Question

If $\alpha and\beta$ are two different values of $\theta$ lying between 0 and 2$\pi$ which satisfy $3\cos \theta +4\sin \theta =6$. Find the value of $\sin$ $(\alpha +\beta )$.

Answer 1

Arrange as a quadratic in tan($\theta /$2) as above $9ta{n}^2\left(\theta /2\right)-8tan\left(\theta /2\right)+3=0$ . Its root are complex as
64 - 4.3.9 = -ive i.e., $b^2$ - 4ac < 0.
Hence the given equation is wrong . If we choose 9/2 in place of 6 in R.H.S., then it becomes
$15ta{n}^2\left(\theta /2\right)-16tan\left(\theta /2\right)+3=0$
$\therefore tan\left(\alpha /2\right)+tan\left(\beta /2\right)=16/15$
$tan\left(\alpha /2\right)\cdot tan\left(\beta /2\right)=1/5$
$\therefore tan(\alpha +\beta )/2=\frac{tan\left(\alpha /2\right)+tan\left(\beta /2\right)}{1-tan\left(\alpha /2\right)tan\left(\beta /2\right)}$
$=\frac{16/15}{1-\left(1/5\right)}=\frac{4}{3}$
Now $sin(\alpha +\beta )=\frac{2tan(\alpha +\beta )/2}{1+ta{n}^2(\alpha +\beta )/2}$
$=\frac{2.\left(4/3\right)}{1+\left(16/9\right)}=\frac{24}{25}$