Question

ABC are interior angles of a triangle provetan( A+B/2)=cot(C/2)

Answer 1

We know that the sum of all interior angles of any triangle is 180° and in question it is given that A, B and C are the interior angles of a triangle.




According to the angle property of triangles, sum of A, B and C will be equal to 180°




$\bold{ = > A + B + C = 18 0 \degree} \\ \\ \bold{ = > A + B = 180 \degree - C }$



Divide both sides by 2 ,



$\bold{= > \dfrac{A + B}{2}= \dfrac{180-C}{2} }\\ \\ \\ \bold{= > \dfrac{A +B}{2}= 90° - \dfrac{ C}{2}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - - - - - ( \: 1 \: )$




Therefore,


$\bold{tan( \frac{A+B}{2}) }$



Putting the value of ( A + B ) / 2 from ( 1 ) ,



$\bold{tan( 90 \degree - \frac{C}{2}) }$



We know tan( 90 - ∅ ) = cot∅



$\bold{cot( \frac{C}{2}) }$

$\bold{ Hence, mproved \: that \: \: tan( \frac{A+B}{2}) = cot( \frac{C}{2}) }$

Answer 2

We know, sum of all interior angles of any triangle = 180°, so the sum of A , B and C will be 180°.


A + B + C = 180°

A + B = 180° - C


Divide by 2 on both sides,


( A + B ) / 2 = ( 180° - C ) / 2

( A + B ) / 2 = 90° - C / 2 -- 1




To Proove : tan( A + B ) / 2 = cot( C / 2 )


Proove : tan( A + B ) / 2



• Putting the value of ( A + B ) / 2


=> tan( 90° - C / 2 )

=> cot( C / 2 ) [ tan( 90 - A ) = cotA. ]




Proved.
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