if ∆ABC is an isosceles right triangle right angled at B, then tan A + Cot C÷cot A + Cot c
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Answer:
bc/ab(2ab+bc/ab)
Step-by-step explanation:
=tan a +cotc/cota+cotc
=bc/ab+(bc/ab*bc/ab)+bc/ab
=bc/ab[1+bc/ab+1]
=bc/ab(2ab+bc/ab)
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