किसी त्रिभुज ABC में यदि `2b^(2)=a^(2)+c^(2)` हो, तो सिद्ध कीजिए कि - `(sin 3B)/(sin B)=((a^(2)-c^(2))/(2ac))^(2)`

Explanation

`because " " 2b^(2)=a^(2)+c^(2) " "` …(1)
`because " " cos B=(a^(2)+c^(2)-b^(2))/(2ac)=(2b^(2)-b^(2))/(2ac)`
`rArr " " cos B = (b^(2))/(2ac)" "` ….(2)
`rArr " " L.H.S. = (sin 3B)/(sin B)=(3 sin B - 4 sin^(3)B)/(sin B)`
`= 3-4sin^(2)B=3-4(1-cos^(2)B)`
`= 4 cos^(2)B-1`
`=4((b^(2))/(2ac))^(2)-1 " "` [समीकरण (2) से]
`= ((2b^(2))^(2)-4a^(2)c^(2))/(4a^(2)c^(2))`
`= ((a^(2)+c^(2))^(2)-4a^(2)c^(2))/(4a^(2)c^(2))" " [because 2b^(2)=a^(2)+c^(2)]`
`=((a^(2)-c^(2))^(2))/(4a^(2)c^(2))=((a^(2)-c^(2))/(2ac))^(2)`
`= R.H.S. " "` इति सिद्धम्