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Popular Trigonometry >

solvefor x,f=arctan(x/(sqrt(1-x^2)))

  • Pre Algebra
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Solution

solvefor

Solution

x=1+tan2(f)​tan(f)​
Solution steps
f=arctan(1−x2​x​)
Switch sidesarctan(1−x2​x​)=f
Apply trig inverse properties
arctan(1−x2​x​)=f
arctan(x)=a⇒x=tan(a)1−x2​x​=tan(f)
1−x2​x​=tan(f)
Solve 1−x2​x​=tan(f):x=1+tan2(f)​tan(f)​
1−x2​x​=tan(f)
Multiply both sides by 1−x2​1−x2​x​1−x2​=tan(f)1−x2​
Simplifyx=tan(f)1−x2​
Square both sides:x2=tan2(f)−x2tan2(f)
x=tan(f)1−x2​
x2=(tan(f)1−x2​)2
Expand (tan(f)1−x2​)2:tan2(f)−x2tan2(f)
(tan(f)1−x2​)2
Apply exponent rule: (a⋅b)n=anbn=tan2(f)(1−x2​)2
(1−x2​)2:1−x2
Apply radical rule: a​=a21​=((1−x2)21​)2
Apply exponent rule: (ab)c=abc=(1−x2)21​⋅2
21​⋅2=1
21​⋅2
Multiply fractions: a⋅cb​=ca⋅b​=21⋅2​
Cancel the common factor: 2=1
=1−x2
=tan2(f)(1−x2)
Expand (1−x2)tan2(f):tan2(f)−x2tan2(f)
(1−x2)tan2(f)
=tan2(f)(1−x2)
Apply the distributive law: a(b−c)=ab−aca=tan2(f),b=1,c=x2=tan2(f)⋅1−tan2(f)x2
=1⋅tan2(f)−x2tan2(f)
Multiply: 1⋅tan2(f)=tan2(f)=tan2(f)−x2tan2(f)
=tan2(f)−x2tan2(f)
x2=tan2(f)−x2tan2(f)
x2=tan2(f)−x2tan2(f)
Solve x2=tan2(f)−x2tan2(f):x=1+tan2(f)​tan(f)​,x=−1+tan2(f)​tan(f)​
x2=tan2(f)−x2tan2(f)
Move x2tan2(f)to the left side
x2=tan2(f)−x2tan2(f)
Add x2tan2(f) to both sidesx2+x2tan2(f)=tan2(f)−x2tan2(f)+x2tan2(f)
Simplifyx2+x2tan2(f)=tan2(f)
x2+x2tan2(f)=tan2(f)
Factor x2+x2tan2(f):x2(1+tan2(f))
x2+x2tan2(f)
Factor out common term x2=x2(1+tan2(f))
x2(1+tan2(f))=tan2(f)
Divide both sides by 1+tan2(f)
x2(1+tan2(f))=tan2(f)
Divide both sides by 1+tan2(f)1+tan2(f)x2(1+tan2(f))​=1+tan2(f)tan2(f)​
Simplifyx2=1+tan2(f)tan2(f)​
x2=1+tan2(f)tan2(f)​
For x2=f(a) the solutions are x=f(a)​,−f(a)​
x=1+tan2(f)tan2(f)​​,x=−1+tan2(f)tan2(f)​​
Simplify 1+tan2(f)tan2(f)​​:1+tan2(f)​tan(f)​
1+tan2(f)tan2(f)​​
Apply radical rule: assuming a≥0,b≥0=1+tan2(f)​tan2(f)​​
Apply radical rule: assuming a≥0tan2(f)​=tan(f)=1+tan2(f)​tan(f)​
Simplify −1+tan2(f)tan2(f)​​:−1+tan2(f)​tan(f)​
−1+tan2(f)tan2(f)​​
Simplify 1+tan2(f)tan2(f)​​:1+tan2(f)​tan(f)​
1+tan2(f)tan2(f)​​
Apply radical rule: assuming a≥0,b≥0=1+tan2(f)​tan2(f)​​
Apply radical rule: assuming a≥0tan2(f)​=tan(f)=1+tan2(f)​tan(f)​
=−tan2(f)+1​tan(f)​
=−1+tan2(f)​tan(f)​
x=1+tan2(f)​tan(f)​,x=−1+tan2(f)​tan(f)​
x=1+tan2(f)​tan(f)​,x=−1+tan2(f)​tan(f)​
Verify Solutions:x=1+tan2(f)​tan(f)​True,x=−1+tan2(f)​tan(f)​False
Check the solutions by plugging them into 1−x2​x​=tan(f)
Remove the ones that don't agree with the equation.
Plug in x=1+tan2(f)​tan(f)​:True
1−(1+tan2(f)​tan(f)​)2​(1+tan2(f)​tan(f)​)​=tan(f)
Simplify 1−(1+tan2(f)​tan(f)​)2​(1+tan2(f)​tan(f)​)​:tan(f)
1−(1+tan2(f)​tan(f)​)2​1+tan2(f)​tan(f)​​
Apply the fraction rule: acb​​=c⋅ab​=1+tan2(f)​1−(1+tan2(f)​tan(f)​)2​tan(f)​
1−(1+tan2(f)​tan(f)​)2​=1+tan2(f)​1​
1−(1+tan2(f)​tan(f)​)2​
(1+tan2(f)​tan(f)​)2=1+tan2(f)tan2(f)​
(1+tan2(f)​tan(f)​)2
Apply exponent rule: (ba​)c=bcac​=(1+tan2(f)​)2tan2(f)​
(1+tan2(f)​)2:1+tan2(f)
Apply radical rule: a​=a21​=((1+tan2(f))21​)2
Apply exponent rule: (ab)c=abc=(1+tan2(f))21​⋅2
21​⋅2=1
21​⋅2
Multiply fractions: a⋅cb​=ca⋅b​=21⋅2​
Cancel the common factor: 2=1
=1+tan2(f)
=1+tan2(f)tan2(f)​
=1−tan2(f)+1tan2(f)​​
Join 1−1+tan2(f)tan2(f)​:1+tan2(f)1​
1−1+tan2(f)tan2(f)​
Convert element to fraction: 1=1+tan2(f)1(1+tan2(f))​=1+tan2(f)1⋅(1+tan2(f))​−1+tan2(f)tan2(f)​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=1+tan2(f)1⋅(1+tan2(f))−tan2(f)​
1⋅(1+tan2(f))−tan2(f)=1
1⋅(1+tan2(f))−tan2(f)
1⋅(1+tan2(f))=1+tan2(f)
1⋅(1+tan2(f))
Multiply: 1⋅(1+tan2(f))=(1+tan2(f))=1+tan2(f)
Remove parentheses: (a)=a=1+tan2(f)
=1+tan2(f)−tan2(f)
Add similar elements: tan2(f)−tan2(f)=0=1
=1+tan2(f)1​
=1+tan2(f)1​​
Apply radical rule: assuming a≥0,b≥0=1+tan2(f)​1​​
Apply rule 1​=1=1+tan2(f)​1​
=tan2(f)+1​1​tan2(f)+1​tan(f)​
Multiply 1+tan2(f)​1+tan2(f)​1​:1
1+tan2(f)​1+tan2(f)​1​
Multiply fractions: a⋅cb​=ca⋅b​=1+tan2(f)​1⋅1+tan2(f)​​
Cancel the common factor: 1+tan2(f)​=1
=1tan(f)​
Apply rule 1a​=a=tan(f)
tan(f)=tan(f)
True
Plugx=−1+tan2(f)​tan(f)​:1−(−1+tan2(f)​tan(f)​)2​−1+tan2(f)​tan(f)​​=tan(f)⇒False
1−(−1+tan2(f)​tan(f)​)2​(−1+tan2(f)​tan(f)​)​=tan(f)
Solve by substitution
1−(−1+tan2(f)​tan(f)​)2​−1+tan2(f)​tan(f)​​=tan(f)
Let: tan(f)=u1−(−1+u2​u​)2​−1+u2​u​​=u
1−(−1+u2​u​)2​−1+u2​u​​=u:True for all u
1−(−1+u2​u​)2​−1+u2​u​​=u
Multiply both sides by 1−(1+u2​u​)2​1−(−1+u2​u​)2​−1+u2​u​​1−(1+u2​u​)2​=u1−(1+u2​u​)2​
Simplify−1+u2​1−(−1+u2​u​)2​1−(1+u2​u​)2​u​=u1−(1+u2​u​)2​
Square both sides:1+u2u2​=1+u2u2​
−1+u2​1−(−1+u2​u​)2​1−(1+u2​u​)2​u​=u1−(1+u2​u​)2​
​−1+u2​1−(−1+u2​u​)2​1−(1+u2​u​)2​u​​2=​u1−(1+u2​u​)2​​2
Expand ​−1+u2​1−(−1+u2​u​)2​1−(1+u2​u​)2​u​​2:1+u2u2​
​−1+u2​1−(−1+u2​u​)2​1−(1+u2​u​)2​u​​2
Apply exponent rule: (−a)n=an,if n is even​−1+u2​1−(−1+u2​u​)2​1−(1+u2​u​)2​u​​2=​1+u2​1−(−1+u2​u​)2​1−(1+u2​u​)2​u​​2=​1+u2​1−(−1+u2​u​)2​1−(1+u2​u​)2​u​​2
Apply exponent rule: (ba​)c=bcac​=(1+u2​1−(−1+u2​u​)2​)2(1−(1+u2​u​)2​u)2​
Apply exponent rule: (a⋅b)n=anbn​1+u2​1−(−1+u2​u​)2​​2=(1+u2​)2​1−(−1+u2​u​)2​​2=(1+u2​)2(1−(−1+u2​u​)2​)2(1−(1+u2​u​)2​u)2​
Apply exponent rule: (a⋅b)n=anbn​1−(1+u2​u​)2​u​2=u2​1−(1+u2​u​)2​​2=(1+u2​)2(1−(−1+u2​u​)2​)2u2(1−(1+u2​u​)2​)2​
(1+u2​)2:1+u2
Apply radical rule: a​=a21​=((1+u2)21​)2
Apply exponent rule: (ab)c=abc=(1+u2)21​⋅2
21​⋅2=1
21​⋅2
Multiply fractions: a⋅cb​=ca⋅b​=21⋅2​
Cancel the common factor: 2=1
=1+u2
=(1+u2)(1−(−1+u2​u​)2​)2(1−(1+u2​u​)2​)2u2​
​1−(1+u2​u​)2​​2:1−(1+u2​u​)2
Apply radical rule: a​=a21​=​(1−(1+u2​u​)2)21​​2
Apply exponent rule: (ab)c=abc=(1−(1+u2​u​)2)21​⋅2
21​⋅2=1
21​⋅2
Multiply fractions: a⋅cb​=ca⋅b​=21⋅2​
Cancel the common factor: 2=1
=1−(1+u2​u​)2
=(1+u2)(1−(−1+u2​u​)2​)2(1−(1+u2​u​)2)u2​
​1−(−1+u2​u​)2​​2:1−(−1+u2​u​)2
Apply radical rule: a​=a21​=​(1−(−1+u2​u​)2)21​​2
Apply exponent rule: (ab)c=abc=(1−(−1+u2​u​)2)21​⋅2
21​⋅2=1
21​⋅2
Multiply fractions: a⋅cb​=ca⋅b​=21⋅2​
Cancel the common factor: 2=1
=1−(−1+u2​u​)2
=(1+u2)(1−(−1+u2​u​)2)(1−(1+u2​u​)2)u2​
Expand (1+u2)(1−(−1+u2​u​)2)(1−(1+u2​u​)2)u2​:1+u2u2​
(1+u2)(1−(−1+u2​u​)2)(1−(1+u2​u​)2)u2​
(1+u2)(1−(−1+u2​u​)2)=(1+u2)(1−(1+u2​u​)2)
(1+u2)(1−(−1+u2​u​)2)
Apply exponent rule: (−a)n=an,if n is even(−u2+1​u​)2=(1+u2​u​)2=(u2+1)(−(u2+1​u​)2+1)
=(u2+1)(−(u2+1​u​)2+1)u2(−(u2+1​u​)2+1)​
Cancel the common factor: 1−(1+u2​u​)2=1+u2u2​
=1+u2u2​
Expand ​u1−(1+u2​u​)2​​2:1+u2u2​
​u1−(1+u2​u​)2​​2
Apply exponent rule: (a⋅b)n=anbn=u2​1−(1+u2​u​)2​​2
​1−(1+u2​u​)2​​2:1−(1+u2​u​)2
Apply radical rule: a​=a21​=​(1−(1+u2​u​)2)21​​2
Apply exponent rule: (ab)c=abc=(1−(1+u2​u​)2)21​⋅2
21​⋅2=1
21​⋅2
Multiply fractions: a⋅cb​=ca⋅b​=21⋅2​
Cancel the common factor: 2=1
=1−(1+u2​u​)2
=u2(1−(1+u2​u​)2)
Expand (1−(1+u2​u​)2)u2:1+u2u2​
(1−(1+u2​u​)2)u2
(1+u2​u​)2=1+u2u2​
(1+u2​u​)2
Apply exponent rule: (ba​)c=bcac​=(1+u2​)2u2​
(1+u2​)2:1+u2
Apply radical rule: a​=a21​=((1+u2)21​)2
Apply exponent rule: (ab)c=abc=(1+u2)21​⋅2
21​⋅2=1
21​⋅2
Multiply fractions: a⋅cb​=ca⋅b​=21⋅2​
Cancel the common factor: 2=1
=1+u2
=1+u2u2​
=u2(−u2+1u2​+1)
=u2(1−1+u2u2​)
Apply the distributive law: a(b−c)=ab−aca=u2,b=1,c=1+u2u2​=u2⋅1−u21+u2u2​
=1⋅u2−1+u2u2​u2
1⋅u2=u2
1⋅u2
Multiply: 1⋅u2=u2=u2
1+u2u2​u2=1+u2u4​
1+u2u2​u2
Multiply fractions: a⋅cb​=ca⋅b​=1+u2u2u2​
u2u2=u4
u2u2
Apply exponent rule: ab⋅ac=ab+cu2u2=u2+2=u2+2
Add the numbers: 2+2=4=u4
=1+u2u4​
=u2−u2+1u4​
Convert element to fraction: u2=1+u2u2(1+u2)​=−1+u2u4​+1+u2u2(1+u2)​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=1+u2−u4+u2(1+u2)​
Expand −u4+u2(1+u2):u2
−u4+u2(1+u2)
Expand u2(1+u2):u2+u4
u2(1+u2)
Apply the distributive law: a(b+c)=ab+aca=u2,b=1,c=u2=u2⋅1+u2u2
=1⋅u2+u2u2
Simplify 1⋅u2+u2u2:u2+u4
1⋅u2+u2u2
1⋅u2=u2
1⋅u2
Multiply: 1⋅u2=u2=u2
u2u2=u4
u2u2
Apply exponent rule: ab⋅ac=ab+cu2u2=u2+2=u2+2
Add the numbers: 2+2=4=u4
=u2+u4
=u2+u4
=−u4+u2+u4
Simplify −u4+u2+u4:u2
−u4+u2+u4
Group like terms=−u4+u4+u2
Add similar elements: −u4+u4=0=u2
=u2
=1+u2u2​
=1+u2u2​
1+u2u2​=1+u2u2​
1+u2u2​=1+u2u2​
Solve 1+u2u2​=1+u2u2​:True for all u
1+u2u2​=1+u2u2​
Subtract 1+u2u2​ from both sides1+u2u2​−1+u2u2​=1+u2u2​−1+u2u2​
Simplify0=0
Both sides are equal
Trueforallu
Trueforallu
Substitute back u=tan(f)Trueforalltan(f)
Trueforalltan(f)
tan(f)=True for all u∈R:f=arctan(Trueforallu∈R)+πn
tan(f)=Trueforallu∈R
Apply trig inverse properties
tan(f)=Trueforallu∈R
General solutions for tan(f)=True for all u∈Rtan(x)=a⇒x=arctan(a)+πnf=arctan(Trueforallu∈R)+πn
f=arctan(Trueforallu∈R)+πn
Combine all the solutionsf=arctan(Trueforallu∈R)+πn
Since the equation is undefined for:arctan(Trueforallu∈R)+πnNoSolutionforf∈R
The solution isx=1+tan2(f)​tan(f)​
x=1+tan2(f)​tan(f)​

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Frequently Asked Questions (FAQ)

  • What is the general solution for solvefor x,f=arctan(x/(sqrt(1-x^2))) ?

    The general solution for solvefor x,f=arctan(x/(sqrt(1-x^2))) is x=(tan(f))/(sqrt(1+tan^2(f)))
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