Question

In any triangle ABC ,angle C=45degree, then prove that (1+cotA) (1+cotB) = 2

Answer 1

Hi, 
C = 45 degree so A+B = 180-45= 135

=> cot(A+ B) = cot(180 -45) 
=> (cotA.cotB - 1)/(cotA + cotB) = -cot 45 = -1 
=> cotA.cotB - 1 = -cot A - cotB 
=>cotA.cotB = 1-cotA -cotB 
cotAcotB+cotA+cotB= 11+cotAcotB+cotA+cotB= 1+1  [Adding 1 on both side](1+cotB)+(cotAcotB+cotA)=2(1+cotB)+cotA(cotB+1)=2(1+cotB)(1+cotA)= 2

Answer 2

Given:

Angle C = 45°

To Prove:

(1+cotA) (1+cotB) = 2

Solution:

Since C = 45°

Thus,

A+B = 180 - 45 = 135

= cot(A+ B) = cot(180 -45)  

= (cotA.cotB - 1)/(cotA + cotB) = -cot 45 = -1  

= cotA.cotB - 1 = -cot A - cotB  

= cotA.cotB = 1-cotA -cotB

= cotA.cotB + cotA +  cotB = 1

1 + cotA.cotB + cotA +  cotB = 1 + 1 ( Adding one on both the sides)

( 1 + cotB) +  cotA.cotB + cotA = 2

( 1 + cotB) + cotA ( cotB + 1) = 2

(1+cotA) (1+cotB) = 2