Question

cos2a=3cos2B-1/3-cos2B​

Answer 1

Answer:

Given : cos 2a = ( 3 cos 2b - 1 ) / ( 3 - cos 2b ) ........ (1)

∴ cos 2a

= { [ 3 ( 1 - tan² b ) / ( 1 + tan² b ) ] - 1 } / { 3 - [ ( 1 - tan² b ) / ( 1 + tan² b ) ] }

= ( 3 - 3tan² b - 1 - tan² b ) / ( 3 + 3tan² b - 1 + tan² b )

= ( 2 - 4tan² b ) / ( 2 + 4tan² b )

= ( 1 - 2tan² b ) / ( 1 + 2tan² b )

∴ cos 2a = ( 1 - tan² a ) / ( 1 + tan² a ) = ( 1 - 2tan² b ) / ( 1 + 2tan² b )

∴ comparing the corresponding terms,

... tan² a = 2 tan² b

∴ tan a = √2· tan b ...........

i think..it is helpful to u