दिया है-
`tan A=(BC)/(AB)=(1)/(sqrt3)`
पाइथागोरस प्रमेय से, `AC^(2)=AB^(2)+BC^(2)=(sqrt3)^(2)+(1)^(2)=4`
`AC=sqrt4=2`
अब `sin A=("लम्ब")/(" कर्ण")=(BC)/(AC)=1/2`
`cos A= ("आधार")/(" कर्ण")=(AB)/(AC)=(sqrt3)/(2)`
`sin C=(AB)/(AC)=(sqrt3)/(2)`
`cos C=(BC)/(AC)=1/2`
अतः
`cos A cos C-sinC=(sqrt3)/(2)xx1/2-1/2xx(sqrt3)/(2)=(sqrt3)/(4)-(sqrt3)/(4)=0`