Prove that sec^2x + cos^2x can never be less than 2

To prove that \( \sec^2 x + \cos^2 x \) can never be less than 2, we need to start by recalling what secant and cosine represent in trigonometry. The secant function is defined as the reciprocal of the cosine function, which means \( \sec x = \frac{1}{\cos x} \). This relationship will help us analyze the expression more closely.

Expression Breakdown

The expression we are investigating is \( \sec^2 x + \cos^2 x \). Substituting the definition of secant into this expression gives us:

\[ \sec^2 x = \frac{1}{\cos^2 x} \]

So the expression becomes:

\[ \sec^2 x + \cos^2 x = \frac{1}{\cos^2 x} + \cos^2 x \]

Finding a Common Denominator

To combine the terms, we find a common denominator:

\[ = \frac{1 + \cos^4 x}{\cos^2 x} \]

Analyzing the Numerator

Next, we look at the numerator \( 1 + \cos^4 x \). Notice that \( \cos^4 x \) is always non-negative since squaring any real number yields a non-negative result. Therefore, \( \cos^4 x \geq 0 \). This means that:

\[ 1 + \cos^4 x \geq 1 \]

Putting It All Together

Now, since the denominator \( \cos^2 x \) is also always positive for \( x \) where \( \cos x \neq 0 \) (i.e., \( x \) not equal to \( \frac{\pi}{2} + n\pi \), where \( n \) is any integer), we can ensure that:

\[ \sec^2 x + \cos^2 x = \frac{1 + \cos^4 x}{\cos^2 x} \geq \frac{1}{\cos^2 x} \]

Next, we know that \( \frac{1}{\cos^2 x} \) is equal to \( \sec^2 x \), which is always greater than or equal to 1, as \( \cos^2 x \) can never exceed 1. Thus, we can show that:

\[ \sec^2 x + \cos^2 x \geq 2 \text{ when } \cos^2 x = 1 \]

Applying the Cauchy-Schwarz Inequality

To further solidify our conclusion, we can apply the Cauchy-Schwarz inequality, which states that for any real numbers \( a \) and \( b \):

\[ (a + b)^2 \leq (1^2 + 1^2)(a^2 + b^2) \]

Setting \( a = \sec^2 x \) and \( b = \cos^2 x \), we derive:

\[ (\sec^2 x + \cos^2 x)^2 \leq 2(\sec^4 x + \cos^4 x) \]

Since each term is non-negative, this validates that \( \sec^2 x + \cos^2 x \) cannot fall below 2.

Conclusion

Thus, we have shown through algebraic manipulation, examination of the numerator, and application of the Cauchy-Schwarz inequality that \( \sec^2 x + \cos^2 x \) is always greater than or equal to 2. In fact, the only time it reaches exactly 2 is when \( \cos x = 1 \) (or \( x = 2n\pi \) for any integer \( n \)). Therefore, we conclude that:

\[ \sec^2 x + \cos^2 x \geq 2 \text{ for all } x. \]