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Chapter 4: Problem 56

Use trigonometric identities to transform the left side of the equation into the right side \((0<\theta<\pi 2)\). $$\frac{\tan \beta+\cot \beta}{\tan \beta}=\csc ^{2} \beta$$

Short Answer

Expert verified
The transformation of \(\frac{\tan \beta+\cot \beta}{\tan \beta}\) to \(\csc ^{2} \beta\) is achieved through the use of reciprocal identities and quotient identities in trigonometry.

Step by step solution

01

Recall Trigonometric Identities

Remember the reciprocal identities: \(\cot \beta = \frac{1}{\tan \beta}\) and \(\csc \beta = \frac{1}{\sin \beta}\). Also recall the quotient identity \(\tan \beta = \frac{\sin \beta}{\cos \beta}\).
02

Simplify the LHS of the Equation Using the Reciprocal Identity

By substituting \(\cot \beta = \frac{1}{\tan \beta}\) into the equation, the LHS becomes \(\frac{\tan \beta+ \frac{1}{\tan \beta}}{\tan \beta}\). We can further simplify this by splitting the fraction to obtain \(\frac{\tan \beta}{\tan \beta} + \frac{1}{\tan^2 \beta} = 1 + \frac{1}{\tan^2 \beta}\).
03

Continue to Simplify using the Quotient Identity

We replace \(\tan \beta\) with \(\frac{\sin \beta}{\cos \beta}\) in \(\frac{1}{\tan^2 \beta}\) to get \(\frac{1}{\left(\frac{\sin \beta}{\cos \beta}\right)^2} = \frac{\cos^2 \beta}{\sin^2 \beta}\).
04

Final Transformation using Reciprocal Identity

Finally, express the denominator in the form \(\frac{1}{\sin^2 \beta}\) as \(\csc^2 \beta\). Thus, our LHS expression \(\frac{\tan \beta+\cot \beta}{\tan \beta}\) simplifies to the RHS expression \(\csc^2 \beta\).

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Key Concepts

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

Reciprocal Identities
Reciprocal identities are fundamental in trigonometry. They allow us to rewrite trigonometric functions in terms of their reciprocal counterparts. For instance, the cosecant \textbf{(csc)} is the reciprocal of the sine function (sin), so we define it as \( \csc \theta = \dfrac{1}{\sin \theta} \). Similarly, the secant \textbf{(sec)} is the reciprocal of cosine (cos), \( \sec \theta = \dfrac{1}{\cos \theta} \) and cotangent \textbf{(cot)} is the reciprocal of tangent (tan), \( \cot \theta = \dfrac{1}{\tan \theta} \).

Quotient Identity
The quotient identity is another vital concept in trigonometry and is particularly useful for transforming expressions. It relates the tangent function to sine and cosine. The quotient identity for tangent is \( \tan \theta = \dfrac{\sin \theta}{\cos \theta} \). This relationship helps to break down more complex expressions into simpler forms, which can then be further manipulated using other identities.
  • Useful for isolating a single trigonometric function from a complex ratio
  • Facilitates transforming and simplifying trigonometric expressions

Simplifying Trigonometric Expressions
Understanding how to simplify trigonometric expressions is crucial when solving equations involving trigonometric functions. The process often involves applying a series of identities to transform the given expression into a more manageable or recognizable form. The key steps involve:
  • Recognizing which identities are applicable
  • Substituting the identities into the expression
  • Algebraically manipulating the expression as needed
For example, expressions with mixed functions like tangent and cotangent can often be simplified by using reciprocal identities to express them in terms of a single function or by using the quotient identity when we need to convert between sine and cosine.

When faced with an exercise like \(\frac{\tan \beta + \cot \beta}{\tan \beta} = \csc^2 \beta \) it's evident that identifying and applying these principles properly will lead to a successful simplification, ultimately showing the equivalence of two seemingly different expressions.

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Most popular questions from this chapter

Graph the functions \(f\) and \(g .\) Use the graphs to make a conjecture about the relationship between the functions. $$f(x)=\sin x-\cos \left(x+\frac{\pi}{2}\right), \quad g(x)=2 \sin x$$

A sprinkler on a golf green sprays water over a distance of 15 meters and rotates through an angle of \(140^{\circ} .\) Draw a diagram that shows the region that the sprinkler can irrigate. Find the area of the region.

Fill in the blank. If not possible, state the reason. $$\text { As } x \rightarrow 1^{-}, \text {the value of } \arcsin x \rightarrow \text{______}$$

Use a graphing utility to graph the function. Use the graph to determine the behavior of the function as \(x \rightarrow c\) (a) \(x \rightarrow\left(\frac{\pi}{2}\right)^{+}\) (b) \(x \rightarrow\left(\frac{\pi}{2}\right)^{-}\) (c) \(x \rightarrow\left(-\frac{\pi}{2}\right)^{+}\) (d) \(x \rightarrow\left(-\frac{\pi}{2}\right)^{-}\) $$f(x)=\sec x$$

Data Analysis The table shows the average sales \(S\) (in millions of dollars) of an outerwear manufacturer for each month \(t,\) where \(t=1\) represents January. $$\begin{array}{|l|c|c|c|c|c|c|} \hline \text { Time, } t & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text { Sales, } S & 13.46 & 11.15 & 8.00 & 4.85 & 2.54 & 1.70 \\\ \hline \end{array}$$ $$\begin{array}{|l|c|c|c|c|c|c|} \hline \text { Time, } t & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline \text { Sales, } S & 2.54 & 4.85 & 8.00 & 11.15 & 13.46 & 14.30 \\ \hline \end{array}$$ (a) Create a scatter plot of the data. (b) Find a trigonometric model that fits the data. Graph the model with your scatter plot. How well does the model fit the data? (c) What is the period of the model? Do you think it is reasonable given the context? Explain your reasoning. (d) Interpret the meaning of the model's amplitude in the context of the problem.
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