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Trigonometry identities and equations

  • Thread starter wei1006
  • Start date
  • #1
6
0
1) Question statement:
Simplify 2sec^2x-2sec^2xsin^2x-sin^2x-cos^2x

2)Relevant equations:
tan A=sinA/cos A
1+tan^2A=sec^A
cot A=1/tanA
cot A=cos A/sinA
sin^2A+cos^2A=1
secA=1/cos A
cosecA=1/sinA
1+cosec^2A= cot^2A
sin2A=2sinAcosA
cos2A=1-2sin^2A=cos^2A-sin^2A=2cos^A-1
tan2A=(2tanA)/1-tan^2A

3) Attempt:
2sec^2x-2sec^2xsin^2x-sin^2x-cos^2x
= 2sec^2x(1-sin^2x)-sin^2x-cos^2x
= 2sec^2x(cos^2x)-sin^2x-cos^2x

I am stuck after this... Will be helpful if some clues are provided. Thank you!
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
1) Question statement:
Simplify 2sec^2x-2sec^2xsin^2x-sin^2x-cos^2x

2)Relevant equations:
tan A=sinA/cos A
1+tan^2A=sec^A
cot A=1/tanA
cot A=cos A/sinA
sin^2A+cos^2A=1
secA=1/cos A
cosecA=1/sinA
1+cosec^2A= cot^2A
sin2A=2sinAcosA
cos2A=1-2sin^2A=cos^2A-sin^2A=2cos^A-1
tan2A=(2tanA)/1-tan^2A

3) Attempt:
2sec^2x-2sec^2xsin^2x-sin^2x-cos^2x
= 2sec^2x(1-sin^2x)-sin^2x-cos^2x
= 2sec^2x(cos^2x)-sin^2x-cos^2x

I am stuck after this... Will be helpful if some clues are provided. Thank you!
Use that sec(x)=1/cos(x).
 
  • #3
130
30
Also, at the end you have: (... -sin^2x - cos^2x) . What could you do with that?
 

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