Please, check my work on Trig Identities

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Homework Help Overview

The original poster presents a series of trigonometric identity problems, seeking verification of their solutions. The problems involve various identities and transformations related to sine, cosine, tangent, cotangent, and secant functions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants review the original poster's work, noting that some identities appear correct while others require further examination. There is discussion about the clarity of the original poster's presentation and the confidence in their solutions.

Discussion Status

Some participants have confirmed the correctness of specific parts of the original poster's work, particularly problems (a) through (e) and (f). However, there is an acknowledgment of the need for confidence in one's mathematical reasoning, suggesting a supportive environment for further exploration.

Contextual Notes

Participants express varying levels of confidence regarding the correctness of the original poster's solutions, indicating that practice and familiarity with the material are important for developing assurance in one's work.

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Please, check my work.

Homework Statement



a) Show that sqrt{[1+tan^2x]/[1+cot^2x]}=tanx

b) Show that [cosx+sinx]/[cosx-sinx]=1+[2tanx]/[1- tanx]

c)Show that cotxcosx+tanxsinx=(cosecx+ secx)(1-sinxcosx)

d) Show that cosec^2x-cosecx=cot^2x/[1+sinx]

e) Show that sin^3x-cos^3x= (sinx-cosx)(1+sinxcosx)

f) Show that [cosx-1]/[secx+tanx]+[cosx+1]/[secx-tanx]=2(1+tanx)

Homework Equations


The Attempt at a Solution



a) sqrt{[1+tan^2x]/[1+cot^2x]}=tanx

=sqrt{[1+tan^2x]/[1+cot^2x]}

= sqrt{[sec^2x]/[cosec^2x]}

= secx/cosecx

= 1/cosx(sinx)

= tanx

b) [cosx+sinx]/[cosx-sinx]=1+[2tanx]/[1- tanx]

=1+[2tanx]/[1- tanx]

= [1-tanx+2tanx]/[1-tanx]

= [1+tanx]/[1-tanx]

= [1+ sinx/cosx]/[1-sinx/cosx]

= [(cosx+sinx)/cosx]/[(cosx-sinx)/cosx]

= [cosx+sinx]/[cosx-sinx)]

c) cotxcosx+tanxsinx=(cosecx+ secx)(1-sinxcosx)

=(cosecx+secx)(1-sinxcosx)

=(cosecx+secx)(sin^2x-sinxcosx+cos^2x)

= sinx+ [sin^2x/cosx]-cosx-sinx+[cos^2x/sinx]+cosx

= [sin^2x/cosx]+[cos^2x/sinx]

= [sinx/cosx](sinx)+[cosx/sinx](cosx)

= tanxsinx+cotxcosx

d) cosec^2x-cosecx=cot^2x/[1+sinx]

=cot^2x/[1+sinx]

= cot^2x/[1+sinx]*[(1-sinx)/(1-sinx)]

= [cot^2x-cot^2x(sinx)]/(1-sin^2x)

= [cot^2x-cot^2x(sinx)]/(cos^2x)

= [cot^2x]/[cos^2x]- [cot^2xsinx]/[cos^2x]

= [cos^2x]/[sin^2x](1/[cos^2x])- [cos^2x]/[sin^2x](1/[cos^2x])(sinx)

= 1/(sin^2x)-sinx/(sin^2x)

= cosec^2x-cosecx

e) sin^3x-cos^3x= (sinx-cosx)(1+sinxcosx)

= (sinx-cosx)(1+sinxcosx)

=( sinx-cosx)(sin^2x+sinxcosx+cos^2x)

= sin^3x-cos^3x

f) [cosx-1]/[secx+tanx]+[cosx+1]/[secx-tanx]=2(1+tanx)

= [cosx-1]/[secx+tanx]+[cosx+1]/[secx-tanx]

= [cosx-1]/[secx+tanx]+{[cosx+1]/[secx-tanx]*[(secx+tanx)/secx+tanx]}

= {(cosx-1)(secx-tanx)+(cosx+1)(secx+tanx)}/[sex^2x-tan^2x]

= 1-sinx-(1/cosx)+(sinx/cosx)+1+sinx+(1/cosx)+sinx/cosx

= 1+(sinx/cosx)+1+sinx/cosx

= 2+2(sinx/cosx)

= 2(1+tanx)

Thank You very much.
 
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The work was a little difficult for me to read, even with the grouping symbols. But looks like (a) through (e) are correct. I didn't get a chance to look at (f).
 
(f) looks fine !
 
Thank you very much for checking my work, people. Much appreciated.
 
You shouldn't really need your work checked. As long as you're confident you didn't break any rules of math and arrive at the final answer it is most likely correct.
 
TimeToShine said:
You shouldn't really need your work checked. As long as you're confident you didn't break any rules of math
I think that was the point of the OP's post - he wasn't confident about his or her work. It takes quite a bit of practice to reach that point of confidence.
TimeToShine said:
and arrive at the final answer it is most likely correct.
 

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