Question

If secA+tanA=m and secA-tanA=n then find the value of mn

Answer 1

mn = (secA +tanA)(secA-tanA)
      = (sec²A - tan²A) 
since, 1 +tan²A = sec²A
⇒sec²A- tan²A = 1
∴mn = 1

Answer 2

Concept

Trigonometric ratios are the ratios of sides of right angled triangle. Sin x is the division of perpendicular to hypotenuse. Cos x is the division of base and hypotenuse. Tan x is the division of perpendicular and base. Secant x is the inverse of cos x. Cosecant x is the inverse of sin x. Cotangent is the inverse of tangent x.

Given

sec A + tan A=m

sec A -tan A=n

To find

Value of m n.

Explanation

We have to find the value of mn and for this we have to multiply both the following equations:

sec A + tan A= m

sec A - tan A= n

multiplying both equations we get

m n=(sec A+ tan A) (sec A -tan A)

m n = $sec^{2}A -tan ^{2} A$      

We know that (A+B)(A-B)=$A^{2} -B^{2}$ and $sec^{2}x -tan^{2}x=1$

m n = 1

Hence the value of m n is equal to 1.

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