Question

If Sec[A]-Tan[A] = Sqrt[3]-Sqrt[2], find the value of Sec[A]+Tan[A]

Answer 1

Rewrite

`\sec(A) - \tan(A) = ((\sec(A) - \tan(A))(\sec(A) + \tan(A)))/(\sec(A) + \tan(A)) = (\sec^2(A) - \tan^2(A))/(\sec(A) + \tan(A))`

Recall that

`\cos^2(x) + \sin^2(x) = 1 \implies 1 + \tan^2(x) = \sec^2(x)`

which means

`\sec^2(A) - \tan^2(A)`

and

`\sec(A) - \tan(A) = \frac1{\sec(A) + \tan(A)}`

Then the value we want is simply the reciprocal of the given value,

`\sec(A) + \tan(A) = \frac1{\sqrt3 - \sqrt2}`

and we can consider rationalizing the denominator to write

`\frac1{\sqrt3 - \sqrt2} = (\sqrt3 + \sqrt2)/(\left(\sqrt3 - \sqrt2\right)\left(\sqrt3 + \sqrt2\right)) = (\sqrt3 + \sqrt2)/(\left(\sqrt3\right)^2 - \left(\sqrt2\right)^2) = \boxed{\sqrt3 + \sqrt2}`