Question

yadi A+B=π/4prove that (cotA-1)(cotB-1)=2​

Answer 1

Answer:

Step-by-step explanation:

tan(A+B)=tan(pi/4)

=>(tanA+tanB)/(1-tanAtanB)=1

=>tanA+tanB=1-tanAtanB)

=>tanA+tanB+tanAtanB+1=2

=>tanA+tanAtanB+1+tanB=2

=>tanA(1+tanB)+(1+tanB)=2

=>(1+tanA)(1+tanB)=2

Again

cot(A+B)=cot(pi/4)

=>(cotAcotB-1)/(cotB+cotA)=1

=>cotAcotB-1=cotB+cotA

=>cotAcotB-cotB-cotA+1=1+1

=>cotB(cot A-1)-1(cotA-1)=2

=>(cot A-1)(cotB-1)=2

Answer 2

Answer:

Step-by-step explanation:

A+B = π/4

//Now take Cot on both sides

Cot(A+B) = Cot (π/4) = Cot45°

//Formula: Cot(A+B) = (CotA.CotB - 1) / (CotA + CotB) & Cot45° = 1

(CotA.CotB - 1) / (CotA + CotB)  = 1

CotA.CotB - 1 = Cot A + Cot B

CotA.CotB - CotA - CotB = 1

CotB(CotA - 1) - CotA  =  1

//Add 1 on both sides

CotB(CotA-1) - CotA +  1  = 1+1

CotB(CotA-1) - (CotA-1) = 2

(CotA-1) (CotB-1) = 2