(sec2A- tan2A)= 1-sinA/ 1+SinA
Answers
Answered by
1
Please see the attachment
Attachments:
Answered by
1
RHS = 1-sinA/ 1+SinA
=[(1-sinA)/(1+sinA)][(1-sinA)/(1-sinA)]
= (1-sinA)²/[1²-sin²A]
= (1+sin²A-2sinA)/(cos²A)
= 1/cos²A+sin²A/cos²A-2sinA/cos²A
= sec²A+tan²A-2tanAsecA
= (secA-tanA)²
= LHS
••••
Similar questions