If (secA+tanA)=m and (secA-tanA) =n,show that mn=1.
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Lhs- (secA +tanA) (secA-tanA)
Since {(a+b) (a-b)} =a^2-b^2
So sec^2A-tan^2A
And from identity
It is equal to one
Since {(a+b) (a-b)} =a^2-b^2
So sec^2A-tan^2A
And from identity
It is equal to one
Answered by
1
:• MN =(SecA + TanA) (SecA-TanA)
=Sec^2 A + Tan^2A
=1(ans)
=Sec^2 A + Tan^2A
=1(ans)
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