Question

[cot^2A(secA-1)]/1+sinA=[sec^2A(1-sinA)/1+secA

Answer 1

trig?

Answer 2

yaaa

Answer 3

did you try multiplying and dividing by
\((\sec A +1)(1-\sin A)\)
on left side

Answer 4

cot²A(secA-1)/1+sinA
cot²A(secA-1)(1-sinA)/((1+sinA)(1-sinA)
cot²A(secA-1)(1-sinA)/(cos²A)
(secA-1)(1-sinA)/(sin²A)
(secA-1)(secA+1)(1-sinA)/(sin²A)(secA+1)‡
(sec²A-1)(1-sinA)/(sin²A)(secA+1)
(tan²A)(1-sinA)/(sin²A)(secA+1)
((sin²A)/(cos²A)(1-sinA))/(sin²A)(secA...‡
(1-sinA)/((secA+1)(cos²A))
sec²A(1-sinA)/(1+secA)=RHS