Verify the identity. 4 tan x + 4 tan y- 4 cot x + 4 cot y 1- tan x tan y cot x cot y - 1 Step 1 First recall the reciprocal identity for tan 0. 1 tan 0 = cot(0) cot (0) Step 2 Now simplify the left side of the given expression. 4 4 tan x + 4 tan y 1- tan x tan y cot x cot y 1 cot x cot y 4 cot y + cot x cot y cot x cot y - cot x cot y 4 cot x + 4 cot y cot x cot y - 1
Verify the identity. 4 tan x + 4 tan y- 4 cot x + 4 cot y 1- tan x tan y cot x cot y - 1 Step 1 First recall the reciprocal identity for tan 0. 1 tan 0 = cot(0) cot (0) Step 2 Now simplify the left side of the given expression. 4 4 tan x + 4 tan y 1- tan x tan y cot x cot y 1 cot x cot y 4 cot y + cot x cot y cot x cot y - cot x cot y 4 cot x + 4 cot y cot x cot y - 1
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE:
1. Give the measures of the complement and the supplement of an angle measuring 35°.
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Equations and Inequations
Equations and inequalities describe the relationship between two mathematical expressions.
Linear Functions
A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.
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Solve ... what will come in the empty blanks that I left... plz
![### Tutorial Exercise
**Verify the identity:**
\[
\frac{4 \tan x + 4 \tan y}{1 - \tan x \tan y} = \frac{4 \cot x + 4 \cot y}{\cot x \cot y - 1}
\]
**Step 1:**
First recall the reciprocal identity for \(\tan \theta\).
\[
\tan \theta = \frac{1}{\cot(\theta)}
\]
(Diagram shows a fraction \(\frac{1}{\cot(\theta)}\) with \(\cot(\theta)\) as the denominator and the expression simplified to show \(\cot(\theta)\).)
**Step 2:**
Now simplify the left side of the given expression.
\[
\frac{4 \tan x + 4 \tan y}{1 - \tan x \tan y} = \frac{\frac{4}{\cot x} + \frac{4}{\cot y}}{1 - \frac{1}{\cot x} \cdot \frac{1}{\cot y}}
\]
This expression becomes:
\[
= \frac{(4 \cot y + \text{________})}{\cot x \cdot \cot y}
\]
\[
= \frac{\left(\cot x \cdot \cot y - \text{________}\right)}{\cot x \cdot \cot y}
\]
\[
= \frac{4 \cot x + 4 \cot y}{\cot x \cot y - 1}
\]
(Blanks indicate parts of the expression to be filled in. The diagram also includes lines grouping and simplifying components of the expression.)
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Transcribed Image Text:### Tutorial Exercise
**Verify the identity:**
\[
\frac{4 \tan x + 4 \tan y}{1 - \tan x \tan y} = \frac{4 \cot x + 4 \cot y}{\cot x \cot y - 1}
\]
**Step 1:**
First recall the reciprocal identity for \(\tan \theta\).
\[
\tan \theta = \frac{1}{\cot(\theta)}
\]
(Diagram shows a fraction \(\frac{1}{\cot(\theta)}\) with \(\cot(\theta)\) as the denominator and the expression simplified to show \(\cot(\theta)\).)
**Step 2:**
Now simplify the left side of the given expression.
\[
\frac{4 \tan x + 4 \tan y}{1 - \tan x \tan y} = \frac{\frac{4}{\cot x} + \frac{4}{\cot y}}{1 - \frac{1}{\cot x} \cdot \frac{1}{\cot y}}
\]
This expression becomes:
\[
= \frac{(4 \cot y + \text{________})}{\cot x \cdot \cot y}
\]
\[
= \frac{\left(\cot x \cdot \cot y - \text{________}\right)}{\cot x \cdot \cot y}
\]
\[
= \frac{4 \cot x + 4 \cot y}{\cot x \cot y - 1}
\]
(Blanks indicate parts of the expression to be filled in. The diagram also includes lines grouping and simplifying components of the expression.)
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