1.

The value of `(sin5alpha-sin3alpha)/(cos5alpha+2cos4alpha+cos3alpha)` is

Answer» Here, we will use,
`sinC - sinD = 2sin((C-D)/2) sin((C+D)/2)`
`cosC + cosD = 2cos((C+D)/2)cos((C-D)/2)`
`L.H.S. = (sin5alpha-sin3alpha)/(cos5alpha+2cos4alpha+cos3alpha)`
`=(2sinalphacos4alpha)/(2cos4alphacosalpha+2cos4alpha)`
`=sinalpha/(1+cos alpha)`
`=(2sin(alpha/2)cos(alpha/2))/(2cos^2(alpha/2))`
`=tan(alpha/2)`
`:. (sin5alpha-sin3alpha)/(cos5alpha+2cos4alpha+cos3alpha) = tan(alpha/2)`


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