Question

Question 5 (x)
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$\left(\frac{1+ta{n}^{2}A}{1+co{t}^{2}A}\right)=(\frac{1-tanA}{1-cotA}{)}^2=ta{n}^{2}A$

Answer 1

$\left(i\right)\frac{1+ta{n}^{2}A}{1+co{t}^{2}A})=\frac{1+ta{n}^{2}A}{1+\frac{1}{ta{n}^{2}A}})=\frac{1+ta{n}^{2}A}{\left[\frac{(1+ta{n}^{2}A)}{ta{n}^{2}A}\right]}=ta{n}^{2}A$
$\left(ii\right)(\frac{1-tanA}{1-cotA}{)}^2=[\frac{1-\frac{sinA}{cosA}}{1-\frac{cosA}{sinA}}{]}^2=\frac{\left(si{n}^{2}A\right)\times (cosA-sinA{)}^2}{\left(co{s}^{2}A\right)\times (sinA-cosA{)}^2}=\frac{\left(si{n}^{2}A\right)}{\left(co{s}^{2}A\right)}\times (-1{)}^2=\frac{si{n}^{2}A}{co{s}^{2}A}=ta{n}^{2}A$
$\therefore$ $\left(\frac{1+ta{n}^{2}A}{1+co{t}^{2}A}\right)=(\frac{1-tanA}{1-cotA}{)}^2=ta{n}^{2}A$.