Question

secA + tanA = m and secA – tanA = n, find the value of mn .

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Answer 1

Given,

m = sec A + tan A

n = sec A - tan A

hence ,

mn = (sec A + tan A) (sec A - tan A)

We know,

$(a+b)(a-b)=(a^{2}-b^{2})$

Thus,

mn = $sec^{2}A-tan^{2}A$

mn = 1

By Trigonometry Equations,

$sec^{2}A-tan^{2}=1$

Answer 2

Answer:

Given,

m = secA + tanA

n = secA - tanA

Hence,

mn = (secA + tanA) (secA - tanA)

We Know,

$(a + b)(a - b) = (a {}^{2} - b {}^{2} )$

Thus,

mn = sec^2A - tan^2A

mn = 1

By Trigonometry Equation

Sec^2A - tan^2 = 1

Step-by-step explanation:

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