Math, asked by NutNiggaUnited, 1 year ago

If A, B and C are interior angles of a triangle ABC, prove that tan[(B+C)/2] = cot(A/2)

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Answers

Answered by Hasti152002

Hey... I think this can be ur answer dear!!

A+B+C=180 (angle sum property)
Dividing by 2
A+B+C/2=90
B+C/2=90 - A/2
Now,multiplying both sides with tan so,
Tan(B+C/2) = tan (90-A/2)
Tan (B+C/2)= cot A/2 (tan and cot are complementary)

Hence proved..

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Answered by mysticd

Hi,

We know that.

A + B + C = 180°

B + C = 180 - A

( B + C ) / 2 = ( 180 - A ) / 2

( B + C ) / 2 = ( 90 - A / 2 )

Tan[( B + C ) /2 ] = tan ( 90 - A /2 )

= Cot A/2

[ Since tan ( 90 - x ) = cotx ]

I hope this helps you.

: )

NutNiggaUnited: Thank you brother

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