Question

How do you simplify $\left[ \dfrac{2+{{\tan }^{2}}x}{{{\sec }^{2}}x} \right]-1$?

Answer 1

To solve the given question, we will need some of the properties of trigonometric ratios as follows. The trigonometric identity states that, $1+{{\tan }^{2}}x={{\sec }^{2}}x$. We should also know the property, $\sec x=\dfrac{1}{\cos x}$. We will use these properties to simplify the given expression.

Complete answer:
The given expression is $\left[ \dfrac{2+{{\tan }^{2}}x}{{{\sec }^{2}}x} \right]-1$. The denominator of the expression has the term ${{\sec }^{2}}x$ We know the trigonometric identity which states that, $1+{{\tan }^{2}}x={{\sec }^{2}}x$. Using this identity in the denominator of the given expression. It can be written as,
$\Rightarrow \left[ \dfrac{2+{{\tan }^{2}}x}{1+{{\tan }^{2}}x} \right]-1$
The numerator of the above expression is $2+{{\tan }^{2}}x$, we can replace 2 with the addition of 1 and 1. Doing this for the above expression we get,
$\Rightarrow \left[ \dfrac{1+1+{{\tan }^{2}}x}{1+{{\tan }^{2}}x} \right]-1$
The above expression can also be written as,
$\Rightarrow \left[ \dfrac{1}{1+{{\tan }^{2}}x}+\dfrac{1+{{\tan }^{2}}x}{1+{{\tan }^{2}}x} \right]-1$
$\Rightarrow \left[ \dfrac{1}{1+{{\tan }^{2}}x}+1 \right]-1$
$\Rightarrow \dfrac{1}{1+{{\tan }^{2}}x}$
Again, using the identity which states that $1+{{\tan }^{2}}x={{\sec }^{2}}x$ in the denominator of the above expression, it can be written as
$\Rightarrow \dfrac{1}{{{\sec }^{2}}x}$
We know the trigonometric relationship between $\sec x$and $\cos x$ as $\sec x=\dfrac{1}{\cos x}$. This property can also be expressed as $\cos x=\dfrac{1}{\sec x}$. Using this property in the denominator of the above expression, we get
$\Rightarrow \dfrac{1}{{{\sec }^{2}}x}={{\cos }^{2}}x$
Hence, the given expression can be written in simplified form as, ${{\cos }^{2}}x$.
$\Rightarrow \left[ \dfrac{2+{{\tan }^{2}}x}{{{\sec }^{2}}x} \right]-1={{\cos }^{2}}x$

Note: The trigonometric identities and the trigonometric properties should be remembered to solve these types of problems. The above problem can also be solved by applying the identity $1+{{\tan }^{2}}x={{\sec }^{2}}x$ on the numerator. This can be done as, $\left[ \dfrac{2+{{\tan }^{2}}x}{{{\sec }^{2}}x} \right]-1$
Using the property $1+{{\tan }^{2}}x={{\sec }^{2}}x$ on the numerator of the expression we get, $\Rightarrow \left[ \dfrac{1+{{\sec }^{2}}x}{{{\sec }^{2}}x} \right]-1$
The above expression can also be written as,
$\Rightarrow \left[ \dfrac{1}{{{\sec }^{2}}x}+\dfrac{{{\sec }^{2}}x}{{{\sec }^{2}}x} \right]-1$
$\Rightarrow \left[ \dfrac{1}{{{\sec }^{2}}x}+1 \right]-1=\dfrac{1}{{{\sec }^{2}}x}$
Using the property $\cos x=\dfrac{1}{\sec x}$, the above expression can be expressed as,
$\Rightarrow \dfrac{1}{{{\sec }^{2}}x}={{\cos }^{2}}x$
By both methods, we are getting the same answer.