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Please, check my work.
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)
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.
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.