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Algebra Examples
tan(θ)+cot(θ)=sec(θ)csc(θ)tan(θ)+cot(θ)=sec(θ)csc(θ)
Step 1
Step 1.1
Simplify each term.
Step 1.1.1
Rewrite tan(θ)tan(θ) in terms of sines and cosines.
sin(θ)cos(θ)+cot(θ)=sec(θ)csc(θ)sin(θ)cos(θ)+cot(θ)=sec(θ)csc(θ)
Step 1.1.2
Rewrite cot(θ)cot(θ) in terms of sines and cosines.
sin(θ)cos(θ)+cos(θ)sin(θ)=sec(θ)csc(θ)sin(θ)cos(θ)+cos(θ)sin(θ)=sec(θ)csc(θ)
sin(θ)cos(θ)+cos(θ)sin(θ)=sec(θ)csc(θ)sin(θ)cos(θ)+cos(θ)sin(θ)=sec(θ)csc(θ)
sin(θ)cos(θ)+cos(θ)sin(θ)=sec(θ)csc(θ)sin(θ)cos(θ)+cos(θ)sin(θ)=sec(θ)csc(θ)
Step 2
Step 2.1
Simplify sec(θ)csc(θ)sec(θ)csc(θ).
Step 2.1.1
Rewrite sec(θ)sec(θ) in terms of sines and cosines.
sin(θ)cos(θ)+cos(θ)sin(θ)=1cos(θ)csc(θ)sin(θ)cos(θ)+cos(θ)sin(θ)=1cos(θ)csc(θ)
Step 2.1.2
Rewrite csc(θ)csc(θ) in terms of sines and cosines.
sin(θ)cos(θ)+cos(θ)sin(θ)=1cos(θ)⋅1sin(θ)sin(θ)cos(θ)+cos(θ)sin(θ)=1cos(θ)⋅1sin(θ)
Step 2.1.3
Multiply 1cos(θ)1cos(θ) by 1sin(θ)1sin(θ).
sin(θ)cos(θ)+cos(θ)sin(θ)=1cos(θ)sin(θ)sin(θ)cos(θ)+cos(θ)sin(θ)=1cos(θ)sin(θ)
sin(θ)cos(θ)+cos(θ)sin(θ)=1cos(θ)sin(θ)sin(θ)cos(θ)+cos(θ)sin(θ)=1cos(θ)sin(θ)
sin(θ)cos(θ)+cos(θ)sin(θ)=1cos(θ)sin(θ)sin(θ)cos(θ)+cos(θ)sin(θ)=1cos(θ)sin(θ)
Step 3
Multiply both sides of the equation by cos(θ)cos(θ).
cos(θ)(sin(θ)cos(θ)+cos(θ)sin(θ))=cos(θ)1cos(θ)sin(θ)cos(θ)(sin(θ)cos(θ)+cos(θ)sin(θ))=cos(θ)1cos(θ)sin(θ)
Step 4
Apply the distributive property.
cos(θ)sin(θ)cos(θ)+cos(θ)cos(θ)sin(θ)=cos(θ)1cos(θ)sin(θ)cos(θ)sin(θ)cos(θ)+cos(θ)cos(θ)sin(θ)=cos(θ)1cos(θ)sin(θ)
Step 5
Step 5.1
Cancel the common factor.
cos(θ)sin(θ)cos(θ)+cos(θ)cos(θ)sin(θ)=cos(θ)1cos(θ)sin(θ)
Step 5.2
Rewrite the expression.
sin(θ)+cos(θ)cos(θ)sin(θ)=cos(θ)1cos(θ)sin(θ)
sin(θ)+cos(θ)cos(θ)sin(θ)=cos(θ)1cos(θ)sin(θ)
Step 6
Step 6.1
Combine cos(θ) and cos(θ)sin(θ).
sin(θ)+cos(θ)cos(θ)sin(θ)=cos(θ)1cos(θ)sin(θ)
Step 6.2
Raise cos(θ) to the power of 1.
sin(θ)+cos1(θ)cos(θ)sin(θ)=cos(θ)1cos(θ)sin(θ)
Step 6.3
Raise cos(θ) to the power of 1.
sin(θ)+cos1(θ)cos1(θ)sin(θ)=cos(θ)1cos(θ)sin(θ)
Step 6.4
Use the power rule aman=am+n to combine exponents.
sin(θ)+cos(θ)1+1sin(θ)=cos(θ)1cos(θ)sin(θ)
Step 6.5
Add 1 and 1.
sin(θ)+cos2(θ)sin(θ)=cos(θ)1cos(θ)sin(θ)
sin(θ)+cos2(θ)sin(θ)=cos(θ)1cos(θ)sin(θ)
Step 7
Step 7.1
Factor cos(θ) out of cos(θ)sin(θ).
sin(θ)+cos2(θ)sin(θ)=cos(θ)1cos(θ)(sin(θ))
Step 7.2
Cancel the common factor.
sin(θ)+cos2(θ)sin(θ)=cos(θ)1cos(θ)sin(θ)
Step 7.3
Rewrite the expression.
sin(θ)+cos2(θ)sin(θ)=1sin(θ)
sin(θ)+cos2(θ)sin(θ)=1sin(θ)
Step 8
Subtract 1sin(θ) from both sides of the equation.
sin(θ)+cos2(θ)sin(θ)-1sin(θ)=0
Step 9
Step 9.1
Combine the numerators over the common denominator.
sin(θ)+cos2(θ)-1sin(θ)=0
Step 9.2
Reorder cos2(θ) and -1.
sin(θ)+-1+cos2(θ)sin(θ)=0
Step 9.3
Rewrite -1 as -1(1).
sin(θ)+-1(1)+cos2(θ)sin(θ)=0
Step 9.4
Factor -1 out of cos2(θ).
sin(θ)+-1(1)-1(-cos2(θ))sin(θ)=0
Step 9.5
Factor -1 out of -1(1)-1(-cos2(θ)).
sin(θ)+-1(1-cos2(θ))sin(θ)=0
Step 9.6
Rewrite -1(1-cos2(θ)) as -(1-cos2(θ)).
sin(θ)+-(1-cos2(θ))sin(θ)=0
Step 9.7
Apply pythagorean identity.
sin(θ)+-sin2(θ)sin(θ)=0
Step 9.8
Cancel the common factor of sin2(θ) and sin(θ).
Step 9.8.1
Factor sin(θ) out of -sin2(θ).
sin(θ)+sin(θ)(-sin(θ))sin(θ)=0
Step 9.8.2
Cancel the common factors.
Step 9.8.2.1
Multiply by 1.
sin(θ)+sin(θ)(-sin(θ))sin(θ)⋅1=0
Step 9.8.2.2
Cancel the common factor.
sin(θ)+sin(θ)(-sin(θ))sin(θ)⋅1=0
Step 9.8.2.3
Rewrite the expression.
sin(θ)+-sin(θ)1=0
Step 9.8.2.4
Divide -sin(θ) by 1.
sin(θ)-sin(θ)=0
sin(θ)-sin(θ)=0
sin(θ)-sin(θ)=0
Step 9.9
Subtract sin(θ) from sin(θ).
0=0
0=0
Step 10
Since 0=0, the equation will always be true.
Always true
Step 11
The result can be shown in multiple forms.
Always true
Interval Notation:
(-∞,∞)