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awarded  Great Question
Jun
30
awarded  Nice Question
Jun
27
comment $\cot^{-1}\frac{y}{\sqrt{1-x^2-y^2}} = 2\tan^{-1}\sqrt{\frac{3-4x^2}{4x^2}} - \tan^{-1}\sqrt{\frac{3-4x^2}{x^2}} $
@DavidH Yup, that's correct.
May
19
comment $\cot^{-1}\frac{y}{\sqrt{1-x^2-y^2}} = 2\tan^{-1}\sqrt{\frac{3-4x^2}{4x^2}} - \tan^{-1}\sqrt{\frac{3-4x^2}{x^2}} $
@Mann Converting to $\cos^{-1}$ along with a bit of manipulation does reduce calculation but it's still quite a lot.
May
19
comment $\cot^{-1}\frac{y}{\sqrt{1-x^2-y^2}} = 2\tan^{-1}\sqrt{\frac{3-4x^2}{4x^2}} - \tan^{-1}\sqrt{\frac{3-4x^2}{x^2}} $
@Mann Could you please describe in more detail how the above is useful?
May
19
asked $\cot^{-1}\frac{y}{\sqrt{1-x^2-y^2}} = 2\tan^{-1}\sqrt{\frac{3-4x^2}{4x^2}} - \tan^{-1}\sqrt{\frac{3-4x^2}{x^2}} $
May
19
accepted Sum of the series $\tan^{-1}\frac{4}{4n^2+3}$
May
19
answered Sum of the series $\tan^{-1}\frac{4}{4n^2+3}$
May
19
reviewed Approve Sum of the series $\tan^{-1}\frac{4}{4n^2+3}$
May
19
comment Sum of the series $\tan^{-1}\frac{4}{4n^2+3}$
It was a question on my exam.
May
19
asked Sum of the series $\tan^{-1}\frac{4}{4n^2+3}$
May
15
awarded  Popular Question
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29
awarded  Notable Question
Mar
9
awarded  Popular Question
Dec
25
accepted How to remember trigonometric ratios for allied angles?
Dec
19
awarded  Constituent
Dec
12
comment Rational + irrational = always irrational?
@user156035 no. for example, $e$ and $\ln(2)$ are irrational, but $e^{\ln(2)} = 2$, which is rational.
Dec
10
awarded  Caucus
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1
awarded  Popular Question
Oct
24
awarded  Yearling