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Popular Trigonometry >

tan(t)-tan^2(2)+sec^3(t)>0

  • Pre Algebra
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Solution

tan(t)−tan2(2)+sec3(t)>0

Solution

Falseforallt∈R
Solution steps
tan(t)−tan2(2)+sec3(t)>0
Periodicity of tan(t)−tan2(2)+sec3(t):2π
The compound periodicity of the sum of periodic functions is the least common multiplier of the periodstan(t),sec3(t)
Periodicity of tan(t):π
Periodicity of tan(x)is π=π
Periodicity of sec3(t):2π
Periodicity of secn(x)=Periodicity of sec(x),if n is odd
Periodicity of sec(t):2π
Periodicity of sec(x)is 2π=2π
2π
Combine periods: π,2π
=2π
Express with sin, cos
tan(t)−tan2(2)+sec3(t)>0
Use the basic trigonometric identity: tan(x)=cos(x)sin(x)​cos(t)sin(t)​−tan2(2)+sec3(t)>0
Use the basic trigonometric identity: tan(x)=cos(x)sin(x)​cos(t)sin(t)​−(cos(2)sin(2)​)2+sec3(t)>0
Use the basic trigonometric identity: sec(x)=cos(x)1​cos(t)sin(t)​−(cos(2)sin(2)​)2+(cos(t)1​)3>0
cos(t)sin(t)​−(cos(2)sin(2)​)2+(cos(t)1​)3>0
Simplify cos(t)sin(t)​−(cos(2)sin(2)​)2+(cos(t)1​)3:cos2(2)cos3(t)cos2(2)cos2(t)sin(t)−sin2(2)cos3(t)+cos2(2)​
cos(t)sin(t)​−(cos(2)sin(2)​)2+(cos(t)1​)3
(cos(2)sin(2)​)2=cos2(2)sin2(2)​
(cos(2)sin(2)​)2
Apply exponent rule: (ba​)c=bcac​=cos2(2)sin2(2)​
(cos(t)1​)3=cos3(t)1​
(cos(t)1​)3
Apply exponent rule: (ba​)c=bcac​=cos3(t)13​
Apply rule 1a=113=1=cos3(t)1​
=cos(t)sin(t)​−cos2(2)sin2(2)​+cos3(t)1​
Least Common Multiplier of cos(t),cos2(2),cos3(t):cos2(2)cos3(t)
cos(t),cos2(2),cos3(t)
Lowest Common Multiplier (LCM)
Compute an expression comprised of factors that appear in at least one of the factored expressions=cos2(2)cos3(t)
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 cos2(2)cos3(t)
For cos(t)sin(t)​:multiply the denominator and numerator by cos2(2)cos2(t)cos(t)sin(t)​=cos(t)cos2(2)cos2(t)sin(t)cos2(2)cos2(t)​=cos2(2)cos3(t)sin(t)cos2(2)cos2(t)​
For cos2(2)sin2(2)​:multiply the denominator and numerator by cos3(t)cos2(2)sin2(2)​=cos2(2)cos3(t)sin2(2)cos3(t)​
For cos3(t)1​:multiply the denominator and numerator by cos2(2)cos3(t)1​=cos3(t)cos2(2)1⋅cos2(2)​=cos2(2)cos3(t)cos2(2)​
=cos2(2)cos3(t)sin(t)cos2(2)cos2(t)​−cos2(2)cos3(t)sin2(2)cos3(t)​+cos2(2)cos3(t)cos2(2)​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=cos2(2)cos3(t)sin(t)cos2(2)cos2(t)−sin2(2)cos3(t)+cos2(2)​
=NoSolutionfort∈R
Falseforallt∈R

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