Question

A+B=45°prove that (cotA-1)(cotB-1)=2

Answer 1

Answer:

proved

Step-by-step explanation:

By taking L.H.S

=$cotAcotB-cotA-cotB+1$

=$cot(A+B)=\frac{cotAcotB-1}{cotA+cotB}\\ cot(A+B)*(cotA+cotB) = cotAcotB-1$

=$cot(A+B)(cotA+cotB)+1=cotAcotB$

=$cot(A+B)(cotA+cotB)+1-cotA-cotB+1$

since A+B=45

cot(A+B)=1

finally all terms are cancelled except 1+1

which gives = 2 = R.H.S