Question

Prove that:

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

Answer 1

Solution :-

$\frac{1+ta{n}^{2}A}{1+co{t}^{2}A}=\frac{1+\frac{si{n}^{2}A}{co{s}^{2}A}}{1+\frac{co{s}^{2}A}{si{n}^{2}A}}$

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

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

and $(\frac{1-tanA}{1-cotA}{)}^2=(\frac{1-\frac{sinA}{cosA}}{1-\frac{cosA}{sinA}}{)}^2$

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

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

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