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