Question

Prove $\frac{\cot A\cos A}{\cot A+\cos A}=\frac{\cot A-\cos A}{\cot A\cos A}$.

Answer 1

Simplifying the LHS of $\frac{\cot A\cos A}{\cot A+\cos A}=\frac{\cot A-\cos A}{\cot A\cos A}$.

$\frac{\cot A\cos A}{\cot A+\cos A}=\frac{\frac{\cos A}{\sin A}\cos A}{\frac{\cos A}{\sin A}+\cos A}$

$=\frac{{\cos }^{2}A}{\cos A+\cos A\sin A}$

$=\frac{1-{\sin }^{2}A}{\cos A\left(1+\sin A\right)}$

$=\frac{\left(1-\sin A\right)\left(1+\sin A\right)}{\cos A\left(1+\sin A\right)}$

$=\frac{1-\sin A}{\cos A}$

Now, simplifying the RHS of $\frac{\cot A-\cos A}{\cot A\cos A}=\frac{\cot A-\cos A}{\cot A\cos A}$.

$\frac{\cot A-\cos A}{\cot A\cos A}=\frac{\frac{\cos A}{\sin A}-\cos A}{\frac{\cos A}{\sin A}\times \cos A}$

$=\frac{\cos A-\cos A\sin A}{{\cos }^{2}A}$

$=\frac{\cos A\left(1-\sin A\right)}{{\cos }^{2}A}$

$=\frac{1-\sin A}{\cos A}$

This shows that $LHS=RHS$.