q4) If Sin-1( x-1) =∏/4 then find the value of x.
Inverse Trigonometric
function – Properties
sin-1 (-x) = -sin-1 (x) if x is in [-1,1]
cos-1 (-x) = - cos-1x if x is in [-1,1]
tan-1(-x) = - tan-1x if x is in ( -, )
cot-1(-x) =  - cot-1x if x is in (-,)
cosec-1(-x) = - cosec-1x if x is in (-,-1] U [1,)
sec-1(-x) =  - sec-1x if x is in (-,-1] U [1,)
Other important
properties
sin x+ cos x = /2 ;
-1
-1
if x is in [-1,1]
tan
1
x  tan
1
y  tan
1 
xy 
 1  xy 


If x > 0 , y > 0 and xy < 1
tan
1
tan
1
x  tan
1
x  tan
1
y     tan
y    tan
If x > 0 , y > 0 and xy > 1
1 
xy 
 1  xy  If x<0,y<0 and xy < 1


1 
xy 
 1  xy 


Find the value of
tan
1  1 
 2   tan
 
1  1 
3
 
Solution :
1
1
Let tan1     and tan1    
2
3
1
1
  
 tan     and tan     and ,     , 
2
3
 2 2


1 1



2 3

Now tan       
 1   1 . 1  
2 3

   

 tan       1
Other important properties
sin-1 x+ cos-1 x = /2 ;
if x is in [-1,1]
tan
1
x  tan
1
y  tan
1 
xy 
 1  xy 


If x > 0 , y > 0 and xy < 1
 xy 
tan1 x  tan1 y     tan 1 
If x<0,y<0 and xy < 1

 1  xy 
tan
1
x  tan
1
y    tan
If x > 0 , y > 0 and xy > 1
1 
xy 
 1  xy 


Prove that

tan  1  x   cot  1  x  1   tan  1 x 2  x  1

Solution :
Let tan  1 x   and cot  1  x  1   
1
 tan   x and cot    x  1   tan  
x 1
And L.H.S. of the given identity is +
1
x 1
 tan      
 1 
1  x. 

 x  1
x
tan   tan 
as tan      
1  tan . tan 
Question
−1
Tan1/2 ( sin
2𝑥
1+𝑥2
+ cos
−1
1−𝑦2
1+𝑦2
)
In triangle ABC if A = tan-12 and B =
tan-1 3 , prove that C = 450
Solution :
For triangle ABC , A+B+C = 
tan A  2; tanB  3
tan  A  B   tan    C 
tan A  tanB
  tanC
1  tan A.tanB
  1   tanC  tanC  1

32
  tanC
1  3.2
Inverse Trigonometric
function – Conversion
To convert one inverse function to other
inverse function :
1. Assume given inverse function as some angle
2.
( say  )
Draw a right angled triangle satisfying the angle. Find the third
un known side
3. Find the trigonometric function from the triangle in step 2. Take
its inverse and we will get  = desired inverse function
The value of cot-1 3 + cosec-1  5 is
(a) /3 (b) /2
( c) /4 (d) none
Step 1
Assume given inverse function as some angle (
say  )
Let cot-1 3 + cosec-1  5 = x + y,
Where x = cot-13 ; cot x = 3 and
y = cosec-1  5 ; cosec y =  5
If sin-1 x + sin-1 (1- x) = cos-1x,
the value of x could be
(a) 1, 0 (b) 1,1/2 (c) 0,1/2
Solution :

 2x  x2
(d) 1, -1/2
 1  x   2x  x   0
2
2
 x .  2  x  1  2x    0
1
 x  0 , ,2
2
x  2 as x  sin 
1
 x 0,
2
If cos-1 x + cos-1 y + cos-1z = ,
Then prove that x2+y2+z2 = 1 - 2xyz
Solution :
Let cos 1x  A , cos 1 y  B and cos 1 z  C
 cos A  x , cos B  y and cos C  z
and given : A+B+C = 
Now, L.H.S. = cos2A + cos2B +cos2C
= cos2A + 1- sin2B +cos2C
= 1+(cos2A - sin2B) +cos2C