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## 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.