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Popular Trigonometry >

((1-tan^2(a)))/((tan(a)))=2cot^2(a)

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Solution

(tan(a))(1−tan2(a))​=2cot2(a)

Solution

a=2.15228…+πn
+1
Degrees
a=123.31684…∘+180∘n
Solution steps
(tan(a))(1−tan2(a))​=2cot2(a)
Subtract 2cot2(a) from both sidestan(a)1−tan2(a)​−2cot2(a)=0
Simplify tan(a)1−tan2(a)​−2cot2(a):tan(a)1−tan2(a)−2cot2(a)tan(a)​
tan(a)1−tan2(a)​−2cot2(a)
Convert element to fraction: 2cot2(a)=tan(a)2cot2(a)tan(a)​=tan(a)1−tan2(a)​−tan(a)2cot2(a)tan(a)​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=tan(a)1−tan2(a)−2cot2(a)tan(a)​
tan(a)1−tan2(a)−2cot2(a)tan(a)​=0
g(x)f(x)​=0⇒f(x)=01−tan2(a)−2cot2(a)tan(a)=0
Rewrite using trig identities
1−tan2(a)−2cot2(a)tan(a)
Use the basic trigonometric identity: tan(x)=cot(x)1​=1−(cot(a)1​)2−2cot2(a)cot(a)1​
Simplify 1−(cot(a)1​)2−2cot2(a)cot(a)1​:1−cot2(a)1​−2cot(a)
1−(cot(a)1​)2−2cot2(a)cot(a)1​
(cot(a)1​)2=cot2(a)1​
(cot(a)1​)2
Apply exponent rule: (ba​)c=bcac​=cot2(a)12​
Apply rule 1a=112=1=cot2(a)1​
2cot2(a)cot(a)1​=2cot(a)
2cot2(a)cot(a)1​
Multiply fractions: a⋅cb​=ca⋅b​=cot(a)1⋅2cot2(a)​
Multiply the numbers: 1⋅2=2=cot(a)2cot2(a)​
Cancel the common factor: cot(a)=2cot(a)
=1−cot2(a)1​−2cot(a)
=1−cot2(a)1​−2cot(a)
1−cot2(a)1​−2cot(a)=0
Solve by substitution
1−cot2(a)1​−2cot(a)=0
Let: cot(a)=u1−u21​−2u=0
1−u21​−2u=0:u≈−0.65729…
1−u21​−2u=0
Multiply both sides by u2
1−u21​−2u=0
Multiply both sides by u21⋅u2−u21​u2−2uu2=0⋅u2
Simplify
1⋅u2−u21​u2−2uu2=0⋅u2
Simplify 1⋅u2:u2
1⋅u2
Multiply: 1⋅u2=u2=u2
Simplify −u21​u2:−1
−u21​u2
Multiply fractions: a⋅cb​=ca⋅b​=−u21⋅u2​
Cancel the common factor: u2=−1
Simplify −2uu2:−2u3
−2uu2
Apply exponent rule: ab⋅ac=ab+cuu2=u1+2=−2u1+2
Add the numbers: 1+2=3=−2u3
Simplify 0⋅u2:0
0⋅u2
Apply rule 0⋅a=0=0
u2−1−2u3=0
u2−1−2u3=0
u2−1−2u3=0
Solve u2−1−2u3=0:u≈−0.65729…
u2−1−2u3=0
Write in the standard form an​xn+…+a1​x+a=0−2u3+u2−1=0
Find one solution for −2u3+u2−1=0 using Newton-Raphson:u≈−0.65729…
−2u3+u2−1=0
Newton-Raphson Approximation Definition
f(u)=−2u3+u2−1
Find f′(u):−6u2+2u
dud​(−2u3+u2−1)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=−dud​(2u3)+dud​(u2)−dud​(1)
dud​(2u3)=6u2
dud​(2u3)
Take the constant out: (a⋅f)′=a⋅f′=2dud​(u3)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=2⋅3u3−1
Simplify=6u2
dud​(u2)=2u
dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=2u2−1
Simplify=2u
dud​(1)=0
dud​(1)
Derivative of a constant: dxd​(a)=0=0
=−6u2+2u−0
Simplify=−6u2+2u
Let u0​=−1Compute un+1​ until Δun+1​<0.000001
u1​=−0.75:Δu1​=0.25
f(u0​)=−2(−1)3+(−1)2−1=2f′(u0​)=−6(−1)2+2(−1)=−8u1​=−0.75
Δu1​=∣−0.75−(−1)∣=0.25Δu1​=0.25
u2​=−0.66666…:Δu2​=0.08333…
f(u1​)=−2(−0.75)3+(−0.75)2−1=0.40625f′(u1​)=−6(−0.75)2+2(−0.75)=−4.875u2​=−0.66666…
Δu2​=∣−0.66666…−(−0.75)∣=0.08333…Δu2​=0.08333…
u3​=−0.65740…:Δu3​=0.00925…
f(u2​)=−2(−0.66666…)3+(−0.66666…)2−1=0.03703…f′(u2​)=−6(−0.66666…)2+2(−0.66666…)=−4u3​=−0.65740…
Δu3​=∣−0.65740…−(−0.66666…)∣=0.00925…Δu3​=0.00925…
u4​=−0.65729…:Δu4​=0.00010…
f(u3​)=−2(−0.65740…)3+(−0.65740…)2−1=0.00042…f′(u3​)=−6(−0.65740…)2+2(−0.65740…)=−3.90792…u4​=−0.65729…
Δu4​=∣−0.65729…−(−0.65740…)∣=0.00010…Δu4​=0.00010…
u5​=−0.65729…:Δu5​=1.51148E−8
f(u4​)=−2(−0.65729…)3+(−0.65729…)2−1=5.90512E−8f′(u4​)=−6(−0.65729…)2+2(−0.65729…)=−3.90684…u5​=−0.65729…
Δu5​=∣−0.65729…−(−0.65729…)∣=1.51148E−8Δu5​=1.51148E−8
u≈−0.65729…
Apply long division:u+0.65729…−2u3+u2−1​=−2u2+2.31459…u−1.52137…
−2u2+2.31459…u−1.52137…≈0
Find one solution for −2u2+2.31459…u−1.52137…=0 using Newton-Raphson:No Solution for u∈R
−2u2+2.31459…u−1.52137…=0
Newton-Raphson Approximation Definition
f(u)=−2u2+2.31459…u−1.52137…
Find f′(u):−4u+2.31459…
dud​(−2u2+2.31459…u−1.52137…)
Apply the Sum/Difference Rule: (f±g)′=f′±g′=−dud​(2u2)+dud​(2.31459…u)−dud​(1.52137…)
dud​(2u2)=4u
dud​(2u2)
Take the constant out: (a⋅f)′=a⋅f′=2dud​(u2)
Apply the Power Rule: dxd​(xa)=a⋅xa−1=2⋅2u2−1
Simplify=4u
dud​(2.31459…u)=2.31459…
dud​(2.31459…u)
Take the constant out: (a⋅f)′=a⋅f′=2.31459…dudu​
Apply the common derivative: dudu​=1=2.31459…⋅1
Simplify=2.31459…
dud​(1.52137…)=0
dud​(1.52137…)
Derivative of a constant: dxd​(a)=0=0
=−4u+2.31459…−0
Simplify=−4u+2.31459…
Let u0​=1Compute un+1​ until Δun+1​<0.000001
u1​=0.28397…:Δu1​=0.71602…
f(u0​)=−2⋅12+2.31459…⋅1−1.52137…=−1.20678…f′(u0​)=−4⋅1+2.31459…=−1.68540…u1​=0.28397…
Δu1​=∣0.28397…−1∣=0.71602…Δu1​=0.71602…
u2​=1.15391…:Δu2​=0.86993…
f(u1​)=−2⋅0.28397…2+2.31459…⋅0.28397…−1.52137…=−1.02537…f′(u1​)=−4⋅0.28397…+2.31459…=1.17867…u2​=1.15391…
Δu2​=∣1.15391…−0.28397…∣=0.86993…Δu2​=0.86993…
u3​=0.49614…:Δu3​=0.65777…
f(u2​)=−2⋅1.15391…2+2.31459…⋅1.15391…−1.52137…=−1.51356…f′(u2​)=−4⋅1.15391…+2.31459…=−2.30105…u3​=0.49614…
Δu3​=∣0.49614…−1.15391…∣=0.65777…Δu3​=0.65777…
u4​=3.11809…:Δu4​=2.62195…
f(u3​)=−2⋅0.49614…2+2.31459…⋅0.49614…−1.52137…=−0.86532…f′(u3​)=−4⋅0.49614…+2.31459…=0.33003…u4​=3.11809…
Δu4​=∣3.11809…−0.49614…∣=2.62195…Δu4​=2.62195…
u5​=1.76452…:Δu5​=1.35357…
f(u4​)=−2⋅3.11809…2+2.31459…⋅3.11809…−1.52137…=−13.74928…f′(u4​)=−4⋅3.11809…+2.31459…=−10.15778…u5​=1.76452…
Δu5​=∣1.76452…−3.11809…∣=1.35357…Δu5​=1.35357…
u6​=0.99203…:Δu6​=0.77249…
f(u5​)=−2⋅1.76452…2+2.31459…⋅1.76452…−1.52137…=−3.66430…f′(u5​)=−4⋅1.76452…+2.31459…=−4.74350…u6​=0.99203…
Δu6​=∣0.99203…−1.76452…∣=0.77249…Δu6​=0.77249…
u7​=0.27025…:Δu7​=0.72177…
f(u6​)=−2⋅0.99203…2+2.31459…⋅0.99203…−1.52137…=−1.19348…f′(u6​)=−4⋅0.99203…+2.31459…=−1.65353…u7​=0.27025…
Δu7​=∣0.27025…−0.99203…∣=0.72177…Δu7​=0.72177…
u8​=1.11489…:Δu8​=0.84464…
f(u7​)=−2⋅0.27025…2+2.31459…⋅0.27025…−1.52137…=−1.04192…f′(u7​)=−4⋅0.27025…+2.31459…=1.23356…u8​=1.11489…
Δu8​=∣1.11489…−0.27025…∣=0.84464…Δu8​=0.84464…
u9​=0.44970…:Δu9​=0.66519…
f(u8​)=−2⋅1.11489…2+2.31459…⋅1.11489…−1.52137…=−1.42683…f′(u8​)=−4⋅1.11489…+2.31459…=−2.14500…u9​=0.44970…
Δu9​=∣0.44970…−1.11489…∣=0.66519…Δu9​=0.66519…
u10​=2.16551…:Δu10​=1.71580…
f(u9​)=−2⋅0.44970…2+2.31459…⋅0.44970…−1.52137…=−0.88496…f′(u9​)=−4⋅0.44970…+2.31459…=0.51576…u10​=2.16551…
Δu10​=∣2.16551…−0.44970…∣=1.71580…Δu10​=1.71580…
Cannot find solution
The solution isu≈−0.65729…
u≈−0.65729…
Verify Solutions
Find undefined (singularity) points:u=0
Take the denominator(s) of 1−u21​−2u and compare to zero
Solve u2=0:u=0
u2=0
Apply rule xn=0⇒x=0
u=0
The following points are undefinedu=0
Combine undefined points with solutions:
u≈−0.65729…
Substitute back u=cot(a)cot(a)≈−0.65729…
cot(a)≈−0.65729…
cot(a)=−0.65729…:a=arccot(−0.65729…)+πn
cot(a)=−0.65729…
Apply trig inverse properties
cot(a)=−0.65729…
General solutions for cot(a)=−0.65729…cot(x)=−a⇒x=arccot(−a)+πna=arccot(−0.65729…)+πn
a=arccot(−0.65729…)+πn
Combine all the solutionsa=arccot(−0.65729…)+πn
Show solutions in decimal forma=2.15228…+πn

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