Question

guys plz help me.........​

Answer 1

Step-by-step explanation:

Use the formulae:

(i) $2cosA cosB= cos(A+B) +cos(A-B)$

(ii)$2sinAsinB = cos(A-B) - cos(A+B)$

$=\frac{cos2\theta cos3\theta - cos2\theta cos7\theta + cos \theta cos10\theta}{sin4\theta sin3\theta - sin2\theta sin5\theta + sin4\theta sin7\theta} \\=\frac{cos5\theta +cos\theta -cos5\theta-cos9\theta+cos9\theta+cos11\theta}{cos\theta - cos7\theta -cos3\theta +cos7\theta + cos3\theta - cos11\theta}\\=\frac{cos\theta + cos11\theta}{cos\theta - cos11\theta}$

Again use the same formulae in reverse.

$=\frac{2cos6\theta cos5\theta}{2sin5\theta sin6\theta}\\=cot6\theta . cot5\theta$