Question

Prove:$\frac{1+{\tan }^{2}A}{1+co{t}^{2}A}={\left[\frac{1-\tan A}{1-cotA}\right]}^2={\tan }^{2}A.$

Answer 1

We know that

$$\begin{gathered} 1+{\tan }^{2}A=se{c}^{2}A \\ 1+co{t}^{2}A=\cos e{c}^{2}A \end{gathered}$$

Consider LHS

$$\begin{gathered} \frac{1+{\tan }^{2}A}{1+co{t}^{2}A} \\ =\frac{se{c}^{2}A}{\cos e{c}^{2}A}[u\sin gidentities] \\ =\frac{{\displaystyle \frac{1}{{\cos }^{2}A}}}{{\displaystyle \frac{1}{{\sin }^{2}A}}}[\because secA=\frac{1}{\cos A},\cos ecA=\frac{1}{\sin A}] \\ =\frac{{\sin }^{2}A}{{\cos }^{2}A} \\ ={\tan }^{2}A \end{gathered}$$

Consider RHS

$$\begin{gathered} {\left[\frac{1-\tan A}{1-cotA}\right]}^2 \\ ={\left[\frac{1-{\displaystyle \frac{\sin A}{\cos A}}}{1-{\displaystyle \frac{\cos A}{\sin A}}}\right]}^2 \\ ={\left[\frac{{\displaystyle \frac{\cos A-\sin A}{\cos A}}}{{\displaystyle \frac{\sin A-\cos A}{\sin A}}}\right]}^2 \\ =\frac{{\displaystyle \frac{{\left(\cos A-\sin A\right)}^2}{{\cos }^{2}A}}}{{\displaystyle \frac{{\left(\sin A-\cos A\right)}^2}{{\sin }^{2}A}}} \\ =\frac{{\displaystyle \frac{{\left(\cos A-\sin A\right)}^2}{{\cos }^{2}A}}}{{\displaystyle \frac{{\left(\cos A-\sin A\right)}^2}{{\sin }^{2}A}}} \\ =\frac{{\displaystyle \frac{1}{{\cos }^{2}A}}}{{\displaystyle \frac{1}{{\sin }^{2}A}}} \\ =\frac{{\sin }^{2}A}{{\cos }^{2}A} \\ ={\tan }^{2}A \end{gathered}$$

Since, the values of LHS and RHS are the same.

Hence proved.