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sin A cos A - $\dfrac{\text{sin A cos (90° - A) cos A}}{\text{sec (90° - A)}} - \dfrac{\text{cos A sin (90° - A) sin A}}{\text{cosec (90° - A)}}$ = 0

By formula,

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

$\Rightarrow \text{sin A cos A} - \dfrac{\text{sin A sin A cos A}}{\text{cosec A}} - \dfrac{\text{cos A cos A sin A}}{\text{sec A}} \\[1em] \Rightarrow \text{sin A cos A} - \dfrac{\text{sin}^2 A \text{cos A}}{\dfrac{1}{\text{sin A}}} - \dfrac{\text{cos}^2 A \text{sin A}}{\dfrac{1}{\text{cos A}}} \\[1em] \Rightarrow \text{sin A cos A} - \text{sin}^3 A \text{ cos A} - \text{cos}^3 A \text{sin A} \\[1em] \Rightarrow \text{sin A cos A} - \text{sin A cos A}(\text{sin}^2 A + \text{cos}^2 A)$

By formula,

sin² A + cos² A = 1.

$\Rightarrow \text{sin A cos A} - \text{sin A cos A} \\[1em] \Rightarrow 0.$

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

Hence, proved that sin A cos A - $\dfrac{\text{sin A cos (90° - A) cos A}}{\text{sec (90° - A)}} - \dfrac{\text{cos A sin (90° - A) sin A}}{\text{cosec (90° - A)}}$ = 0.