Question

If tan⁻¹yz/xr+tan⁻¹zx/yr+tan⁻¹xy/xr= π/2, then prove that x²+y²+z²=r²

Answer 1

Given, tan^-1(yz/xr) + tan^-1(zx/yr) + tan^-1(xy/zr) = π/2

tan^-1(yz/xr) + tan^-1(zx/yr) = π/2 - tan^-1(xy/zr)

we know, tan^-1m + tan^-1n = tan^-1(m + n)/(1 - mn)

so, tan^-1(yz/xr) + tan^-1(zx/yr) = tan^-1{(yz/xr) + (zx/yr)}/{1 - yz/xr × zx/yr}

= tan^-1{(y²z + zx²)/{(xyr² - yz²x)}

= tan^-1{zr(x² + y²)/xy(r² - z²)}

now, tan^-1{zr(x² + y²)/xy(r² - z²)} = π/2 - tan^-1(xy/zr)

zr(x² + y²)/xy(r² - z²) = tan[π/2 - tan^-1(xy/zr)]

zr(x² + y²)/xy(r² - z²) = cot{tan^-1(xy/zr)}

zr(x² + y²)/xy(r² - z²) = tan{tan^-1(zr/xy)}

[ we know, cot(tan^-1x) = cot{cot^-1(1/x)}]

zr(x² + y²)/xy(r² - z²) = zr/xy

now, (x² + y²)/(r² - z²) = 1

x² + y² + z² = r²

hence proved

Answer 2

HELLO DEAR,



GIVEN:-
tan-¹(yz/xr) + tan-¹(zx/yr) + tan-¹(xy/zr) = π/2

tan-¹(yz/xr) + tan-¹(zx/yr) = π/2 - tan-¹(xy/zr)

we may know:-
tan-¹a + tan-¹b = tan-¹(a + b)/(1 - ab)

using here,

so, tan-¹{(yz/xr) + (zx/yr)}/{1 - yz/xr * zx/yr} = tan-¹{zr(x² + y²)/xy(r² - z²)} = π/2 - tan-¹(xy/zr)

=> tan-¹{(y²z + zx²)/{(xyr² - yz²x)} = tan-¹{zr(x² + y²)/xy(r² - z²)} = π/2 - tan-¹(xy/zr)

=> tan-¹{zr(x² + y²)/xy(r² - z²)} = tan-¹{zr(x² + y²)/xy(r² - z²)} = π/2 - tan-¹(xy/zr)

taking tangent both side,

zr(x² + y²)/xy(r² - z²) = tan[π/2 - tan-¹(xy/zr)]

[ tan (π/2 - A) = cotA ]

zr(x² + y²)/xy(r² - z²) = cot{tan-¹(xy/zr)}

zr(x² + y²)/xy(r² - z²) = tan{tan-¹(zr/xy)}

[ as, cot(tan-¹x) = cot{cot-¹(1/x)}]

zr(x² + y²)/xy(r² - z²) = zr/xy

(x² + y²)/(r² - z²) = 1

x² + y² + z² = r²


I HOPE ITS HELP YOU DEAR,
THANKS