/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 104 Verify that equation is an ident... [FREE SOLUTION] | 91Ó°ÊÓ

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Verify that equation is an identity. \((\sec \alpha+\csc \alpha)(\cos \alpha-\sin \alpha)=\cot \alpha-\tan \alpha\)

Short Answer

Expert verified
The equation is verified as an identity.

Step by step solution

01

Expand the Left Side of the Equation

Expand the left side of the equation \((\sec \alpha + \csc \alpha)(\cos \alpha - \sin \alpha)\). Apply the distributive property to get:\[\sec \alpha \cdot \cos \alpha - \sec \alpha \cdot \sin \alpha + \csc \alpha \cdot \cos \alpha - \csc \alpha \cdot \sin \alpha\] This results in:\[1 - \sec \alpha \cdot \sin \alpha + \csc \alpha \cdot \cos \alpha - 1\]Which simplifies to:\[- \sec \alpha \cdot \sin \alpha + \csc \alpha \cdot \cos \alpha\]
02

Simplify Using Trigonometric Identities

Recall the identities for secant and cosecant:\[ \sec \alpha = \frac{1}{\cos \alpha}, \quad \csc \alpha = \frac{1}{\sin \alpha} \]Substitute these identities into the simplified expression:\[- \frac{1}{\cos \alpha} \sin \alpha + \frac{1}{\sin \alpha} \cos \alpha\]This can be rewritten as:\[- \frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha} \] Which simplifies to: \[-\tan \alpha + \cot \alpha\]
03

Simplify the Right Side of the Equation

The expression for the right side of the equation is already given as:\[\cot \alpha - \tan \alpha\] This is the same as the simplified form of the left side found in Step 2.
04

Conclusion

Since both sides of the equation simplify to the same expression, \(\cot \alpha - \tan \alpha\), the original equation \((\sec \alpha + \csc \alpha)(\cos \alpha - \sin \alpha) = \cot \alpha - \tan \alpha\) is verified as an identity.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Secant Function
The secant function is a vital part of trigonometry, often abbreviated as \( \sec \). It is the reciprocal of the cosine function. This means that to find the secant of an angle \( \alpha \), you divide 1 by the cosine of that angle. The formula is:
  • \( \sec \alpha = \frac{1}{\cos \alpha} \)
If the cosine of an angle is close to zero, the secant value becomes very large, which is an important property when analyzing trigonometric identities and equations. Remember that the domain of the secant function excludes angles where the cosine is zero, such as \( 90^{\circ} \) and \( 270^{\circ} \), where the function is undefined.
To visualize \( \sec \alpha \), think of it as projecting outward from the unit circle along the tangent line at the point \( (\cos \alpha, \sin \alpha) \). This projection helps in understanding its behavior in trigonometric equations.
Cosecant Function
The cosecant function, represented as \( \csc \), is the reciprocal of the sine function. For any angle \( \alpha \), it is expressed as:
  • \( \csc \alpha = \frac{1}{\sin \alpha} \)
Similar to the secant function, the cosecant can become very large when the sine of an angle is near zero. This characteristic influences the graph's spikes upwards or downwards, making the function undefined at multiples of \( 180^{\circ} \), where the sine is zero.
Visually, the \( \csc \alpha \) can also be thought of as the length from the origin to where the point on the circle intercepts a line perpendicular to the \( y \)-axis. This geometrical interpretation makes it easier to grasp its relation to other trigonometric functions in identities.
Tangent Function
Tangent is another foundational trigonometric function and is frequently represented as \( \tan \). Unlike secant and cosecant, tangent involves both sine and cosine. The tangent of an angle \( \alpha \) is the ratio of the sine to the cosine of that angle:
  • \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \)
This function is vital since it ties together sine and cosine and shows dramatic behavior changes near angles where cosine approaches zero (like \( 90^{\circ} \)). These angles cause tangent to spike or drop suddenly to infinity or negative infinity.
In the unit circle, tangent represents the slope of the line that connects the origin with the point on the circle corresponding to angle \( \alpha \). Mastery of tangent function helps in solving identities, as seen in transforming expressions like \(-\tan \alpha + \cot \alpha\) into equivalent forms.
Cotangent Function
The cotangent function, denoted as \( \cot \), is closely linked to tangent. It is essentially its reciprocal. For an angle \( \alpha \), the cotangent is given by:
  • \( \cot \alpha = \frac{\cos \alpha}{\sin \alpha} \)
Cotangent has unique properties with behaviors similar yet opposite to tangent. It is undefined where the sine of the angle is zero, like at \( 0^{\circ} \), \( 180^{\circ} \), etc.
At these points, while tangent becomes infinite, cotangent "switches off," creating a critical understanding of its variance with respect to angle increments. Visually, \( \cot \alpha \) often denotes the angle where a line perpendicular to the \( x \)-axis and the radius at \( \alpha \) meets the x-axis itself. Like tangent, cotangent is crucial in simplifying identities or transforming expressions such as \( \cot \alpha - \tan \alpha \), as demonstrated in verifying complex trigonometric identities.

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Most popular questions from this chapter

Solve each equation ( \(x\) in radians and \(\theta\) in degrees) for all exact solutions where appropriate. Round approximale values in radians to four decimal places and approximate values in degrees to the nearest tenth. Write answers using the least possible nonnegative angle measures. $$\sin \theta-\sin 2 \theta=0$$

Solve each problem. Nautical Mile The British nautical mile is defined as the length \(L\) of a minute of arc on any meridian. Since Earth is flatter at its poles, the nautical mile varies with latitude and, in feet, is $$ L=6077-31 \cos 2 \theta $$ where \(\theta\) is the latitude in degrees. (See the figure.) (Figure can't copy) (a) Find the latitude between \(0^{\circ}\) and \(90^{\circ}\) at which the nautical mile is 6074 feet. (b) At what latitude between \(0^{\circ}\) and \(90^{\circ}\) (inclusive) is the nautical mile 6108 feet? (c) In the United States, the nautical mile is defined everywhere as 6080.2 feet. At what latitude between \(0^{\circ}\) and \(90^{\circ}\) does this agree with the British nautical mile?

Write each expression as an algebraic expression in \(u, u>0\). $$\tan (\arccos u)$$

Solve each equation ( \(x\) in radians and \(\theta\) in degrees) for all exact solutions where appropriate. Round approximale values in radians to four decimal places and approximate values in degrees to the nearest tenth. Write answers using the least possible nonnegative angle measures. $$3 \csc ^{2} \frac{x}{2}=2 \sec x$$

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