Question

\(\sec x=a+\frac{1}{4 a}\) then \(\sec x+\tan x\) is: (a) \(\frac{2}{\mathrm{a}}\) (b) a (c) \(\frac{1}{\mathrm{a}}\) (d) \(\frac{1}{2 \mathrm{a}}\)

Short answer

Based on the step-by-step solution above, we found that: $$ \sec x + \tan x = \frac{1 + \sqrt{1 - \left(\frac{1}{a + \frac{1}{4a}}\right)^2}}{a + \frac{1}{4a}} $$ However, none of the given options (a), (b), (c), and (d) match this expression. So, the correct answer is: None of the given options.

Step-by-step solution

Express \(\sec x\) in terms of \(\cos x\)

Given that \(\sec x = a + \frac{1}{4a}\), we can write \(\sec x\) in terms of \(\cos x\) using the identity \(\sec x = \frac{1}{\cos x}\), as follows: $$ \frac{1}{\cos x} = a + \frac{1}{4a} $$

Solve for \(\cos x\)

Next, solve for \(\cos x\) in the equation we found in the previous step: $$ \cos x = \frac{1}{a + \frac{1}{4a}} $$

Find \(\sin x\) using the Pythagorean identity

Now that we have found \(\cos x\), we can find \(\sin x\) using the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\): $$ \sin^2 x = 1 - \cos^2 x $$ $$ \sin^2 x = 1 - \left(\frac{1}{a + \frac{1}{4a}}\right)^2 $$ $$ \sin x = \sqrt{1-\left(\frac{1}{a +\frac{1}{4a}}\right)^2} $$

Find \(\tan x\) using the trigonometric identity

Now that we have both \(\cos x\) and \(\sin x\), we can find \(\tan x\) using the identity \(\tan x = \frac{\sin x}{\cos x}\): $$ \tan x = \frac{\sqrt{1 - \left(\frac{1}{a + \frac{1}{4a}}\right)^2}}{\frac{1}{a + \frac{1}{4a}}} $$

Find \(\sec x + \tan x\)

Finally, add \(\sec x\) and \(\tan x\) to get the expression we want to find: $$ \sec x + \tan x = a + \frac{1}{4a} + \frac{\sqrt{1 - \left(\frac{1}{a + \frac{1}{4a}}\right)^2}}{\frac{1}{a + \frac{1}{4a}}} $$ However, this expression is not what we need to find. To get the right expression, we need to simplify the expression further and spot the given options within the simplification. $$ \sec x + \tan x = \frac{1}{\cos x}+\frac{\sin x}{\cos x} = \frac{1+\sin x}{\cos x} $$ Thus, we have: $$ \sec x + \tan x = \frac{1 + \sqrt{1 - \left(\frac{1}{a + \frac{1}{4a}}\right)^2}}{a + \frac{1}{4a}} $$ Now let's compare this expression with the given options: (a) \(\frac{2}{a}\) (b) a (c) \(\frac{1}{a}\) (d) \(\frac{1}{2a}\) None of these options match the simplified expression we found for \(\sec x + \tan x\). So, there is no correct answer among the given options.

Key concepts

Secant Function

The secant function, denoted as \( \sec x \), is an important trigonometric function. It is defined as the reciprocal of the cosine function. Mathematically, this can be expressed as:
\[ \sec x = \frac{1}{\cos x} \]This tells us that wherever the cosine function is defined and non-zero, the secant function will be as well. Being a reciprocal, the secant function will have values equal in magnitude to 1 over the cosine value of a given angle. If \( \cos x \) is small, \( \sec x \) becomes large and vice versa.
The domain of the secant function excludes angles where the cosine is zero since division by zero is undefined. These angles are of the form \( x = (2n+1)\frac{\pi}{2} \), where \( n \) is an integer. The range of \( \sec x \) is \(( -\infty, -1 ] \cup [ 1, \infty )\).

• Secant is positive in the first and fourth quadrants, where cosine is also positive.
• It helps in solving equations involving \( \sec x \), especially when combined with identities related to sine and cosine.

Tangent Function

The tangent function, often written as \( \tan x \), expresses the ratio of sine to cosine. It is defined by:
\[ \tan x = \frac{\sin x}{\cos x} \]Tangent can be difficult to visualize, but thinking of it in terms of sine and cosine can make it easier to understand. It indicates the slope of a line drawn from a point on the unit circle to the origin.
The tangent function is periodic with a period of \( \pi \), meaning it repeats every \( \pi \) units. Its domain excludes points where \( \cos x = 0 \), similar to where secant is undefined, because division by zero is indeterminate. These points occur at \( x = (2n+1)\frac{\pi}{2} \) for integers \( n \).

• Tangent is positive in the first and third quadrants, where sine and cosine have the same sign.
• The range of \( \tan x \) is all real numbers.
• This function is vital in solving trigonometric equations and can also describe angles in triangles and slopes in calculus.

Pythagorean Identity

The Pythagorean identity is a fundamental relation in trigonometry, connecting the squares of sine and cosine to 1. Expressed mathematically, it states:
\[ \sin^2 x + \cos^2 x = 1 \]This identity is rooted in the Pythagorean theorem and is always valid for any angle \( x \). It can be rearranged to find either \( \sin^2 x \) or \( \cos^2 x \) if one is known:

• \( \sin^2 x = 1 - \cos^2 x \)
• \( \cos^2 x = 1 - \sin^2 x \)

Understanding this identity allows us to express one trigonometric function in terms of another. This is especially useful in solving complex problems, where simplifying expressions is necessary.
Additionally, a derivation of the identity helps define tangent and secant identities:

• By dividing through by \( \cos^2 x \), you get \( 1 + \tan^2 x = \sec^2 x \).
• This directly relates to the problem at hand, providing a link between secant and tangent values.

This interconnectedness of trigonometric functions through identities makes them powerful tools in mathematical analysis and problem-solving.