Correct Answer: $\frac{1}{41}$

Solution : Given: $\sin \theta+\cos \theta=\frac{1}{29}$
We know that
$\sin \theta+\cos \theta=\frac{1}{29}$
Squaring both sides,
$(\sin \theta+\cos \theta)^2=\frac{1}{29^2}$
⇒ $1+2\sin \theta\cos \theta=\frac{1}{841}$
⇒ $2\sin \theta\cos \theta=\frac{1}{841}-1=-\frac{840}{841}$...(i)
Now,
$(\sin \theta-\cos \theta)^2=1-2\sin \theta\cos \theta$
From equation (i)
$(\sin \theta-\cos \theta)^2=1-(-\frac{840}{841})=\frac{1681}{841}$
⇒ $(\sin \theta-\cos \theta)=\sqrt{(\frac{1681}{841})}=\frac{41}{29}$
⇒ $(\sin \theta-\cos \theta) =\frac{41}{29}$
⇒ $\frac{\operatorname{sin} \theta+\operatorname{cos} \theta}{\operatorname{sin} \theta-\operatorname{cos} \theta}=\frac{1}{29}×\frac{29}{41}=\frac{1}{41}$
Hence, the correct answer is $\frac{1}{41}$.