Prove that SinA-cosA+1/sinA+ cosA-1 =1/secA-tanA
Answers
Answered by
3
Step-by-step explanation:
LHS
devide cosA in both numerator & denominator
tanA-1+secA/tanA+1-secA
=(tanA+secA)-1/(tanA-secA+1) ×(tanA-secA)/(tanA-secA)
=tan²A-sec²A-(tanA-secA)/ (tanA-secA+1)(tanA-secA)
=-1-tanA+secA/(tanA-secA+1)(tanA-secA)
=-(tanA-secA+1)/(tanA-secA+1) (tanA-secA)
=-1/tanA-secA
=-1/-(secA-tanA)
=1/secA-tanA
Attachments:
Similar questions