Question

If ${\sec ^2}\theta + {\tan ^2}\theta + 1 = 2$, then find the value of $\sec \left( { - \theta } \right)$:
A. $- 2$
B. $- \dfrac{1}{2}$
C. $1$
D. $\pm 1$

Answer 1

Hint : Solve using trigonometric identities.

Given that: ${\sec ^2}\theta + {\tan ^2}\theta + 1 = 2$
Converting the above equation in the terms of Sin and Cos, we get

$\Rightarrow \dfrac{1}{{{{\cos }^2}\theta }} + \dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} + 1 = 2{\text{ }}\left( {\because \tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}{\text{ and }}\sec \theta = \dfrac{1}{{\cos \theta }}} \right) \\\ \Rightarrow \dfrac{{1 + {{\sin }^2}\theta + {{\cos }^2}\theta }}{{{{\cos }^2}\theta }} = 2 \\\ \Rightarrow \dfrac{{1 + 1}}{{{{\cos }^2}\theta }} = 2{\text{ }}\left( {\because {{\sin }^2}\theta + {{\cos }^2}\theta = 1} \right) \\\ \Rightarrow \dfrac{2}{{{{\cos }^2}\theta }} = 2 \\\ \Rightarrow {\cos ^2}\theta = 1 \\\ \Rightarrow \cos \theta = \pm 1{\text{ }} \ldots \, \ldots \left( 1 \right) \\\$
We know that, $\cos \left( { - \theta } \right) = \cos \left( \theta \right)$
$\therefore \sec \left( { - \theta } \right) = \sec \left( \theta \right) = \dfrac{1}{{\cos \theta }}$
Put the value of $\cos \theta$ from equation $\left( 1 \right)$, we get
$\sec \left( { - \theta } \right) = \pm 1$

Note: In these types of problems, where there is no direct formula for the given trigonometric terms, one should always try to convert them to some trigonometric terms which have some relation using trigonometric relations and identities so as to make the problem easier to calculate.