Prove the following :

$\dfrac{\text{cos (90° - A) sin (90° - A)}}{\text{tan} (90° - \text{ A})} = 1 - \text{cos}^ 2 \text{ A}$

We know that,

cos (90 - θ) = sin θ, tan (90 - θ) = cot θ and sin (90 - θ) = cos θ.

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

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

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

Hence, proved that $\dfrac{\text{cos (90° - A) sin (90° - A)}}{\text{tan} (90° - \text{ A})} = 1 - \text{cos}^ 2 \text{ A}$.