Solutions
Integral CalculatorDerivative CalculatorAlgebra CalculatorMatrix CalculatorMore...
Graphing
Line Graph CalculatorExponential Graph CalculatorQuadratic Graph CalculatorSin graph CalculatorMore...
Calculators
BMI CalculatorCompound Interest CalculatorPercentage CalculatorAcceleration CalculatorMore...
Geometry
Pythagorean Theorem CalculatorCircle Area CalculatorIsosceles Triangle CalculatorTriangles CalculatorMore...
Tools
NotebookGroupsCheat SheetsWorksheetsPracticeVerify
en
English
Español
Português
Français
Deutsch
Italiano
Русский
中文(简体)
한국어
日本語
Tiếng Việt
עברית
العربية
Popular Trigonometry >

(cot(θ)+csc(θ))/(sec(θ)+1)=sin(θ)

  • Pre Algebra
  • Algebra
  • Pre Calculus
  • Calculus
  • Functions
  • Linear Algebra
  • Trigonometry
  • Statistics
  • Physics
  • Chemistry
  • Finance
  • Economics
  • Conversions

Solution

sec(θ)+1cot(θ)+csc(θ)​=sin(θ)

Solution

θ=0.90455…+2πn,θ=2π−0.90455…+2πn
+1
Degrees
θ=51.82729…∘+360∘n,θ=308.17270…∘+360∘n
Solution steps
sec(θ)+1cot(θ)+csc(θ)​=sin(θ)
Subtract sin(θ) from both sidessec(θ)+1cot(θ)+csc(θ)​−sin(θ)=0
Simplify sec(θ)+1cot(θ)+csc(θ)​−sin(θ):sec(θ)+1cot(θ)+csc(θ)−sin(θ)(sec(θ)+1)​
sec(θ)+1cot(θ)+csc(θ)​−sin(θ)
Convert element to fraction: sin(θ)=sec(θ)+1sin(θ)(sec(θ)+1)​=sec(θ)+1cot(θ)+csc(θ)​−sec(θ)+1sin(θ)(sec(θ)+1)​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=sec(θ)+1cot(θ)+csc(θ)−sin(θ)(sec(θ)+1)​
sec(θ)+1cot(θ)+csc(θ)−sin(θ)(sec(θ)+1)​=0
g(x)f(x)​=0⇒f(x)=0cot(θ)+csc(θ)−sin(θ)(sec(θ)+1)=0
Express with sin, cos
cot(θ)+csc(θ)−(1+sec(θ))sin(θ)
Use the basic trigonometric identity: cot(x)=sin(x)cos(x)​=sin(θ)cos(θ)​+csc(θ)−(1+sec(θ))sin(θ)
Use the basic trigonometric identity: csc(x)=sin(x)1​=sin(θ)cos(θ)​+sin(θ)1​−(1+sec(θ))sin(θ)
Use the basic trigonometric identity: sec(x)=cos(x)1​=sin(θ)cos(θ)​+sin(θ)1​−(1+cos(θ)1​)sin(θ)
Simplify sin(θ)cos(θ)​+sin(θ)1​−(1+cos(θ)1​)sin(θ):sin(θ)cos(θ)cos(θ)(cos(θ)+1)−sin2(θ)(cos(θ)+1)​
sin(θ)cos(θ)​+sin(θ)1​−(1+cos(θ)1​)sin(θ)
Combine the fractions sin(θ)cos(θ)​+sin(θ)1​:sin(θ)cos(θ)+1​
Apply rule ca​±cb​=ca±b​=sin(θ)cos(θ)+1​
=sin(θ)cos(θ)+1​−sin(θ)(cos(θ)1​+1)
Join 1+cos(θ)1​:cos(θ)cos(θ)+1​
1+cos(θ)1​
Convert element to fraction: 1=cos(θ)1cos(θ)​=cos(θ)1⋅cos(θ)​+cos(θ)1​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=cos(θ)1⋅cos(θ)+1​
Multiply: 1⋅cos(θ)=cos(θ)=cos(θ)cos(θ)+1​
=sin(θ)cos(θ)+1​−cos(θ)cos(θ)+1​sin(θ)
Multiply cos(θ)cos(θ)+1​sin(θ):cos(θ)sin(θ)(cos(θ)+1)​
cos(θ)cos(θ)+1​sin(θ)
Multiply fractions: a⋅cb​=ca⋅b​=cos(θ)(cos(θ)+1)sin(θ)​
=sin(θ)cos(θ)+1​−cos(θ)(cos(θ)+1)sin(θ)​
Least Common Multiplier of sin(θ),cos(θ):sin(θ)cos(θ)
sin(θ),cos(θ)
Lowest Common Multiplier (LCM)
Compute an expression comprised of factors that appear either in sin(θ) or cos(θ)=sin(θ)cos(θ)
Adjust Fractions based on the LCM
Multiply each numerator by the same amount needed to multiply its
corresponding denominator to turn it into the LCM sin(θ)cos(θ)
For sin(θ)cos(θ)+1​:multiply the denominator and numerator by cos(θ)sin(θ)cos(θ)+1​=sin(θ)cos(θ)(cos(θ)+1)cos(θ)​
For cos(θ)(cos(θ)+1)sin(θ)​:multiply the denominator and numerator by sin(θ)cos(θ)(cos(θ)+1)sin(θ)​=cos(θ)sin(θ)(cos(θ)+1)sin(θ)sin(θ)​=sin(θ)cos(θ)sin2(θ)(cos(θ)+1)​
=sin(θ)cos(θ)(cos(θ)+1)cos(θ)​−sin(θ)cos(θ)sin2(θ)(cos(θ)+1)​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=sin(θ)cos(θ)(cos(θ)+1)cos(θ)−sin2(θ)(cos(θ)+1)​
=sin(θ)cos(θ)cos(θ)(cos(θ)+1)−sin2(θ)(cos(θ)+1)​
cos(θ)sin(θ)(1+cos(θ))cos(θ)−(1+cos(θ))sin2(θ)​=0
g(x)f(x)​=0⇒f(x)=0(1+cos(θ))cos(θ)−(1+cos(θ))sin2(θ)=0
Factor (1+cos(θ))cos(θ)−(1+cos(θ))sin2(θ):(1+cos(θ))(cos(θ)−sin2(θ))
(1+cos(θ))cos(θ)−(1+cos(θ))sin2(θ)
Factor out common term (1+cos(θ))=(1+cos(θ))(cos(θ)−sin2(θ))
(1+cos(θ))(cos(θ)−sin2(θ))=0
Solving each part separately1+cos(θ)=0orcos(θ)−sin2(θ)=0
1+cos(θ)=0:θ=π+2πn
1+cos(θ)=0
Move 1to the right side
1+cos(θ)=0
Subtract 1 from both sides1+cos(θ)−1=0−1
Simplifycos(θ)=−1
cos(θ)=−1
General solutions for cos(θ)=−1
cos(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​cos(x)123​​22​​21​0−21​−22​​−23​​​xπ67π​45π​34π​23π​35π​47π​611π​​cos(x)−1−23​​−22​​−21​021​22​​23​​​​
θ=π+2πn
θ=π+2πn
cos(θ)−sin2(θ)=0:θ=arccos(2−1+5​​)+2πn,θ=2π−arccos(2−1+5​​)+2πn
cos(θ)−sin2(θ)=0
Rewrite using trig identities
cos(θ)−sin2(θ)
Use the Pythagorean identity: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=cos(θ)−(1−cos2(θ))
−(1−cos2(θ)):−1+cos2(θ)
−(1−cos2(θ))
Distribute parentheses=−(1)−(−cos2(θ))
Apply minus-plus rules−(−a)=a,−(a)=−a=−1+cos2(θ)
=cos(θ)−1+cos2(θ)
−1+cos(θ)+cos2(θ)=0
Solve by substitution
−1+cos(θ)+cos2(θ)=0
Let: cos(θ)=u−1+u+u2=0
−1+u+u2=0:u=2−1+5​​,u=2−1−5​​
−1+u+u2=0
Write in the standard form ax2+bx+c=0u2+u−1=0
Solve with the quadratic formula
u2+u−1=0
Quadratic Equation Formula:
For a=1,b=1,c=−1u1,2​=2⋅1−1±12−4⋅1⋅(−1)​​
u1,2​=2⋅1−1±12−4⋅1⋅(−1)​​
12−4⋅1⋅(−1)​=5​
12−4⋅1⋅(−1)​
Apply rule 1a=112=1=1−4⋅1⋅(−1)​
Apply rule −(−a)=a=1+4⋅1⋅1​
Multiply the numbers: 4⋅1⋅1=4=1+4​
Add the numbers: 1+4=5=5​
u1,2​=2⋅1−1±5​​
Separate the solutionsu1​=2⋅1−1+5​​,u2​=2⋅1−1−5​​
u=2⋅1−1+5​​:2−1+5​​
2⋅1−1+5​​
Multiply the numbers: 2⋅1=2=2−1+5​​
u=2⋅1−1−5​​:2−1−5​​
2⋅1−1−5​​
Multiply the numbers: 2⋅1=2=2−1−5​​
The solutions to the quadratic equation are:u=2−1+5​​,u=2−1−5​​
Substitute back u=cos(θ)cos(θ)=2−1+5​​,cos(θ)=2−1−5​​
cos(θ)=2−1+5​​,cos(θ)=2−1−5​​
cos(θ)=2−1+5​​:θ=arccos(2−1+5​​)+2πn,θ=2π−arccos(2−1+5​​)+2πn
cos(θ)=2−1+5​​
Apply trig inverse properties
cos(θ)=2−1+5​​
General solutions for cos(θ)=2−1+5​​cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnθ=arccos(2−1+5​​)+2πn,θ=2π−arccos(2−1+5​​)+2πn
θ=arccos(2−1+5​​)+2πn,θ=2π−arccos(2−1+5​​)+2πn
cos(θ)=2−1−5​​:No Solution
cos(θ)=2−1−5​​
−1≤cos(x)≤1NoSolution
Combine all the solutionsθ=arccos(2−1+5​​)+2πn,θ=2π−arccos(2−1+5​​)+2πn
Combine all the solutionsθ=π+2πn,θ=arccos(2−1+5​​)+2πn,θ=2π−arccos(2−1+5​​)+2πn
Since the equation is undefined for:π+2πnθ=arccos(2−1+5​​)+2πn,θ=2π−arccos(2−1+5​​)+2πn
Show solutions in decimal formθ=0.90455…+2πn,θ=2π−0.90455…+2πn

Graph

Sorry, your browser does not support this application
View interactive graph

Popular Examples

1/(cos(2x))+tan(2x)=3cos(2x),0<x<90sin(x)= 4/5 ,0<= x<2pi7sin^2(θ)-5sin(θ)=2sec(2x)=-(2/(sqrt(3)))(e^{-ln(-(sin(θ))/(cos(θ)))})/2*sin(θ)=0

Frequently Asked Questions (FAQ)

  • What is the general solution for (cot(θ)+csc(θ))/(sec(θ)+1)=sin(θ) ?

    The general solution for (cot(θ)+csc(θ))/(sec(θ)+1)=sin(θ) is θ=0.90455…+2pin,θ=2pi-0.90455…+2pin
Study ToolsAI Math SolverPopular ProblemsWorksheetsStudy GuidesPracticeCheat SheetsCalculatorsGraphing CalculatorGeometry CalculatorVerify Solution
AppsSymbolab App (Android)Graphing Calculator (Android)Practice (Android)Symbolab App (iOS)Graphing Calculator (iOS)Practice (iOS)Chrome ExtensionSymbolab Math Solver API
CompanyAbout SymbolabBlogHelp
LegalPrivacyTermsCookie PolicyCookie SettingsDo Not Sell or Share My Personal InfoCopyright, Community Guidelines, DSA & other Legal ResourcesLearneo Legal Center
Social Media
Symbolab, a Learneo, Inc. business
© Learneo, Inc. 2024