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Prove that $ \tan^{-1}\large\frac{yz}{xr}+ \tan^{-1}\frac{zx}{yr}+ \tan^{-1}\frac{xy}{zr}=\frac{\pi}{2}\;where\;x^2+y^2+z^2=r^2$

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  • \( tan^{-1}x+tan^{-1}y=tan^{-1} \bigg( \large\frac{x+y}{1-xy} \bigg) \)
L.H.S.
Using the above formula we get
\(tan^{-1}\large\frac{yz}{xr}+tan^{-1}\large\frac{zr}{yr}+tan^{-1}\large\frac{xy}{zr} \)
\( = tan^{-1}\large\frac{\large\frac{yz}{xr}+\large\frac{zx}{yr}}{1-\large\frac{z^2xy}{r^2xy}}+tan^{-1}\large\frac{xy}{zr}\)
\(=tan^{-1}\bigg(\large\frac{y^2z+x^2z}{xyr^2}.\large\frac{r^2xy}{r^2xy-z^2xy}\bigg)+tan^{-1}\large\frac{xy}{zr}\)
\(=tan^{-1}\bigg(\large\frac{z(y^2+x^2)}{xy(r^2-z^2)}\bigg)\)\(+tan^{-1}\large\frac{xy}{zr}\)
use the formula once again and get
\( = tan^{-1} \bigg(\large\frac{\large\frac{z(y^2+x^2)}{xy(r^2-z^2)}+\large\frac{xy}{zr}}{1-\large\frac{xyz(y^2+x^2)}{xyzr(r^2-z^2)}}\bigg)\)
\(=tan^{-1}\bigg(\large\frac{\large\frac{yz}{xr}+\large\frac{xz}{yr}+\frac{xy}{zr}-\large\frac{xyz}{r^3}}{1-\big(\large\frac{x^2+y^2+z^2}{r^2}\big)}\bigg)\)
given \( x^2+y^2+z^2=r^2\)
\( \Rightarrow 1-\large\frac{x^2+y^2+z^2}{r^2}=0\)
\(\Rightarrow\: L.H.S.= tan^{-1} \infty = \large\frac{\pi}{2}\)
=R.H.S.
answered Feb 19, 2013 by thanvigandhi_1
edited Jan 30, 2014 by balaji
 

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