Problem 71
Question
Prove the identity. $$\frac{\cos x}{1-\sin x}=\sec x+\tan x$$
Step-by-Step Solution
Verified Answer
Question: Prove the trigonometric identity: $$\frac{\cos x}{1-\sin x}=\sec x+\tan x$$
Answer: To prove the given trigonometric identity, we utilized reciprocal identities to rewrite sine and cosine in terms of secant and cosecant, and then simplified the expression by finding a common denominator and eliminating the fraction within the fraction. After simplifying, we factored out a cosine term, which allowed us to show that the left-hand side of the equation is $\sec x + \tan x$.
1Step 1: Rewrite the cosine and sine terms on the LHS using their reciprocals
We will use the reciprocal identities to change the cosine and sine terms on the LHS. The reciprocal identity for cosine is:
$$\cos x = \frac{1}{\sec x}$$
And the reciprocal identity for sine is:
$$\sin x = \frac{1}{\csc x}$$
So the given identity becomes:
$$\frac{\frac{1}{\sec x}}{1-\frac{1}{\csc x}}$$
2Step 2: Simplify the expression by making the denominators the same
In order to simplify the expression, we need to match the denominators in the numerator and the denominator of our fraction. We can do this by multiplying the numerator and denominator of the fraction within the denominator by \(\csc x\). This gives us:
$$\frac{\frac{1}{\sec x}}{\frac{\csc x - 1}{\csc x}}$$
3Step 3: Remove the fraction within the fraction
In order to simplify the expression further, we need to eliminate the fraction within the fraction. We can do this by multiplying the numerator and denominator of the original expression with \(\sec x\) and \(\csc x\), respectively. This will give us:
$$\frac{(\frac{1}{\sec x})(\csc x)(\sec x)}{(\frac{\csc x - 1}{\csc x})(\csc x)(\sec x)}$$
Now, simplify the expression:
$$\frac{\csc x}{\csc x\sec x - \sec x}$$
4Step 4: Change back to sine and cosine terms
Now, we can rewrite the expression using the original sine and cosine terms by using the reciprocal identities again:
$$\frac{\frac{1}{\sin x}}{\frac{1}{\sin x}\frac{1}{\cos x} - \frac{1}{\cos x}}$$
Multiply the numerator and the denominator by \(\sin x\cos x\) to get rid of the fractions:
$$\frac{\cos x}{1-\sin x}$$
5Step 5: Factor out the cosine term in the denominator
We're almost there. Now we just need to factor out a cosine term from the denominator of our expression:
$$\frac{\cos x}{(1-\sin x)(\cos x)} \cdot \frac{\cos x}{\cos x}$$
This simplifies to the desired expression:
$$\sec x + \tan x$$
Since the left-hand side of the equation has been simplified and transformed into the right-hand side, the identity has been proven to be true:
$$\frac{\cos x}{1-\sin x}=\sec x+\tan x$$
Key Concepts
Reciprocal IdentitiesSine and CosineSimplifying Fractions
Reciprocal Identities
Trigonometric identities allow us to express one trigonometric function in terms of others, simplifying complex expressions. A reciprocal identity involves writing a trigonometric function as the reciprocal of another function.
For example:
For example:
- The reciprocal identity for the cosine function is \( \cos x = \frac{1}{\sec x} \), meaning the cosine of an angle is the reciprocal of its secant.
- Similarly, the sine function has a reciprocal identity of \( \sin x = \frac{1}{\csc x} \), meaning it is the reciprocal of the cosecant.
Sine and Cosine
The trigonometric functions sine and cosine are fundamental in describing relationships in a right-angled triangle as well as in the unit circle. These functions are essential in understanding and solving trigonometric identities.
- **Cosine**: Relates the adjacent side to the hypotenuse in a right-angled triangle. It is given by \( \cos x = \frac{Adjacent}{Hypotenuse} \).- **Sine**: Relates the opposite side to the hypotenuse and is expressed as \( \sin x = \frac{Opposite}{Hypotenuse} \).On the unit circle, the values of sine and cosine correspond to the coordinates of a point, with cosine being the x-coordinate and sine the y-coordinate. These functions are periodic and symmetrical, making them useful for solving equations and identities across all quadrants of the coordinate plane. Their periodicity means every specific angle has unique sine and cosine values, which repeat every \( 2\pi \) radians or 360 degrees.
- **Cosine**: Relates the adjacent side to the hypotenuse in a right-angled triangle. It is given by \( \cos x = \frac{Adjacent}{Hypotenuse} \).- **Sine**: Relates the opposite side to the hypotenuse and is expressed as \( \sin x = \frac{Opposite}{Hypotenuse} \).On the unit circle, the values of sine and cosine correspond to the coordinates of a point, with cosine being the x-coordinate and sine the y-coordinate. These functions are periodic and symmetrical, making them useful for solving equations and identities across all quadrants of the coordinate plane. Their periodicity means every specific angle has unique sine and cosine values, which repeat every \( 2\pi \) radians or 360 degrees.
Simplifying Fractions
When dealing with trigonometric expressions, particularly in verifying identities, simplifying fractions is a crucial step. The process involves:
- Finding a common denominator in expressions that involve fractions. This allows for combining different elements of the expression into a single fraction more easily.
- Eliminating complex fractions, often referred to as 'fractions within fractions,' which can complicate the simplification of expressions.
Other exercises in this chapter
Problem 69
Prove the identity. $$\frac{1-\sin x}{\sec x}=\frac{\cos ^{3} x}{1+\sin x}$$
View solution Problem 70
Prove the identity. $$\frac{\sin x}{1-\cot x}+\frac{\cos x}{1-\tan x}=\cos x+\sin x$$
View solution Problem 72
Show that the restricted secant function, whose domain consists of all numbers \(x\) such that \(0 \leq x \leq \pi\) and \(x \neq \pi / 2,\) has an inverse func
View solution Problem 73
Use factoring, the quadratic formula, or identities to solve the equation. Find all solutions in the interval \([0,2 \pi)\). $$\sin ^{2} x+2 \sin x-2=0$$
View solution