Question

inverse trigonometry ......solve this question​

Answer 1

Given:

x²+y²+z²=r²

To Prove:

$\sf{tan^{-1}\dfrac{yz}{xr}+tan^{-1}\dfrac{zx}{yr}+tan^{-1}\dfrac{xy}{zr}=\dfrac{\pi}{2}}$

Solution:

We know that,

$\sf{tan^{-1}x+tan^{-1}y+tan^{-1}z=tan^{-1}\bigg(\dfrac{x+y+z-xyz}{1-(xy+yz+zx)}\bigg)}$

$\sf{tan^{-1}(\infty)=\dfrac{\pi}{2}}$

It is given that,

x²+y²+z²=r²

$\sf{\dfrac{x^{2}+y^{2}+z^{2}}{r^{2}}=1}-------(1)$

Now,

$\sf{L.H.S.=tan^{-1}\dfrac{yz}{xr}+tan^{-1}\dfrac{zx}{yr}+tan^{-1}\dfrac{xy}{zr}}$

$\sf{L.H.S.=tan^{-1}\Bigg(\dfrac{\dfrac{yz}{xr}+\dfrac{zx}{yr}+\dfrac{xy}{zr}-\Big(\dfrac{yz}{xr}.\dfrac{zx}{yr}.\dfrac{xy}{zr}\Big)}{1-\Big(\dfrac{yz}{xr}.\dfrac{zx}{yr}+\dfrac{zx}{yr}.\dfrac{xy}{zr}+\dfrac{xy}{zr}.\dfrac{yz}{xr}\Big)}\Bigg)}$

$\sf{L.H.S.=tan^{-1}\Bigg(\dfrac{\dfrac{yz}{xr}+\dfrac{zx}{yr}+\dfrac{xy}{zr}-\Big(\dfrac{xyz}{r^{3} }\Big)}{1-\Big(\dfrac{z^{2}}{r^{2}}+\dfrac{y^{2}}{r^{2}}+\dfrac{x^{2}}{r^{2}}\Big)}\Bigg)}$

$\sf{L.H.S.=tan^{-1}\Bigg(\dfrac{\dfrac{yz}{xr}+\dfrac{zx}{yr}+\dfrac{xy}{zr}-\Big(\dfrac{xyz}{r^{3} }\Big)}{1-\Big(\dfrac{x^{2}+y^{2}+z^{2}}{r^{2}}\Big)}\Bigg)}$

From (1),

$\sf{L.H.S.=tan^{-1}\Bigg(\dfrac{\dfrac{yz}{xr}+\dfrac{zx}{yr}+\dfrac{xy}{zr}-\Big(\dfrac{xyz}{r^{3} }\Big)}{1-(1)}\Bigg)}$

$\sf{L.H.S.=tan^{-1}\Bigg(\dfrac{\dfrac{yz}{xr}+\dfrac{zx}{yr}+\dfrac{xy}{zr}-\Big(\dfrac{xyz}{r^{3} }\Big)}{1-1}\Bigg)}$

$\sf{L.H.S.=tan^{-1}\Bigg(\dfrac{\dfrac{yz}{xr}+\dfrac{zx}{yr}+\dfrac{xy}{zr}-\Big(\dfrac{xyz}{r^{3} }\Big)}{0}\Bigg)}$

We know that something upon 0 is '∞'.

Therefore,

$\sf{L.H.S.=tan^{-1}(\infty)}$

$\sf{L.H.S.=\dfrac{\pi}{2} }$

$\sf{L.H.S.=R.H.S.}$

Hence Proved