यहाँ `3 cotA=4`
`impliescot A=4/3=("आधार")/(" लम्ब")=(AC)/(BC)`
`DeltaABC` में पाइथागोरस प्रमेय से, `AB^(2)=AC^(2)+BC^(2)`
`=(4)^(2)+(3)^(2)=16+9=25`
` impliesAB=5`
`therefore tanA=(BC)/(AC)=3/4`
अब `L.H. S. =(1-tan ^(2)A)/(1+tan^(2)A)=(1-((3)/(4))^(2))/(1+((3)/(4))^(2))=(1-(9)/(16))/(1+(9)/(16))=7/25`
`R.H.S. =cos ^(2)A-sin^(2)A=((4)/(5))^(2)-((3)/(5))^(2)=7/25`
अतः `(1-tan^(2)A)/(1+tan^(2)A)=cos^(2)A-sin^(2)A.`