Question

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

• A. True
• B. False

Answer 1

The correct option is A

True

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