Question

pleaseee help me in this question

Answer 1

In Triangle ABC

A+B+C=180

A/2+B/2+C/2=90 OR (180/2)

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

by taking lcm in rhs


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

TAN [(B+C)/2]=TAN [90-A/2]
TAN [(B+C)/2] = COT A /2





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Answer 2

hey sisso,,,, ✌✌✌

follow me here...

given:: interior angle= A, B, C,,

such that A+B+C= 180

B+C= 180°-A

divide equation by 2..

B+ C/2 = 180°-A/2
APPLY TAN FUNCTION on both sids...

we get..

tan B+C/2 = tan(180°-A)/2

tanB+C/2= tan 180°-tanA/2

tanB+C/2 = 0 - tanA/2. since.. tan 180= sin180/cos180 = 0/1= 0..

tanB+C/2 = cot A/2,, here reciprocal it to remove negative sign..

hope it helps u sisso,, and sorry for delay