Sum it Square it

From the question, we get $P=tanA+tanB$ and $Q=tanAtanB$ .

Hence, $tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}=\frac{P}{1-Q}$ .

By Pythagorean identity, $sin^2\theta+cos^2\theta=1$ .

Also, $\frac{sin^2\theta}{cos^2\theta}=tan^2\theta$ .

Combining the two equations, we get $sin^2\theta=\frac{tan^2\theta}{tan^2\theta+1}$ .

Therefore, $sin^2(A+B)=\frac{tan^2(A+B)}{tan^2(A+B)+1}=\frac{(\frac{P}{1-Q})^2}{(\frac{P}{1-Q})^2+1}=\frac{P^2}{P^2+(1-Q)^2}$ .

Here tan (a) +tan (b)=p and tan (a)×tan( b)=q . Substituting tan (a) by sin (a)/cos (a) and tan (b )by sin (b)/cos (b) in tan (a)+tan (b)=p, we get: sin (a) cos (b)+ cos (a) sin (b)= p×cos (a)×cos (b). Squaring both sides and replacing cos (a) by 1/sec (a) and cos (b) by 1/sec (b) we get: sin^2 (a+b) =p^2/(sec^2 (a)×sec^2 (b)). Now sec^2 (a)×sec^2 (b) = (1+tan^2 (a))(1+tan^2 (b))= 1+tan^2 (a)+tan^2 (b)+tan^2 (a)×tan^2 (b). Now: tan^2 (a)+tan^2 (b) =(tan (a)+tan (b))^2 - 2×tan (a)×tan (b)=p^2-2q and tan^2 a×tan^2 b=q^2. Therefore, sin^2 (a+b)=p^2/(1+p^2-2q+q^2) = p^2/(p^2+(1-q)^2).

Start with vieta's $\tan a +\tan b =p,\tan a \tan b =q$ use an identity $\tan (a+b)=\dfrac{\tan a+\tan b}{1-\tan a \tan b}$ now, substitute from vieta's $\tan (a+b) = \dfrac{p}{1-q}\longrightarrow a+b=\tan^{-1}(\dfrac{p}{1-q})$ now, get both sides in $\sin^2$ $\sin^2(a+b)=\sin^2(\tan^{-1}(\dfrac{p}{1-q})$ use the identity $\sin^2 \theta =\dfrac{1}{2}(1-\cos 2\theta )$ and we get $\sin^2(\tan^{-1}(\dfrac{p}{1-q}))=\dfrac{1}{2}(1-\cos (2\tan^{-1}(\dfrac{p}{1-q} ))$ use the identity $\cos(2\tan^{-1}\theta)=\dfrac{1-\theta^2}{\theta^2+1}$ $\dfrac{1}{2}(1-\cos (2\tan^{-1}(\dfrac{p}{1-q} ))=\dfrac{1}{2}(1-( \dfrac{1-(\dfrac{p}{1-q})^2}{1+(\dfrac{p}{1-q})^2})$ upon some expansion, $\sin^2(a+b)=\boxed{\dfrac{p^2}{p^2+(1-q)^2}}$

It is not reduced the generalization, can assume that P=3 and Q=2. So this equation has two roots: $x_1=1, x_2=2$ Applying to this problem, it has: $tan(A)=2, tan(B)=1 \Rightarrow sin(A)=2cos(A), sin(B)=cos(B)$ $cos(A)=\frac{1}{\sqrt{5}}, sin(A)=\frac{2}{\sqrt{5}}$ $cos(B)=sin(A)=\frac{1}{\sqrt{2}}$ $sin(A)cos(B)+sin(B)cos(A)=\frac{3}{\sqrt{10}}$ $\Rightarrow sin^{2}(A+B)=\boxed{\frac{P^{2}}{P^{2}+(1-Q)^{2}}}$

Tan(A+B) = Tan A + tan B /{1-tan A tanB}

But tanA +tanB = P ........TanA tanB = Q Frm here we get tan A+B Acc.to Mr , phythagoras

sinA+B get opt a