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Popular Trigonometry >

prove (2sec^2(x)-2tan^2(x))/(csc(x))=sin(2x)sec(x)

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Solution

prove csc(x)2sec2(x)−2tan2(x)​=sin(2x)sec(x)

Solution

True
Solution steps
csc(x)2sec2(x)−2tan2(x)​=sin(2x)sec(x)
Manipulating left sidecsc(x)2sec2(x)−2tan2(x)​
Express with sin, cos
csc(x)2sec2(x)−2tan2(x)​
Use the basic trigonometric identity: sec(x)=cos(x)1​=csc(x)2(cos(x)1​)2−2tan2(x)​
Use the basic trigonometric identity: tan(x)=cos(x)sin(x)​=csc(x)2(cos(x)1​)2−2(cos(x)sin(x)​)2​
Use the basic trigonometric identity: csc(x)=sin(x)1​=sin(x)1​2(cos(x)1​)2−2(cos(x)sin(x)​)2​
Simplify sin(x)1​2(cos(x)1​)2−2(cos(x)sin(x)​)2​:cos2(x)sin(x)(2−2sin2(x))​
sin(x)1​2(cos(x)1​)2−2(cos(x)sin(x)​)2​
Apply the fraction rule: cb​a​=ba⋅c​=1(2(cos(x)1​)2−2(cos(x)sin(x)​)2)sin(x)​
2(cos(x)1​)2=cos2(x)2​
2(cos(x)1​)2
(cos(x)1​)2=cos2(x)1​
(cos(x)1​)2
Apply exponent rule: (ba​)c=bcac​=cos2(x)12​
Apply rule 1a=112=1=cos2(x)1​
=2⋅cos2(x)1​
Multiply fractions: a⋅cb​=ca⋅b​=cos2(x)1⋅2​
Multiply the numbers: 1⋅2=2=cos2(x)2​
2(cos(x)sin(x)​)2=cos2(x)2sin2(x)​
2(cos(x)sin(x)​)2
(cos(x)sin(x)​)2=cos2(x)sin2(x)​
(cos(x)sin(x)​)2
Apply exponent rule: (ba​)c=bcac​=cos2(x)sin2(x)​
=2⋅cos2(x)sin2(x)​
Multiply fractions: a⋅cb​=ca⋅b​=cos2(x)sin2(x)⋅2​
=1sin(x)(cos2(x)2​−cos2(x)2sin2(x)​)​
Combine the fractions cos2(x)2​−cos2(x)2sin2(x)​:cos2(x)2−2sin2(x)​
Apply rule ca​±cb​=ca±b​=cos2(x)2−2sin2(x)​
=1(cos2(x)−2sin2(x)+2​)sin(x)​
Remove parentheses: (a)=a=1cos2(x)2−sin2(x)⋅2​sin(x)​
Multiply cos2(x)2−sin2(x)⋅2​sin(x):cos2(x)sin(x)(−2sin2(x)+2)​
cos2(x)2−sin2(x)⋅2​sin(x)
Multiply fractions: a⋅cb​=ca⋅b​=cos2(x)(2−sin2(x)⋅2)sin(x)​
=1cos2(x)sin(x)(−2sin2(x)+2)​​
Apply the fraction rule: 1a​=a=cos2(x)(2−sin2(x)⋅2)sin(x)​
=cos2(x)sin(x)(2−2sin2(x))​
=cos2(x)(2−2sin2(x))sin(x)​
Rewrite using trig identities
cos2(x)(2−2sin2(x))sin(x)​
Use the Pythagorean identity: cos2(x)+sin2(x)=1cos2(x)=1−sin2(x)=1−sin2(x)(2−2sin2(x))sin(x)​
Simplify 1−sin2(x)(2−2sin2(x))sin(x)​:2sin(x)
1−sin2(x)(2−2sin2(x))sin(x)​
Factor 2−2sin2(x):2(1−sin2(x))
2−2sin2(x)
Rewrite as=2⋅1−2sin2(x)
Factor out common term 2=2(1−sin2(x))
=1−sin2(x)2(1−sin2(x))sin(x)​
Cancel the common factor: 1−sin2(x)=2sin(x)
=2sin(x)
=2sin(x)
Manipulating right sidesin(2x)sec(x)
Express with sin, cos
sec(x)sin(2x)
Use the basic trigonometric identity: sec(x)=cos(x)1​=cos(x)1​sin(2x)
Simplify cos(x)1​sin(2x):cos(x)sin(2x)​
cos(x)1​sin(2x)
Multiply fractions: a⋅cb​=ca⋅b​=cos(x)1⋅sin(2x)​
Multiply: 1⋅sin(2x)=sin(2x)=cos(x)sin(2x)​
=cos(x)sin(2x)​
=cos(x)sin(2x)​
Rewrite using trig identities
cos(x)sin(2x)​
Use the Double Angle identity: sin(2x)=2sin(x)cos(x)=cos(x)2sin(x)cos(x)​
Cancel the common factor: cos(x)=2sin(x)
=2sin(x)
We showed that the two sides could take the same form⇒True

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Frequently Asked Questions (FAQ)

  • Is (2sec^2(x)-2tan^2(x))/(csc(x))=sin(2x)sec(x) ?

    The answer to whether (2sec^2(x)-2tan^2(x))/(csc(x))=sin(2x)sec(x) is True
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