Question

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

Answer 1

Hint: 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.