Question

prove that sinA+cosA-1÷sinA-cosA+1=secA+tanA

Answer 1

LHS=
sinA+cosA(sin²A-cos²A)÷sinA-cosA+1
sinA+cosA{(sinA+cosA)(sinA-cosA)}÷sinA-cosA+1
sinA+cosA(1+sinA-cosA)÷sinA-cosA+1
sinA+cosA÷1
sinA+cosA÷sin²A+cos²A. :- 1= sin²A+cos²A
1÷sinA+cosA. :-1=sin²A+cos²A
sin²A+cos²A÷sinA+cosA
sin²A÷sin+cosA + cos²A÷sinA+cosA
sin/cos=tanA+cosA/sin=cotA
tanA+cotA÷1. :-1 tan²A+sec²A
tanA+1/tan÷tan²A-sec²A
1÷sec²A-tan²A=
-sec²A+tan²A
-{(sec²+tan²)}=
secA+tanA RHS proved

Answer 2

Hope this will Help u Dear .