Prove that \tan^{-1} \frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}=\frac{\pi}{4}+\frac{1}{2}\cos^{-1}x^2? Let the above expression be equal to

Francisca Rodden

Francisca Rodden

Answered question

2022-01-16

Prove that tan11+x2+1x21+x21x2=π4+12cos1x2?
Let the above expression be equal to ϕ
tanϕ+1tanϕ1=1+x21x2
1+tan2ϕ+2tanϕ1+tan2ϕ2tanϕ=1+x21x2
1+tan2ϕ2tanϕ=1x2
sin2ϕ=x2
ϕ=π412cos1x2
Where am I going wrong?

Answer & Explanation

encolatgehu

encolatgehu

Beginner2022-01-18Added 27 answers

A bit late answer but I thought worth mentioning it.
First note that we can substitute y=x2 and consider 0<y1. Furthermore, the argument of arctan can be simplified as follows:
1+x2+1x21+x21x2=1+1y2y
Now, setting y=cost for t[0,π2) to show is only
arctan1+sintcost=π4+t2
At this point half-angle formulas come into mind:
1+sintcost=(cost2+sint2)2cos2t2sin2t2=cost2+sint2cost2sint2
=1+tant21tant2=tan(π4+t2)
Done.

RizerMix

RizerMix

Expert2022-01-20Added 656 answers

Because for x0 and 1x1 easy to see that: 0<π4+12cos1x2<π2 and we obtain: tan(π4+12cos1x2)=1+tan12arccosx21tan12arccosx2= =cos12arccosx2+sin12arccosx2cos12arccosx2sin12arccosx2=1+x22+1x221+x221x22=1+x2+1x21+x21x2 Your mistake in the last line. Indeed, since 1+x2+1x21+x21x2>1 and from here π4<ϕ<π2, we obtain: 2ϕ=πarcsinx2=π2+arccosx2.
alenahelenash

alenahelenash

Expert2022-01-24Added 556 answers

Your mistake
siny=atosin1(siny)=(2nπ+y,yI,IV quadrant),((2n1)πy,yII,III quadrant):=sin1a

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