Correct Answer: –1

Solution : Given: $\frac{3\left(\cot ^2 47^{\circ}-\sec ^2 43^{\circ}\right)-2\left(\tan ^2 23^{\circ}-\operatorname{cosec}^2 67^{\circ}\right)}{\operatorname{cosec}^2\left(68^{\circ}+\theta\right)-\tan \left(\theta+61^{\circ}\right)-\tan ^2\left(22^{\circ}-\theta\right)+\cot \left(29^{\circ}-\theta\right)}$
= $\frac{3\left(\cot ^2(90^{\circ}- 43^{\circ})-\sec ^2 43^{\circ}\right)-2\left(\tan ^2 23^{\circ}-\operatorname {cosec^2} (90^{\circ}-23^{\circ})\right)}{\sec^2\left(90^{\circ}-(68^{\circ}+\theta)\right)-\cot \left(90^{\circ}-(\theta+61^{\circ})\right)-\tan ^2\left(22^{\circ}-\theta\right)+\cot \left(29^{\circ}-\theta\right)}$
= $\frac{3\left(\tan ^243^{\circ}-\sec ^2 43^{\circ}\right)-2\left(\tan ^2 23^{\circ}-\sec^2 23^{\circ}\right)}{\sec^2\left(22^{\circ}-\theta\right)-\cot \left(29^{\circ}-\theta\right)-\tan ^2\left(22^{\circ}-\theta\right)+\cot \left(29^{\circ}-\theta\right)}$
= $\frac{(3× -1)-(2×(-1))}{1}$
= $\frac{-3+2}{1}$
= $-1$
Hence, the correct answer is –1.