Question

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

$\left(ii\right)\frac{1+{\tan }^{2}A}{1+co{t}^{2}A}={\left(\frac{1-\tan A}{1-cotA}\right)}^2={\tan }^{2}A$

Answer 1

Proof:

Consider the LHS side of the given expression.

$$\begin{gathered} \mathrm{LHS}=\frac{1+{\tan }^{2}A}{1+co{t}^{2}A} \\ \Rightarrow =\frac{1+{\tan }^{2}A}{1+{\displaystyle \frac{1}{{\tan }^{2}A}}}\left[\because cotA=\frac{1}{\tan A}\right] \\ \Rightarrow ={\tan }^{2}A \end{gathered}$$

and

$$\begin{gathered} {\left(\frac{1-\tan A}{1-cotA}\right)}^2={\left(\frac{1-\tan A}{1-{\displaystyle \frac{1}{\tan A}}}\right)}^2\left[\because cotA=\frac{1}{\tan A}\right] \\ \Rightarrow ={\left(-\frac{1-\tan A}{1-\tan A}·\tan A\right)}^2 \\ \Rightarrow ={\tan }^{2}A \\ \Rightarrow =\mathrm{RHS} \end{gathered}$$

Hence proved.