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If tanalpha,tanbeta are the roots of the equation x^(2)+px+q=0(pne0),then |
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Answer» `sin^(2)(alpha+beta)+p sin(alpha+beta)cos(alpha+beta)+qcos^(2)(alpha+beta)=q` `thereforetanalpha+tanbeta=-p,tanalphatanbeta=q` `thereforetan(alpha+beta)=(tanalpha+tanbeta)/(1-tanalphatanbeta)=(p)/(q-1)` Also ,when `tan(alpha+beta)=(p)/(q-1)`. L.H.S. of the EXPRESSION given in `=cos^(2)(alpha+beta)[tan^(2)(alpha+beta)+ptan(alpha+beta)+q]` `=(1)/(1+tan^(2)(alpha+beta))[(p^(2))/((q-1)^(2))+(p^(2))/(q-1)+q]` `=((q-1)^(2))/((q-1)^(2)-p^(2))[(p^(2)+p^(2)(q-1)+q(q-1)^(2))/((q-1)^(2))]` `=(q{p^(2)+(q-1)^(1)})/(p^(2)+(q-1)^(2))` =q = R.H.S.of i.e., Relation given in (a) is also SATISFIED. |
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