Question

Prove that:

\[\frac{\sin A}{\sin (90^\circ - A)} + \frac{\cos A}{\cos (90^\circ - A)} = \sec A \operatorname{cosec} A\]

Answer 1

By formula,

\[\sin (90^\circ - A) = \cos A\] and \[\cos (90^\circ - A) = \sin A.\]

Substituting above values in L.H.S.:

\[\Rightarrow \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}\]

\[\Rightarrow \frac{\sin^2 A + \cos^2 A}{\sin A \cos A}\]

By formula,

\[\sin^2 A + \cos^2 A = 1.\]

\[\Rightarrow \frac{1}{\sin A \cos A}\]

\[\Rightarrow \frac{1}{\sin A} \times \frac{1}{\cos A}\]

\[\Rightarrow \operatorname{cosec} A \sec A.\]

Since, L.H.S. = R.H.S.