Please, check my work on Trig Identities

AI Thread Summary
The discussion focuses on verifying solutions for various trigonometric identities. The original poster presented multiple problems, including showing relationships between tangent, cotangent, and secant functions. Responses indicate that most of the work is correct, particularly for problems a through e, while problem f also appears to be accurate. Participants emphasize the importance of confidence in solving math problems and suggest that practice leads to greater assurance in one's work. Overall, the thread highlights the collaborative effort to confirm the correctness of trigonometric identities.
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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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