Question

How do you write $(\sec \theta - 1)(\sec \theta + 1)$ in terms of sine and cosine?

Answer 1

Hint: Here we are asked to express $(\sec \theta - 1)(\sec \theta + 1)$ in terms of sine and cosine. We make use of the algebraic identity to simplify $(\sec \theta - 1)(\sec \theta + 1)$. We use the identity $({a^2} - {b^2}) = (a - b)(a + b)$. After that we use the trigonometry identity ${\tan ^2}\theta {\text{ + 1}} = {\sec ^2}\theta $. At last we will make use of $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$.

Complete step by step solution:
Here we are given $(\sec \theta - 1)(\sec \theta + 1)$ ……(1)
We are asked to express $(\sec \theta - 1)(\sec \theta + 1)$ in terms of sine and cosine. So we use some basic trigonometric identities to convert the given problem into cosine and sine function.
Note that the given problem is of the form $(a - b)(a + b)$.
We have the algebraic formula given by $(a - b)(a + b) = ({a^2} - {b^2})$ ……(2)
Here $a = \sec \theta $ and $b = 1.$
Hence substituting the values of a and b in the equation (2) we get,
$ \Rightarrow (\sec \theta - 1)(\sec \theta + 1) = ({\sec ^2}\theta - {1^2})$
$ \Rightarrow (\sec \theta - 1)(\sec \theta + 1) = ({\sec ^2}\theta - 1)$ ……(3)
We have the trigonometric identity given by,
${\tan ^2}\theta {\text{ + 1}} = {\sec ^2}\theta $
Taking 1 to the right hand side, we get,
${\tan ^2}\theta = {\sec ^2}\theta {\text{ - 1}}$
Remember that when we transfer any number or variable to the other side, the signs of the same will be changed to the opposite sign.
Now substituting ${\tan ^2}\theta = {\sec ^2}\theta {\text{ - 1}}$ in the equation (3), we get,
$ \Rightarrow (\sec \theta - 1)(\sec \theta + 1) = {\tan ^2}\theta $ ……(4)
We know that $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
Squaring this on both sides we get,
${\tan ^2}\theta = \dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}$
Therefore substituting this valve in the equation (4), we get,
$ \Rightarrow (\sec \theta - 1)(\sec \theta + 1) = \dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}$.

Hence $(\sec \theta - 1)(\sec \theta + 1)$ in terms of sine and cosine is given by,
$(\sec \theta - 1)(\sec \theta + 1) = \dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}$.

Note:
Alternative method :
Here we are given $(\sec \theta - 1)(\sec \theta + 1)$
We covert $(\sec \theta - 1)(\sec \theta + 1)$ in terms of sine and cosine.
Firstly, multiply the expression and try to simplify it.
$ \Rightarrow (\sec \theta - 1)(\sec \theta + 1) = {\sec ^2}\theta + \sec \theta - \sec \theta - 1$
Combining the like terms $\sec \theta - \sec \theta = 0$
Hence the above expression becomes,
$ \Rightarrow (\sec \theta - 1)(\sec \theta + 1) = {\sec ^2}\theta + 0 - 1$
$ \Rightarrow (\sec \theta - 1)(\sec \theta + 1) = {\sec ^2}\theta - 1$
We know that secant is the inverse of cosine function.
So, we convert secant into cosine function.
So we have $\sec \theta = \dfrac{1}{{\cos \theta }}$
Squaring on both sides we get,
$ \Rightarrow {\sec ^2}\theta = \dfrac{{{1^2}}}{{{{\cos }^2}\theta }}$
$ \Rightarrow {\sec ^2}\theta = \dfrac{1}{{{{\cos }^2}\theta }}$
Hence our expression becomes,
$ \Rightarrow (\sec \theta - 1)(\sec \theta + 1) = \dfrac{1}{{{{\cos }^2}\theta }} - 1$
Taking LCM on the right hand side, we get,
$ \Rightarrow (\sec \theta - 1)(\sec \theta + 1) = \dfrac{{1 - {{\cos }^2}\theta }}{{{{\cos }^2}\theta }}$
We have the trigonometric identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$
Taking ${\cos ^2}\theta $ to the other side we get,
${\sin ^2}\theta = 1 - {\cos ^2}\theta $
Therefore substituting this in the above expression we get,
$ \Rightarrow (\sec \theta - 1)(\sec \theta + 1) = \dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}$.
Hence $(\sec \theta - 1)(\sec \theta + 1)$ in terms of sine and cosine is given by,
$(\sec \theta - 1)(\sec \theta + 1) = \dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}$.