If $$\sin \theta - \cos \theta = \frac{1}{29}$$ find the value of $$\sin \theta + \cos \theta$$.

Given,  $$\sin\theta-\cos\theta=\frac{1}{29}$$

$$\Rightarrow$$  $$\left(\sin\theta-\cos\theta\right)^2=\left(\frac{1}{29}\right)^2$$

$$\Rightarrow$$  $$\sin^2\theta+\cos^2\theta-2\sin\theta\ \cos\theta\ =\frac{1}{841}$$

$$\Rightarrow$$  $$1-2\sin\theta\ \cos\theta\ =\frac{1}{841}$$

$$\Rightarrow$$  $$2\sin\theta\ \cos\theta\ =1-\frac{1}{841}$$

$$\Rightarrow$$  $$2\sin\theta\ \cos\theta\ =\frac{840}{841}$$

$$\Rightarrow$$  $$1+2\sin\theta\ \cos\theta\ =1+\frac{840}{841}$$

$$\Rightarrow$$  $$\sin^2\theta\ +\cos^2\theta\ +2\sin\theta\ \cos\theta\ =\frac{1681}{841}$$

$$\Rightarrow$$  $$\left(\sin\theta\ +\cos\theta\ \right)^2=\frac{1681}{841}$$

$$\Rightarrow$$  $$\sin\theta\ +\cos\theta\ =\sqrt{\frac{1681}{841}}$$

$$\Rightarrow$$  $$\sin\theta\ +\cos\theta\ =\frac{41}{29}$$

Hence, the correct answer is Option D