Question

if a,b,c are the interior angles of the triangle ,prove that tan[b+c/2]=cotA/2

Answer 1

Given---> If A , B , C are the interior angles of the triangle.

To prove ---> tan { ( B + C ) / 2 } = Cot A / 2

Proof ---> A, B and C are interior angle of triangle,

So , by anglesum property ,

A + B + C = 180°

=> B + C = 180° - A

Now ,

LHS= tan { ( B + C ) / 2 }

Putting , B + C = 180° - A

= tan { ( 180° - A ) / 2 }

= tan { ( 180° / 2 ) - ( A / 2 ) }

= tan { ( 90° ) - A / 2 }

We know that,

tan( 90° - θ ) = Cotθ , using it here we get

= Cot A / 2 = RHS

Additional information---->

(1) Sin ( 90° -θ ) = Cosθ

(2) Cos ( 90° - θ ) = Sinθ

(3) Cot ( 90° - θ ) = tanθ

(4) Sec ( 90° - θ ) = Cosecθ

(5) Cosec ( 90° - θ ) = Secθ