If secA+tanA=m and secA+tanA=n,prove that mn=1.
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If secA+tanA=m and secA+tanA=n,prove that mn=1.
Please don't answer if you don't know
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Answer:
mn = (secA +tanA)(secA-tanA)
= (sec²A - tan²A)
since, 1 +tan²A = sec²A
⇒sec²A- tan²A = 1
∴mn = 1
Hence proved mn = 1.
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Question:-
If secA+tanA=m and secA+tanA=n,prove that mn=1.
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mn = (secA +tanA)(secA-tanA)
= (sec²A - tan²A)
since, 1 +tan²A = sec²A
⇒sec²A- tan²A = 1
∴mn = 1
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
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