#cot^-1x + cot^-1(1+x)=pi/4#?

1 Answer
iceman
May 10, 2018

#x=-1, 2#

Explanation:

#cot^(-1)(x)+cot^(-1)(1+x)=pi/4#

Take #tan# of both sides:

#tan[cot^(-1)(x)+cot^(-1)(1+x)]=tan(pi/4)#

Expand the left side using the sum formula:

#tan(x+y) = [tan(x)+tan(y)]/[1-tan(x)tan(y)]#

So in this case:

#[tan (cot^(-1)(x))+tan(cot^(-1)(1+x))]/[1-tan(cot^(-1)(x))*tan(cot^(-1)(1+x))]=1#

Simplify:

#[1/x+1/(x+1)]/[1-1/x*1/(x+1)]=1#

#(x+1+x)/[x(x+1)]= 1-1/(x^2+x)#

#(2x+1)/(x^2+x)=(x^2+x-1)/(x^2+x)#

#x^2-x-2=0#

#(x+1)(x-2)=0#

#x=-1, 2#

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