Question

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

Answer 1

Given that a + b = 45°
Cot(a+b) = Cot 45°
⇒ Cot(a+b) = 1
Using identity Cot(a+b) = Cot(a)Cot(b) - 1 / Cot(a) + Cot(b)
⇒ ( Cot(a)Cot(b) - 1 )/ Cot(a) + Cot(b) = 1
⇒Cot(a)Cot(b) - 1 = Cot(a) + Cot(b)
⇒ Cot(a)Cot(b) - Cot(a) - Cot(b) = 1
Adding 1 on both sides, we get
Cot(a)Cot(b) - Cot(a) - Cot(b) + 1 = 2
⇒Cot(a) [Cot(b) - 1 ] -1[Cot(b) - 1] = 2
⇒(Cot(a) - 1 ) (Cot(b) - 1 ) = 2
Hence proved

Answer 2

identity. b=45° where cotb=cot 45°.. where cot45°=1
then ..
(cotA-1) (cotB-1) =2
(cotA-1)( 1-1)=2
(cotA-1)=2
cotA=2+1
cotA=3