Question

How do you simplify \[\dfrac{{1 - {{\tan }^2}\left( x \right)}}{{1 + {{\tan }^2}\left( x \right)}}\].

Answer 1

Hint: In the given question, we have been given a trigonometric expression. This expression is a fraction. The numerator and denominator are similar – have one trigonometric variable each and one constant each and the value of the two terms are equal; there is only one difference – the sign between the two terms. We have to simplify this expression. To do that, we are going to use the basic identity to transform the quantity in the denominator. Then we are going to separate the terms in the numerator, use identities on the separated terms, simplify them, and then combine them to solve for the answer.

Formula Used:
We are going to use the relation between tangent and secant in this question, which is,
\[{\sec ^2}x = 1 + {\tan ^2}x\]

Complete step by step answer:
The expression to be simplified is:
\[p = \dfrac{{1 - {{\tan }^2}\left( x \right)}}{{1 + {{\tan }^2}\left( x \right)}}\]
or \[p = \dfrac{{1 - {{\tan }^2}\left( x \right)}}{{{{\sec }^2}\left( x \right)}}\]
Separating the terms,
$\Rightarrow$ \[p = \dfrac{1}{{{{\sec }^2}x}} - \dfrac{{{{\tan }^2}x}}{{{{\sec }^2}x}}\]
Now, we know
$\Rightarrow$ \[{\cos ^2}x = \dfrac{1}{{{{\sec }^2}x}}\]
and \[{\tan ^2}x = \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}\]
Hence,
$\Rightarrow$ \[p = {\cos ^2}x - \dfrac{{{{\sin }^2}x}}{{{{{{\cos }^2}x}}}} \times {{{{\cos }^2}x}}\]

Thus, \[p = \dfrac{{1 - {{\tan }^2}\left( x \right)}}{{1 + {{\tan }^2}\left( x \right)}} = {\cos ^2}x - {\sin ^2}x\].

Note: In the given question, we had to simplify a trigonometric expression. We did that by using our knowledge of the basic identities, formulae used in trigonometry to simplify the terms and combine them to get the answer. Hence, it is very important that we know the exact formulae which are used in trigonometry and where, how, and when to use them so that we get the correct answer.