Description for Correct answer:
\( \Large sin \theta +cos \theta =1 \)

On squaring both sides, we get

\( \Large (sin \theta +cos \theta)^{2} =1 \)

\( \Large sin^{2} \theta +cos^{2} \theta +2 \ sin \theta cos \theta =1 \)

\( \Large 1+2sin \theta cos \theta =1 \)

\( \Large 2sin \theta cos \theta =0 \)

\( \Large sin \theta cos \theta =0 \)