Prove the following :

$\dfrac{\text{sin (90° - A)}}{\text{cosec (90° - A)}} + \dfrac{\text{cos (90° - A)}}{\text{sec (90° - A)}}$ = 1

We know that,

cos (90 - θ) = sin θ, sec (90 - θ) = cosec θ, sin (90 - θ) = cos θ and cosec (90 - θ) = sec θ.

Solving L.H.S. of the equation, we get :

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

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

Hence, proved that $\dfrac{\text{sin (90° - A)}}{\text{cosec (90° - A)}} + \dfrac{\text{cos (90° - A)}}{\text{sec (90° - A)}}$ = 1.