Trigonometry and Properties of Quadratic Equation

If $\displaystyle \tan A$ and $\displaystyle \tan B$ are the roots of the equation $\displaystyle {{x}^{2}}-px+q=0$, where $\displaystyle p$ and $\displaystyle q$ are real constants, find the value of $\displaystyle {{\sin }^{2}}(A+B)$ in terms of $\displaystyle p$ and  $\displaystyle q$.

Solution

             $\displaystyle \ \ \ \ \tan A$ and $\displaystyle \tan B$ are the roots of the equation $\displaystyle {{x}^{2}}-px+q=0$,

             $\displaystyle \therefore {{x}^{2}}-px+q=(x-\tan A)(x-\tan B)$

             $\displaystyle \therefore {{x}^{2}}-px+q={{x}^{2}}-(\tan A+\tan B)x+\tan A\cdot \tan B$
         
             $\displaystyle \therefore \tan A+\tan B=p$ and $\displaystyle \tan A\cdot \tan B=q$

             $\displaystyle \therefore \frac{{\tan A+\tan B}}{{1-\tan A\tan B}}=\frac{p}{{1-q}}$

             $\displaystyle \therefore \tan (A+B)=\frac{p}{{1-q}}$

             $\displaystyle \therefore \frac{{\sin (A+B)}}{{\cos (A+B)}}=\frac{p}{{1-q}}$

             $\displaystyle \therefore \cos (A+B)=\frac{{1-q}}{p}\cdot \sin (A+B)$

             $\displaystyle \ \ \ $ Since $\displaystyle {{\sin }^{2}}(A+B)+{{\cos }^{2}}(A+B)=1$

             $\displaystyle \ \ \ \ {{\sin }^{2}}(A+B)+{{\left( {\frac{{1-q}}{p}} \right)}^{2}}{{\sin }^{2}}(A+B)=1$

             $\displaystyle \ \ \ \ {{\sin }^{2}}(A+B)\left( {1+\frac{{{{{\left( {1-q} \right)}}^{2}}}}{{{{p}^{2}}}}} \right)=1$

             $\displaystyle \therefore {{\sin }^{2}}(A+B)=\frac{{{{p}^{2}}}}{{{{p}^{2}}+{{{\left( {1-q} \right)}}^{2}}}}$

စာဖတ်သူ၏ အမြင်ကို လေးစားစွာစောင့်မျှော်လျက်!