If A+B=π/4 Prove that (1+tanA)(1+tanB)=2 and (cotA-1)(cotB-1)=2?

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Answer 1 · 2017-12-10T15:53:21

#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 · 2017-12-10T16:00:19

Please see below.

We use the identity #tan(A+B)=(tanA+tanB)/(1-tanAtanB)#

as #A+B=pi/4#, we have

#(tanA+tanB)/(1-tanAtanB)=tan(A+B)=tan(pi/4)=1#

or #tanA+tanB=1-tanAtanB#

or #tanA+tanB+tanAtanB=1#

or #1+tanA+tanB+tanAtanB=2#

or #1(1+tanA)+tanB(1+tanA)=2#

or #(1+tanA)(1+tanB)=2#

For other use identity #cot(A+B)=(cotAcotB-1)/(cotA+cotB)#

and as #A+B=pi/4#,

#(cotAcotB-1)/(cotA+cotB)=cot(A+B)=cot(pi/4)=1#

or #cotAcotB-1=cotA+cotB#

or #cotAcotB-cotA-cotB+1=2#

or #cotA(cotB-1)-1(cotB-1)=2#

or #(cotA-1)(cotB-1)=2#