Question

Simplify $\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}.$

Answer 1

Hint : Using some basic trigonometric identities like we can simplify the above expression.
$\dfrac{1}{{\sec \theta }} = \cos \theta \\\ \tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} \\\ 2{\cos ^2}\theta - 1 = \cos 2\theta \\\ 1 + {\tan ^2}\theta = {\sec ^2}\theta \;$
Such that in order to solve and simplify the given expression we have to use the above identities and express our given expression in that form and thereby simplify it.

Complete step-by-step answer :
Given
$\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}.........................\left( i \right)$
Now to simplify this expression we have to use basic trigonometric identities. We have to convert the given equation in the form of identities known to us and thereby simplify it.
So we know that
$1 + {\tan ^2}\theta = {\sec ^2}\theta$
Substitute this in (i), we get:
$\Rightarrow \dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }} = \dfrac{{1 - {{\tan }^2}\theta }}{{{{\sec }^2}\theta }}.................\left( {ii} \right)$
Now to do any further operations we have to convert our given expression in any known accessible form, so splitting the numerator into two terms:
$\Rightarrow \dfrac{{1 - {{\tan }^2}\theta }}{{{{\sec }^2}\theta }} = \dfrac{1}{{{{\sec }^2}\theta }} - \dfrac{{{{\tan }^2}\theta }}{{{{\sec }^2}\theta }}........................\left( {iii} \right)$
Also we know that
$\dfrac{1}{{\sec \theta }} = \cos \theta \,\,{\text{and}}\,\,\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
So using this basic identity and substituting it in (iii):
We get,
$\Rightarrow \dfrac{1}{{{{\sec }^2}\theta }} - \dfrac{{{{\tan }^2}\theta }}{{{{\sec }^2}\theta }} = {\cos ^2}\theta - \left( {\dfrac{1}{{{{\sec }^2}\theta }} \times \left( {\dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \right)} \right)...........\left( {iv} \right)$
Now again use and substitute $\dfrac{1}{{\sec \theta }} = \cos \theta$ in (iv):

$${\cos ^2}\theta - \left( {\dfrac{1}{{{{\sec }^2}\theta }} \times \left( {\dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \right)} \right) = {\cos ^2}\theta - \left( {{{\cos }^2}\theta \times \left( {\dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \right)} \right) \\\ = {\cos ^2}\theta - {\sin ^2}\theta ........................\left( v \right) \;$$

Now in order to simplify (v) we have to convert the equation to a single term equation i.e. it should be expressed either only in sine terms or cosine terms only.
So let’s convert the equation to a cosine term equation by using:
${\sin ^2}\theta = 1 - {\cos ^2}\theta$
So substituting this in (v), we get:
${\cos ^2}\theta - {\sin ^2}\theta = {\cos ^2}\theta - \left( {1 - {{\cos }^2}\theta } \right) \\\ = {\cos ^2}\theta - 1 + {\cos ^2}\theta \\\ = 2{\cos ^2}\theta - 1...........................\left( {vi} \right) \\\$
Now we know that $2{\cos ^2}\theta - 1 = \cos 2\theta$ .
Therefore our final answer is:
$\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }} = \cos 2\theta$
So, the correct answer is “ $\cos 2\theta$ ”.

Note : Some other equations needed for solving these types of problem are:

$${\sin \left( {2\theta } \right) = 2\sin \left( \theta \right)\cos \left( \theta \right)} \\\ {\cos \left( {2\theta } \right) = {{\cos }^2}\left( \theta \right)-{{\sin }^2}\left( \theta \right) = 1-2{\text{ }}{{\sin }^2}\left( \theta \right) = 2{\text{ }}{{\cos }^2}\left( \theta \right)-1} \;$$

Also while approaching a trigonometric problem one should keep in mind that one should work with one side at a time and manipulate it to the other side. The most straightforward way to do this is to simplify one side to the other directly, but we can also transform both sides to a common expression if we see no direct way to connect the two.