Question

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

$(\cos ecA–\sin A)(secA–\cos A)=\frac{1}{(\tan A+cotA)}$

Answer 1

Proof:

Step1: Simplify LHS

Consider the LHS of the given equation.

$$\begin{gathered} \mathrm{LHS}=(\cos ecA–\sin A)(secA–\cos A) \\ =\left(\frac{1}{\sin A}–\sin A\right)\left(\frac{1}{\cos A}–\cos A\right)\left[\mathrm{Use}\cos ecA=\frac{1}{\sin A}\mathrm{and}secA=\frac{1}{\cos ecA}\right] \\ =\left(\frac{1-{\sin }^{2}A}{\sin A}\right)\left(\frac{1-{\cos }^{2}A}{\cos A}\right) \\ =\frac{{\cos }^{2}A}{\sin A}·\frac{{\sin }^{2}A}{\cos A}\left[\mathrm{Use}{\sin }^{2}A+{\cos }^{2}A=1\right] \\ =\sin A·\cos A \end{gathered}$$

Step2: Simplify RHS

Now consider the RHS of the given equation.

$$\begin{gathered} \mathrm{RHS}=\frac{1}{(\tan A+cotA)} \\ =\frac{1}{{\displaystyle \frac{\sin A}{\cos A}}+{\displaystyle \frac{\cos A}{\sin A}}}\left[\mathrm{Use}\tan A=\frac{\sin A}{\cos A}\mathrm{and}cotA=\frac{\cos A}{\sin A}\right] \\ =\frac{\sin A·\cos A}{{\sin }^{2}A+{\cos }^{2}A}\left[\because {\sin }^{2}A+{\cos }^{2}A=1\right] \\ =\sin A·\cos A \end{gathered}$$

$\mathrm{LHS}=\mathrm{RHS}$

Hence proved.