Letting #x=pi/7#, we have,
#"The R.H.S.="(3-tan^2x)/(1-tan^2x)#,
#=(3cos^2x-sin^2x)/(cos^2x-sin^2x)#,
#={3*1/2(1+cos2x)-1/2(1-cos2x)}/(cos2x)#,
#=(1+2cos2x)/(cos2x)#,
#=(1+2cos2x)/(cos2x)xx(2sinx)/(2sinx)#,
#={2(sinx+2cos2xsinx)}/(2cos2xsinx)#,
#={2(sinx+sin3x-sinx)}/(2cos2xsinx)#,
#=(2sin3x)/(2cos2xsinx)#,
#=(2sin3x)/(2cos2xsinx)xxcosx/cosx#,
#=(2sin3xcosx)/{(cos2x)(2sinxcosx)}#,
#=(2sin3xcosx)/{(cos2x)(sin2x)}xx2/2#,
#=(4sin3xcosx)/(2cos2xsin2x)#.
# rArr" The R.H.S.="(4sin3xcosx)/(sin4x)," whereas, "#
#"The L.H.S="lambdacosx#.
#"Therefore, the given eqn. is, "(4sin3xcosx)/(sin4x)=lambdacosx#.
# rArr (4sin3x)/(sin4x)=lambda....[because, cosx=cos(pi/7)!=0]#.
#:. lambda=(4sin(3pi/7))/(sin(4pi/7))#,
#=(4sin(3pi/7))/(sin{(7-3)pi/7})#,
#=(4sin(3pi/7))/sin{(7/7-3/7)pi}#,
#=(4sin(3pi/7))/sin(pi-3pi/7)#,
#=(4sin(3pi/7))/(sin(3pi/7))#.
# rArr lambda=4#.
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