Respuesta :
Answer:
Follows are the solution to this question:
Step-by-step explanation:
Given:
[tex]\to P+Q+R=180[/tex]
prove that [tex](\frac{\cos p}{\sin q \cdot \sin r}) +(\frac{\cos q}{\sin r \cdot \sin p})+ (\frac{\cos r}{\sin p \cdot \sin q})=2[/tex]
Solving the L.H.S part:
[tex]\to (\frac{\cos p}{\sin q \cdot \sin r}) +(\frac{\cos q}{\sin r \cdot \sin p})+ (\frac{\cos r}{\sin p \cdot \sin q})\\\\[/tex]
Solve the above value by taking L.C.M:
[tex]\to (\frac{\cos p \sin p+\cos q \sin q +\cos r \sin r}{ \sin p \cdot \sin q \cdot \sin r}) \\\\\to (\frac{\cos p \sin p+\cos q \sin q +\cos r \sin r}{ \sin p \cdot \sin q \cdot \sin r}) \\\\[/tex]
multiply the above value by [tex]\frac{2}{2}[/tex]:
[tex]\to (\frac{\cos p \sin p+\cos q \sin q +\cos r \sin r}{ \sin p \cdot \sin q \cdot \sin r}) \times \frac{2}{2} \\\\\to \frac{2 \cos p \sin p+ 2 \cos q \sin q + 2 \cos r \sin r}{2 \sin p \cdot \sin q \cdot \sin r} \\\\\therefore \sin 2A = 2 \sin A \cdot \cos A\\\\\because\\\\\to \frac{ \sin 2p+ \sin 2q + \sin 2r}{2 \sin p \cdot \sin q \cdot \sin r} \\\\\therefore \ \\ \bold{\sin 2p+ \sin 2q + \sin 2r= 4 \sin p \cdot \sin q \cdot \sin r}\\[/tex]
[tex]\to \frac{4 \sin p \cdot \sin q \cdot \sin r}{2 \sin p \cdot \sin q \cdot \sin r} \\\\\to \frac{2 \sin p \cdot \sin q \cdot \sin r}{ \sin p \cdot \sin q \cdot \sin r} \\\\\to 2[/tex]
So, L.H.S=R.H.S
