Question

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 1

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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Answer 2

Question:-

If secA+tanA=m and secA+tanA=n,prove that mn=1.

${\red{\underline{\underline{\bold{Answer:-}}}}}$

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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