Question

How will you solve the expression ${\sec ^2}x - 1$ ?

Answer 1

Hint:
Use the trigonometric identity, ${\sin ^2}x + {\cos ^2}x = 1$. Then, divide the ${\cos ^2}x$ in the whole trigonometric identity. The inverse of $\cos x$ is known as $\sec x$ and division of $\sin x$ and $\cos x$ is known as $\tan x$.

Complete step by step solution:
We have to find out the value of the expression ${\sec ^2}x - 1$.
So, using the trigonometric identity, we know that –
${\sin ^2}x + {\cos ^2}x = 1 \cdots \left( 1 \right)$
Now dividing the equation (1) on the both sides by ${\cos ^2}x$, we get –
$\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + \dfrac{{{{\cos }^2}x}}{{{{\cos }^2}x}} = \dfrac{1}{{{{\cos }^2}x}}$
Now, we know that when we divide the $\sin x$ and $\cos x$ then, it is known as $\tan x$. Also, we know that the inverse of $\cos x$ is known as $\sec x$. Therefore –
$\dfrac{{\sin x}}{{\cos x}} = \tan x \cdots \left( 2 \right)$ , and
$\dfrac{1}{{\cos x}} = \sec x \cdots \left( 3 \right)$
Using the equation (2) and equation (3) in the equation (1), we get –
${\tan ^2}x + 1 = {\sec ^2}x \cdots \left( 4 \right)$
Subtracting 1 from the equation (4) on the both sides, we get –
$ \Rightarrow {\sec ^2}x - 1 = {\tan ^2}x$

Therefore, from the above we got the value of ${\sec ^2}x - 1$ equal to ${\tan ^2}x$ which is the required value for the question.

Note:
We can also simplify the expression ${\sec ^2}x - 1$ by using the right - angled triangle.
Let $y$ be the side opposite to $\theta $, $x$ be the side adjacent to $\theta $ and mark hypotenuse as $r$. Therefore, by using the Pythagoras theorem, we get –
${r^2} = {x^2} + {y^2}$

We can read the trigonometric definitions right from the triangle as corresponding ratios of sides.
Specifically, for angle $\theta $, $\tan \theta = \dfrac{y}{x}$ and $\sec \theta = \dfrac{r}{x}$. Therefore, we can write as –
$
  1 + {\tan ^2}\theta = 1 + {\left( {\dfrac{y}{x}} \right)^2} \\
   \Rightarrow 1 + {\tan ^2}\theta = 1 + \dfrac{{{y^2}}}{{{x^2}}} \\
   \Rightarrow 1 + {\tan ^2}\theta = \dfrac{{{x^2} + {y^2}}}{{{x^2}}} \\
 $
By using the Pythagoras theorem, we get –
$
   \Rightarrow 1 + {\tan ^2}\theta = \dfrac{{{r^2}}}{{{x^2}}} \\
   \Rightarrow 1 + {\tan ^2}\theta = {\sec ^2}\theta \\
 $