cosec^2A+tan^2A/cosec^2A-tan^2A=1+tan^2A/1-tan^2A
Answers
Answered by
0
upload image.............
Answered by
0
Answer:
LHS,
=cosec^2A + tan^2A/cosec^2A - tan^2A
=[1/sin^2A + sin^2A/cos^2A] ÷ [1/sin^2A - sin^2A/cos^2A]
=(cos^2A +sin^2A)/cos^2Asin^2A ÷ (cos^2A -sin^2A)/cos^2Asin^2A
=1/cos^2A -sin^2A
{ cos^2A +sin^2A = 1}
RHS,
=1+tan^2A/1-tan^2A
=[1+sin^2A/cos^2A]÷[1-sin^2A/cos^2A]
=(cos^2A +sin^2A)/cos^2Asin^2A ÷ (cos^2A -sin^2A)/cos^2Asin^2A)
=1//cos^2A -sin^2A
{ cos^2A +sin^2A = 1}
LHS = RHS
hence proved
Similar questions