The value of $$\frac{\cos29^{\circ}\operatorname{cosec}61^{\circ}\tan45^{\circ}+2\sin35^{\circ}\sec55^{\circ}}{3\sin^242^{\circ}+3\sin^248^{\circ}}$$ is

$$\frac{\cos29^{\circ}\operatorname{cosec}61^{\circ}\tan45^{\circ}+2\sin35^{\circ}\sec55^{\circ}}{3\sin^242^{\circ}+3\sin^248^{\circ}}=\frac{\cos29^{\circ}\frac{1}{\sin61^{\circ\ }}\left(1\right)+2\sin35^{\circ}\frac{1}{\cos55^{\circ\ }}}{3\sin^242^{\circ}+3\left[\sin\left(90-42^{\circ}\right)\right]^2}$$

$$=\frac{\cos29^{\circ}\frac{1}{\sin\left(90-29^{\circ\ }\right)}+2\sin35^{\circ}\frac{1}{\cos\left(90-35^{\circ\ }\right)}}{3\sin^242^{\circ}+3\left[\cos42^{\circ\ }\right]^2}$$

$$=\frac{\cos29^{\circ}\frac{1}{\cos29^{\circ\ }}+2\sin35^{\circ}\frac{1}{\sin35^{\circ\ }}}{3\sin^242^{\circ}+3\cos^242^{\circ\ }}$$

$$=\frac{1\ +2}{3\left(\sin^242^{\circ}+\cos^242^{\circ\ }\right)}$$

$$=\frac{3}{3}$$

$$=1$$

Hence, the correct answer is Option B