\(\triangle ABC\) मध्ये, \( \angle ABC = 90^\circ \), \( \angle C = \theta^\circ \).
\[
AB^2 + BC^2 = AC^2 (\text{पायथागोरसचे प्रमेय})
\]
दोन्ही बाजू AC\(^2\) ने भागून,
\[
\frac{AB^2}{AC^2} + \frac{BC^2}{AC^2} = 1
\]
पुढे,
\[
\left( \frac{AB}{AC} \right)^2 + \left( \frac{BC}{AC} \right)^2 = 1
\]
आणि,
\[
\frac{AB}{AC} = \sin \theta \text{आणि} \frac{BC}{AC} = \cos \theta
\]
तर,
\[
\sin^2 \theta + \cos^2 \theta = 1
\]