Prove that tan−11+x2+1−x21+x2−1−x2=π4+12cos−1x2? Let the above expression be equal to ϕ tan⁡ϕ+1tan⁡ϕ−1=1+x21−x2 1+tan2ϕ+2tan⁡ϕ1+tan2ϕ−2tan⁡ϕ=1+x21−x2 1+tan2ϕ2tan⁡ϕ=1x2 sin⁡2ϕ=x2 ϕ=π4−12cos−1x2 Where am I going wrong?

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Prove that tan−11+x2+1−x21+x2−1−x2=π4+12cos−1x2?  Let the above expression be equal to ϕ  tan⁡ϕ+1tan⁡ϕ−1=1+x21−x2   1+tan2ϕ+2tan⁡ϕ1+tan2ϕ−2tan⁡ϕ=1+x21−x2   1+tan2ϕ2tan⁡ϕ=1x2   sin⁡2ϕ=x2   ϕ=π4−12cos−1x2   Where am I going wrong?

encolatgehu · Accepted Answer

A bit late answer but I thought worth mentioning it. First note that we can substitute y=x2 and consider 0&amp;lt;y≥1. Furthermore, the argument of arctan⁡ can be simplified as follows: 1+x2+1−x21+x2−1−x2=1+1−y2y Now, setting y=cos⁡t for t∈[0,π2) to show is only arctan⁡1+sin⁡tcos⁡t=π4+t2 At this point half-angle formulas come into mind: 1+sin⁡tcos⁡t=(cos⁡t2+sin⁡t2)2cos2t2−sin2t2=cos⁡t2+sin⁡t2cos⁡t2−sin⁡t2 =1+tan⁡t21−tan⁡t2=tan⁡(π4+t2) Done.

RizerMix · Accepted Answer

Because for x≠0 and −1≤x≤1 easy to see that: 0&amp;lt;π4+12cos−1⁡x2&amp;lt;π2 and we obtain: tan⁡(π4+12cos1⁡x2)=1+tan⁡12arccos⁡x21tan⁡12arccos⁡x2= =cos⁡12arccos⁡x2+sin⁡12arccos⁡x2cos⁡12arccos⁡x2−sin⁡12arccos⁡x2=1+x22+1−x221+x22−1−x22=1+x2+1−x21+x2−1−x2 Your mistake in the last line. Indeed, since 1+x2+1−x21+x2−1−x2&amp;gt;1 and from here π4&amp;lt;ϕ&amp;lt;π2, we obtain: 2ϕ=πarcsin⁡x2=π2+arccos⁡x2.

alenahelenash · Accepted Answer

Your mistake sin⁡y=a→tosin−1⁡(sin⁡y)=(2nπ+y,y∈I,IV quadrant),((2n−1)π−y,y∈II,III quadrant):=sin−1⁡a

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

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