If A, B and C are interior angles of a triangle ABC, then show that tan (A + B)/2 = cot C/2
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Hi ,
It is given that ,
A , B and C are interior angles of
a triangle ABC ,
A + B + C = 180°
A + B = 180° - C
( A + B )/2 = ( 180° - C )/2
( A + B )/2 = 90° - C/2
Now ,
tan ( A + B )/2 = tan ( 90 - C/2 )
tan ( A + B )/2 = Cot C/2
Hence proved .
I hope this helps you.
: )
It is given that ,
A , B and C are interior angles of
a triangle ABC ,
A + B + C = 180°
A + B = 180° - C
( A + B )/2 = ( 180° - C )/2
( A + B )/2 = 90° - C/2
Now ,
tan ( A + B )/2 = tan ( 90 - C/2 )
tan ( A + B )/2 = Cot C/2
Hence proved .
I hope this helps you.
: )
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