Question

If sin^-1x+sin^-1y+sin^-1z=pi then prove that x✓1-x^2 +y✓1-y^2 +z✓1-z^2= 2 xyz

Answer 1

let sin^-1x=A, sin^-1y=B, sin^-1z=c
.. therefore sinA=x ,sinB=y, sinC=z.. now sin^-1x+sin^-1y+sin^-1z=pi

.. or A+B+C=pi ,, left hand= x rootover 1-x^2+y rootover 1-y^2+ z rootover 1-z^2
= sinA rootover 1-sin^2A+ sinB rootover 1-sin^2B+ sinC rootover 1-sin^2C
= sinAcosA+sinBcosB+sinCcosC
=1/2 (sin2A+sin2B)+sinCcosC
=1/2× 2sin(A+B) cos(A-B)+sinCcosC
=sin (pi-C)cos (A-B)+sinCcosC
=sinCcos (A-B)+sinCcosC
=sinC [cos (A-B)+cos {pi-(A+B)}]
=sinC [cos (A-B)-cos (A+B)]
=sinC [2sinAsinB]
=2sinAsinBsinC=2xyz (proved)