JEE Mathematics/ Lakshmikanta Satapathy/ Inverse trigonometry QA part 6/ Questions on Inverse trigonometric functions involving tan inverse function solved with the related concepts

1. Physics Helpline
L K Satapathy
Inverse Trigonometry 6

2. Inverse Trigonometry 6
Physics Helpline
L K Satapathy
Question :
Answer
   1 12 2tan tan ,
4 4 4
x xIf find the value of x
x x
   
 
   1 12 2tan tan
4 4 4
x x
x x
   
 
   
  
1
2 2
4 4
tan
42 21
4 4
x x
x x
x x
x x

 
 
 
 
 
1 1 1
tan tan tan
1
x y
x y
xy
    
   
1 ( 2)( 4) ( 2)( 4)
tan
( 4)( 4) ( 2)( 2) 4
x x x x
x x x x
     
 
    
2 2
1
2 2
2 8 2 8tan
4( 16) ( 4)
x x x x
x x
      
  

3. Inverse Trigonometry 6
Physics Helpline
L K Satapathy
2
2 16 tan 1
12 4
x   

2
2 4x 
2
1 2 16tan
12 4
x   

1
tan tanx x 
    
2
2 16 12x   
[2 ]nsx A  

4. Inverse Trigonometry 6
Physics Helpline
L K Satapathy
Question :
Answer
1 11 1 1tan cos
4 21 1
x xShow that x
x x
       
   
1 1 1tan
1 1
x xLHS
x x
      
   
1 1 cos2 1 cos2tan
1 cos2 1 cos2
 
 
      
   
 cos2Put x 
1 2 cos 2 sintan
2 cos 2 sin
 
 
    
 
2 2
1
2 2
2cos 2sintan
2cos 2sin
 
 

    
 
2
1 cos2 2cos    
2
1 cos2 2sin    

5. Inverse Trigonometry 6
Physics Helpline
L K Satapathy
 1 cos sintan
cos sin
LHS  
 
  

4
  
11cos2 cos
2
x x     
  
 1 1 tantan
1 tan


 

1
tan tan
4tan
1 tan .tan
4
 
 

 
 
  
 
 
  1
tan tan
4
 
 
[Dividing numerator & denominator by cos]
tan 1
4
 
  
11 cos
4 4 2
x RHS  
   
 1
tan tan 
   

6. Physics Helpline
L K Satapathy
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