If tan A and tan B are the roots of the quadratic equation x^2-ax+b=0 then find the value of sin^2(A+B)?

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Answer 1 · 2018-06-03T15:58:41

#sin^2(A+B)=a^2/(a^2+b^2-2b+1)#

We know that,

#"If"# #alpha and beta# #"are the roots of the quadratic equation :"#

#color(red)(Px^2+Qx+R=0,then,alpha+beta=-Q/Pand alpha*beta=R/P#

We have,

#x^2-ax+b=0=>P=1 , Q=-a and R=b#

The roots are: #alpha=tanA and beta=tanB#

So,

#tanA+tanB=-Q/P=-(-a)/1=a#

#tanA*tanB=R/P=b/1=b#

Now,

#tan(A+B)=(tanA+tanB)/(1-tanAtanB)=a/(1-b)#

#=>tan^2(A+B)=a^2/(1-2b+b^2)#

#=>cot^2(A+B)=(1-2b+b^2)/a^2..to[becausecottheta=1/tantheta]#

#=>1+cot^2(A+B)=1+(1-2b+b^2)/a^2to#[add. bothsides #1#]

#=>csc^2(A+B)=(a^2+1-2b+b^2)/a^2#

#=>sin^2(A+B)=a^2/(a^2+1-2b+b^2)..to[becausesintheta=1/(csctheta) ]#

Note :
We have take the quadratic equn.#Px^2+Qx+R=0#,because
#A, B, a, b,# are used in the question.