Question

Prove that $\left(\frac{1+{\tan }^{2}A}{1+{\cot }^{2}A}\right)={\left(\frac{1-\tan A}{1-\cot A}\right)}^2={\tan }^{2}A$.

Answer 1

LHS =$\frac{(1+ta{n}^{2}A)}{(1+co{t}^{2}A)}$

=$\frac{se{c}^{2}A}{cose{c}^{2}A}$

=$\frac{si{n}^{2}A}{co{s}^{2}A}$

=$ta{n}^{2}A$

Now,$(\frac{1-tanA}{1-cotA}{)}^2$

=$\frac{1+ta{n}^{2}A-2tanA}{1+co{t}^{2}A-2cotA}$

=$\frac{se{c}^{2}A-2tanA}{cose{c}^{2}A-2cotA}$

=$\frac{1}{co{s}^{2}A}$$-$ $\frac{2sinA}{cosA}$$\times \frac{1}{\frac{1}{si{n}^{2}A}-\frac{2cosA}{sinA}}$

=$\frac{1-2sinAcosA}{co{s}^{2}A}$$\times \frac{si{n}^{2}A}{1-2cosAsinA}$

= $\frac{si{n}^{2}A}{co{s}^{2}A}$

= $ta{n}^{2}A$

Hence,$\frac{1+ta{n}^{2}A}{1+co{t}^{2}A}$ = $(\frac{1-tanA}{1-cotA}{)}^2=$$ta{n}^{2}A$