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Popular Trigonometry >

(csc(b))/(csc(b))= 1/(sqrt(1-sin^2(b)))

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Solution

csc(b)csc(b)​=1−sin2(b)​1​

Solution

NoSolutionforb∈R
Solution steps
csc(b)csc(b)​=1−sin2(b)​1​
csc(b)csc(b)​=1
csc(b)csc(b)​
Apply rule aa​=1=1
1=1−sin2(b)​1​
Multiply both sides by 1−sin2(b)​1⋅1−sin2(b)​=1−sin2(b)​1​1−sin2(b)​
Simplify1−sin2(b)​=1
Square both sides:1−sin2(b)=1
1−sin2(b)​=1
(1−sin2(b)​)2=12
Expand (1−sin2(b)​)2:1−sin2(b)
(1−sin2(b)​)2
Apply radical rule: a​=a21​=((1−sin2(b))21​)2
Apply exponent rule: (ab)c=abc=(1−sin2(b))21​⋅2
21​⋅2=1
21​⋅2
Multiply fractions: a⋅cb​=ca⋅b​=21⋅2​
Cancel the common factor: 2=1
=1−sin2(b)
Expand 12:1
12
Apply rule 1a=1=1
1−sin2(b)=1
1−sin2(b)=1
Solve 1−sin2(b)=1:sin(b)=0
1−sin2(b)=1
Move 1to the right side
1−sin2(b)=1
Subtract 1 from both sides1−sin2(b)−1=1−1
Simplify−sin2(b)=0
−sin2(b)=0
Divide both sides by −1
−sin2(b)=0
Divide both sides by −1
−sin2(b)=0
Divide both sides by −1−1−sin2(b)​=−10​
Simplifysin2(b)=0
sin2(b)=0
Apply rule xn=0⇒x=0
sin(b)=0
sin(b)=0
Verify Solutions:sin(b)=0True
Check the solutions by plugging them into 1=1−sin2(b)​1​
Remove the ones that don't agree with the equation.
Plug in sin(b)=0:True
1=1−02​1​
1−02​1​=1
1−02​1​
Apply rule 0a=002=0=1−0​1​
1−0​=1
1−0​
Subtract the numbers: 1−0=1=1​
Apply rule 1​=1=1
=11​
Apply rule 1a​=a=1
1=1
True
The solution issin(b)=0
General solutions for sin(b)=0
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​​​
b=0+2πn,b=π+2πn
b=0+2πn,b=π+2πn
Solve b=0+2πn:b=2πn
b=0+2πn
0+2πn=2πnb=2πn
b=2πn,b=π+2πn
Since the equation is undefined for:2πn,π+2πnNoSolutionforb∈R

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sqrt(3)sec(4x)=23​sec(4x)=2sqrt(2)cos(x)-1=cos(2x)2​cos(x)−1=cos(2x)0=4cos(2θ)0=4cos(2θ)tan^2(x)cos(x)=tan^2(x)tan2(x)cos(x)=tan2(x)cos(x)sin(x)=3sin(x)cos(x)sin(x)=3sin(x)

Frequently Asked Questions (FAQ)

  • What is the general solution for (csc(b))/(csc(b))= 1/(sqrt(1-sin^2(b))) ?

    The general solution for (csc(b))/(csc(b))= 1/(sqrt(1-sin^2(b))) is No Solution for b\in\mathbb{R}
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