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Popular Trigonometry >

(cot^2(x))/(1+sin(x))=(csc(x)-1)/(sin^2(x))

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Solution

1+sin(x)cot2(x)​=sin2(x)csc(x)−1​

Solution

x=2π​+2πn
+1
Degrees
x=90∘+360∘n
Solution steps
1+sin(x)cot2(x)​=sin2(x)csc(x)−1​
Subtract sin2(x)csc(x)−1​ from both sides1+sin(x)cot2(x)​−sin2(x)csc(x)−1​=0
Simplify 1+sin(x)cot2(x)​−sin2(x)csc(x)−1​:sin2(x)(sin(x)+1)cot2(x)sin2(x)−(csc(x)−1)(sin(x)+1)​
1+sin(x)cot2(x)​−sin2(x)csc(x)−1​
Least Common Multiplier of 1+sin(x),sin2(x):sin2(x)(sin(x)+1)
1+sin(x),sin2(x)
Lowest Common Multiplier (LCM)
Compute an expression comprised of factors that appear either in 1+sin(x) or sin2(x)=sin2(x)(sin(x)+1)
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 sin2(x)(sin(x)+1)
For 1+sin(x)cot2(x)​:multiply the denominator and numerator by sin2(x)1+sin(x)cot2(x)​=(1+sin(x))sin2(x)cot2(x)sin2(x)​
For sin2(x)csc(x)−1​:multiply the denominator and numerator by sin(x)+1sin2(x)csc(x)−1​=sin2(x)(sin(x)+1)(csc(x)−1)(sin(x)+1)​
=(1+sin(x))sin2(x)cot2(x)sin2(x)​−sin2(x)(sin(x)+1)(csc(x)−1)(sin(x)+1)​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=sin2(x)(sin(x)+1)cot2(x)sin2(x)−(csc(x)−1)(sin(x)+1)​
sin2(x)(sin(x)+1)cot2(x)sin2(x)−(csc(x)−1)(sin(x)+1)​=0
g(x)f(x)​=0⇒f(x)=0cot2(x)sin2(x)−(csc(x)−1)(sin(x)+1)=0
Rewrite using trig identities
−(−1+csc(x))(1+sin(x))+cot2(x)sin2(x)
Use the Pythagorean identity: 1+cot2(x)=csc2(x)cot2(x)=csc2(x)−1=−(−1+csc(x))(1+sin(x))+(csc2(x)−1)sin2(x)
−(−1+csc(x))(1+sin(x))+(−1+csc2(x))sin2(x)=0
Factor −(−1+csc(x))(1+sin(x))+(−1+csc2(x))sin2(x):(−1+csc(x))(−1−sin(x)+sin2(x)+sin2(x)csc(x))
−(−1+csc(x))(1+sin(x))+(−1+csc2(x))sin2(x)
Factor −1+csc2(x):(csc(x)+1)(csc(x)−1)
−1+csc2(x)
Rewrite 1 as 12=csc2(x)−12
Apply Difference of Two Squares Formula: x2−y2=(x+y)(x−y)csc2(x)−12=(csc(x)+1)(csc(x)−1)=(csc(x)+1)(csc(x)−1)
=−(csc(x)−1)(sin(x)+1)+sin2(x)(csc(x)+1)(csc(x)−1)
Factor out common term (−1+csc(x))=(−1+csc(x))(−(1+sin(x))+(1+csc(x))sin2(x))
Expand sin2(x)(csc(x)+1)−(sin(x)+1):−1−sin(x)+sin2(x)+sin2(x)csc(x)
−(1+sin(x))+(1+csc(x))sin2(x)
=−(1+sin(x))+sin2(x)(1+csc(x))
−(1+sin(x)):−1−sin(x)
−(1+sin(x))
Distribute parentheses=−(1)−(sin(x))
Apply minus-plus rules+(−a)=−a=−1−sin(x)
=−1−sin(x)+(1+csc(x))sin2(x)
Expand sin2(x)(1+csc(x)):sin2(x)+sin2(x)csc(x)
sin2(x)(1+csc(x))
Apply the distributive law: a(b+c)=ab+aca=sin2(x),b=1,c=csc(x)=sin2(x)⋅1+sin2(x)csc(x)
=1⋅sin2(x)+sin2(x)csc(x)
Multiply: 1⋅sin2(x)=sin2(x)=sin2(x)+sin2(x)csc(x)
=−1−sin(x)+sin2(x)+sin2(x)csc(x)
=(csc(x)−1)(sin2(x)+sin2(x)csc(x)−sin(x)−1)
(−1+csc(x))(−1−sin(x)+sin2(x)+sin2(x)csc(x))=0
Solving each part separately−1+csc(x)=0or−1−sin(x)+sin2(x)+sin2(x)csc(x)=0
−1+csc(x)=0:x=2π​+2πn
−1+csc(x)=0
Move 1to the right side
−1+csc(x)=0
Add 1 to both sides−1+csc(x)+1=0+1
Simplifycsc(x)=1
csc(x)=1
General solutions for csc(x)=1
csc(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​csc(x)Undefiend22​323​​1323​​2​2​xπ67π​45π​34π​23π​35π​47π​611π​​csc(x)Undefiend−2−2​−323​​−1−323​​−2​−2​​
x=2π​+2πn
x=2π​+2πn
−1−sin(x)+sin2(x)+sin2(x)csc(x)=0:x=2π​+2πn,x=23π​+2πn
−1−sin(x)+sin2(x)+sin2(x)csc(x)=0
Rewrite using trig identities
−1−sin(x)+sin2(x)+csc(x)sin2(x)
Use the basic trigonometric identity: csc(x)=sin(x)1​=−1−sin(x)+sin2(x)+sin(x)1​sin2(x)
Simplify −1−sin(x)+sin2(x)+sin(x)1​sin2(x):sin2(x)−1
−1−sin(x)+sin2(x)+sin(x)1​sin2(x)
sin(x)1​sin2(x)=sin(x)
sin(x)1​sin2(x)
Multiply fractions: a⋅cb​=ca⋅b​=sin(x)1⋅sin2(x)​
Multiply: 1⋅sin2(x)=sin2(x)=sin(x)sin2(x)​
Cancel the common factor: sin(x)=sin(x)
=−1−sin(x)+sin2(x)+sin(x)
Group like terms=−sin(x)+sin2(x)+sin(x)−1
Add similar elements: −sin(x)+sin(x)=0=sin2(x)−1
=sin2(x)−1
−1+sin2(x)=0
Solve by substitution
−1+sin2(x)=0
Let: sin(x)=u−1+u2=0
−1+u2=0:u=1,u=−1
−1+u2=0
Move 1to the right side
−1+u2=0
Add 1 to both sides−1+u2+1=0+1
Simplifyu2=1
u2=1
For x2=f(a) the solutions are x=f(a)​,−f(a)​
u=1​,u=−1​
1​=1
1​
Apply rule 1​=1=1
−1​=−1
−1​
Apply rule 1​=1=−1
u=1,u=−1
Substitute back u=sin(x)sin(x)=1,sin(x)=−1
sin(x)=1,sin(x)=−1
sin(x)=1:x=2π​+2πn
sin(x)=1
General solutions for sin(x)=1
sin(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​sin(x)021​22​​23​​123​​22​​21​​xπ67π​45π​34π​23π​35π​47π​611π​​sin(x)0−21​−22​​−23​​−1−23​​−22​​−21​​​
x=2π​+2πn
x=2π​+2πn
sin(x)=−1:x=23π​+2πn
sin(x)=−1
General solutions for sin(x)=−1
sin(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​sin(x)021​22​​23​​123​​22​​21​​xπ67π​45π​34π​23π​35π​47π​611π​​sin(x)0−21​−22​​−23​​−1−23​​−22​​−21​​​
x=23π​+2πn
x=23π​+2πn
Combine all the solutionsx=2π​+2πn,x=23π​+2πn
Combine all the solutionsx=2π​+2πn,x=23π​+2πn
Since the equation is undefined for:23π​+2πnx=2π​+2πn

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Frequently Asked Questions (FAQ)

  • What is the general solution for (cot^2(x))/(1+sin(x))=(csc(x)-1)/(sin^2(x)) ?

    The general solution for (cot^2(x))/(1+sin(x))=(csc(x)-1)/(sin^2(x)) is x= pi/2+2pin
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