Math, asked by jayadurai197726, 1 year ago

prove that a+b=45° then(cotA-1)(cotB-1)=2

Answers

Answered by rishimhaske
205
Cot (A+B) = (CotACotB-1)/(CotA + CotB )
We know......Cot(45)=1
so,          1=
(CotACotB-1)/(CotA + CotB )
          
CotACotB-1)=  (CotA + CotB )
          
CotACotB-cotA-cotB=1
adding 1 to both sides.....we get
  
(cotA-1)(cotB-1)=2......proved


Answered by chandujnv002
0

Answer:

We use trigonmetric identities to prove this result.

Step-by-step explanation:

  • Basic trigonmetric identities used:
  1. cot(A+B)=\frac{cotAcotB-1}{cotB+cotA}
  • Now,we proceed with the proof .
  1. Using (1),we have  cot(A+B)=\frac{cotAcotB-1}{cotB-cotA}.
  2. Next,we put,A+B=45°,i.e.,cot(A+B)=1 .
  3. We get,                                    1=\frac{cotAcotB-1}{cotB+cotA} \\\implies cotB+cotA=cotAcotB-1\\\implies cotAcotB-cotB-cotA+1=2\\\implies (cotA-1)(cotB-1)=2

Thus,we are done.

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