Question

高中数学难题求救

sin²A/sin²B+cos²A X cos²C=1，求证：tan²A X cot²B=sin²C
请写出要点，谢谢

Answer 1

sin²A/sin²B+cos²A X cos²C=1
sin²A/sin²B+cos²A X cos²C=(cosA)^2+(sinA)^2
两边同处以(cosA)^2
(tanA)^2/(sinB)^2 + (cosC)^2= 1+(tanA)^2
(tanA)^2/(sinB)^2 -tanA)^2=1-(cosC)^2
(tanA)^2*(1-(sinB)^2)/(sinB)^2=(sinC)^2
(tanA)^2*(cosB)^2/(sinB)^2=(sinC)^2
(tanA)^2 * （cotB)^2=(sinC)^2

Answer 2

sin²A/sin²B+cos²A X cos²C=1
sin²A/sin²B+cos²A-cos²A X sin²C=1
sin²A/sin²B+cos²A-1=cos²A X sin²C
sin²A/sin²B-sin²A=cos²A X sin²C
tan²A/sin²B-tan²A=sin²C
tan²A X cot²B=sin²C