Show that :

$\dfrac{\text{sin A}}{\text{sin(90° - A)}} + \dfrac{\text{cos A}}{\text{cos(90° - A)}}$ = sec A cosec A

To prove:

$\dfrac{\text{sin A}}{\text{sin(90° - A)}} + \dfrac{\text{cos A}}{\text{cos(90° - A)}}$ = sec A cosec A

By formula,

sin (90° - A) = cos A and cos (90° - A) = sin A.

Substituting above values in L.H.S. :

$\Rightarrow \dfrac{\text{sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin A cos A}}$

By formula,

sin² A + cos² A = 1

$\Rightarrow \dfrac{1}{\text{sin A cos A}} \\[1em] \Rightarrow \dfrac{1}{\text{sin A}} \times \dfrac{1}{\text{cos A}} \\[1em] \Rightarrow \text{cosec A sec A}.$

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

Hence, proved that $\dfrac{\text{sin A}}{\text{sin(90° - A)}} + \dfrac{\text{cos A}}{\text{cos(90° - A)}}$ = sec A cosec A.