prove the question in the photo
Attachments:
Answers
Answered by
0
Step-by-step explanation:
sinA-cosA+1/sinA+cosA-1
devide cosA in both numerator & denominator
tanA-1+secA/tanA+1-secA
={(tanA+secA)-1}(tanA+secA)/ {(tanA-secA)+1}(tanA+secA)
=(tanA+secA-1)(tanA+secA)/ tan²A-sec²A+tanA+secA
=(tanA+secA-1)(tanA+secA)/ (-1+tanA+secA)
=tanA+secA
=sinA/cosA+1/cosA
=1+sinA/cosA
Similar questions