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If `(tan(alpha+beta+gamma))/("tan"(alpha-beta-gamma))=(tangamma)/(tanbeta),(beta!=gamma)`then `sin2alpha+s in2beta+s in2gamma=`0 (b) 1(c) 2 (d) |
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Answer» Correct Answer - A `(tan (alpha+beta-gamma))/(tan(alpha-beta+gamma))=(tan gamma)/(tan beta)` `rArr(sin(alpha+beta-gamma)cos(alpha-beta+gamma))/(sin(alpha-beta+gamma)cos(alpha+beta-gamma))=(sin gamma cos beta)/(sin beta cos gamma)` Applying componendo and dividendo, we get `(sin 2alpha)/(sin 2(beta-gamma))=(sin(gamma+beta))/(sin(gamma-beta))` `rArr sin2(beta-gamma)sin(beta+gamma)+sin 2alpha sin(beta-gamma)=0` `rArr sin(beta-gamma)(sin 2alpha+sin2beta+sin2gamma)=0` `rArr sin2alpha+sin 2beta+sin 2y=0` (as `beta ne gamma`) |
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