Prove that
sec^2A-cot^2 (90° - A)
= cos^2 (90° - A) + cos^2 A
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Hey !
LHS= sec²A-cot²(90-A)
= sec²A-tan²A [cot(90-A)=tanA ; 1+tan²A=sec²A]
= 1
RHS=cos²(90-A)+cos²A
=sin²A+cos²A
=1. [cos(90-A)=sinA ;sin²A+cos²A=1]
LHS=RHS
HENCE, Proved.
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