How do you simplify #(1/sqrt(a-1)+sqrt(a+1))/(1/sqrt(a+1)-1/sqrt(a-1))div sqrt(a+1)/((a-1)sqrt(a+1)-(a+1)sqrt(a-1)),# a>1 ?

2 Answers
Sunayan S
Apr 27, 2018

Huge math formatting...

Explanation:

#color(blue)(((1/sqrt(a-1)+sqrt(a+1))/(1/sqrt(a+1)-1/sqrt(a-1)))/ (sqrt(a+1)/((a-1)sqrt(a+1)-(a+1)sqrt(a-1)))#

#=color(red)(((1/sqrt(a-1)+sqrt(a+1))/((sqrt(a-1)-sqrt(a+1))/(sqrt(a+1) cdot sqrt(a-1))))/ (sqrt(a+1)/(sqrt(a-1) cdot sqrt(a-1) cdot sqrt(a+1)-sqrt(a+1) cdot sqrt(a+1)sqrt(a-1)))#

#=color(blue)(((1/sqrt(a-1)+sqrt(a+1))/((sqrt(a-1)-sqrt(a+1))/(sqrt(a+1) cdot sqrt(a-1))))/ (sqrt(a+1)/(sqrt(a+1) cdot sqrt(a-1)(sqrt(a-1)-sqrt(a+1)))#

#=color(red)((1/sqrt(a-1)+sqrt(a+1))/((sqrt(a-1)-sqrt(a+1))/(sqrt(a+1) cdot sqrt(a-1)))xx(sqrt(a+1) cdot sqrt(a-1)(sqrt(a-1)-sqrt(a+1)))/sqrt(a+1)#

#=color(blue)((1/sqrt(a-1)+sqrt(a+1))xx((sqrt(a+1) cdot sqrt(a-1))/(sqrt(a-1)-sqrt(a+1)))xx(cancel((sqrt(a+1))) cdot sqrt(a-1)(sqrt(a-1)-sqrt(a+1)))/cancelsqrt(a+1))#

#=color(red)(((1+sqrt(a+1) cdot sqrt(a-1))/(sqrt(a-1)))xx((sqrt(a+1) cdot sqrt(a-1))/(sqrt(a-1)-sqrt(a+1)))xx sqrt(a-1)cdot(sqrt(a-1)-sqrt(a+1))#

#=color(blue)(((1+sqrt(a+1) cdot sqrt(a-1))/cancel(sqrt(a-1)))xx((sqrt(a+1) cdot cancel((sqrt(a-1))))/color(red)(cancel(color(green)((sqrt(a-1)-sqrt(a+1)))))xx sqrt(a-1)cdot color(red)(cancel color(green)((sqrt(a-1)-sqrt(a+1)))#

#=color(red)(ul(bar(|color(blue)((1+sqrt(a+1) cdot sqrt(a-1))cdot(sqrt((a+1)(a-1))))|#

1s2s2p
Apr 28, 2018

#sqrt(a^2-1)+a^2-1#

Explanation:

To simplify things greatly we will use #u^2=a+1# and #v^2=a-1#, which gives us:
#(v^-1+u)/(u^-1-v^-1)*(uv^2-vu^2)/u=((v^-1+u)(uv^2-vu^2))/(u(u^-1-v^-1))=(uv-u^2+(uv)^2-vu^3)/(1-uv^-1)=(uv(1+uv)-u^2(1+uv))/((v-u)/v)=(uv(1+uv)(v-u))/(v-u)=uv(1+uv)#

#uv(1+uv)=uv+u^2v^2=sqrt(a-1)sqrt(a+1)+(a-1)(a+1)=sqrt(a^2-1)+a^2-1#

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