How do you simplify \[\dfrac{1}{{{{\tan }^2}x + 1}}\]?

Hint:We are to solve this problem involving tangent function. So, we know that tangent function is the ratio of sine function by cosine function. So, we will use this fact and replace the tangent function as the ratio of sine and cosine function and accordingly will use different operations required to convert the trigonometric operation into its simplest form. So, let us see how to solve this problem.


Complete step by step solution:
We are given the trigonometric operation is,
\[\dfrac{1}{{{{\tan }^2}x + 1}}\].
So, we know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$.
Thus, now replacing this in the given equation, we get,
$ = \dfrac{1}{{{{\left( {\dfrac{{\sin x}}{{\cos x}}} \right)}^2} + 1}}$
Opening the brackets, we get,
$ = \dfrac{1}{{\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + 1}}$
Now, taking the LCM in the denominator, we get,
$ = \dfrac{1}{{\dfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\cos }^2}x}}}}$
Now, multiplying both numerator and denominator by ${\cos ^2}x$, we get,
$ = \dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x + {{\cos }^2}x}}$
Now, we know, ${\sin ^2}x + {\cos ^2}x = 1$.
Therefore, using this property in the given trigonometric operation, we get,
$ = \dfrac{{{{\cos }^2}x}}{1}$
$ = {\cos ^2}x$
Therefore, the simplified form of the trigonometric operation \[\dfrac{1}{{{{\tan }^2}x + 1}}\] is ${\cos ^2}x$.

Note:
There is also an alternative way to solve this problem. We can also first substitute the tangent function in terms of secant function. We know that, \[{\tan ^2}x + 1 = {\sec ^2}x\]. So, by using this property we will get the whole function in terms of secant function. Then we know, secant function is the reciprocal of cosine function, that is, $\sec x = \dfrac{1}{{\cos x}}$. Therefore, by using this property and replacing this we will get our answer to the required problem as ${\cos ^2}x$.