QUESTION

# If $\sec A + \tan A = m$ and $\sec A - \tan A = n$ , find the value of $\sqrt {mn}$ .$A.{\text{ }}0 \\ B.{\text{ }} \pm 1 \\ C.{\text{ }} \pm 2 \\ D.{\text{ }} \pm 3 \\$

Hint: In order to find the value of the term given first, solve the functions given in the problem to bring it in the simpler form, something in terms of squares by the use of suitable trigonometric identity.

Given that:
$\sec A + \tan A = m$ ----- (1)
$\sec A - \tan A = n$ ----- (2)
If we visualize in terms of algebra we can see something like $\left( {a + b} \right)\& \left( {a - b} \right)$ in the above terms. In order to bring the terms in squares let us multiply equation (1) and (2).
$\Rightarrow \left( {\sec A + \tan A} \right) \times \left( {\sec A - \tan A} \right) = m \times n$
As we know the algebraic formula $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$
Comparing the above equation with the same formula and proceeding further we get:
$\Rightarrow \left( {\sec A + \tan A} \right) \times \left( {\sec A - \tan A} \right) = m \times n \\ \Rightarrow \left[ {{{\left( {\sec A} \right)}^2} - {{\left( {\tan A} \right)}^2}} \right] = mn \\ \Rightarrow {\sec ^2}A - {\tan ^2}A = mn \\$
As we know the trigonometric relation
$\because 1 + {\tan ^2}\theta = {\sec ^2}\theta \\ \because {\sec ^2}\theta - {\tan ^2}\theta = 1 \\$
Using the relation in the above equation we get:
$\Rightarrow {\sec ^2}A - {\tan ^2}A = mn \\ \Rightarrow mn = 1{\text{ }}\left[ {\because {{\sec }^2}\theta - {{\tan }^2}\theta = 1} \right] \\$
Now taking the square root of the both side of the equation we get:
$\because mn = 1 \\ \Rightarrow \sqrt {mn} = \sqrt 1 = \pm 1 \\ \therefore \sqrt {mn} = \pm 1 \\$
Hence, for the above relation the value of $\sqrt {mn} = \pm 1$ .
So, option B is the correct option.

Note: The problem can also be solved by direct substitution of the value given in the problem statement directly into the required function. Students must remember the trigonometric identities in order to solve such types of problems. Also the students must relate the trigonometric terms in terms of some algebraic terms for simplification.