Question

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

• A. True
• B. False

Answer 1

The correct option is B False

Given

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

Considering the term $\left(\frac{1+{\tan }^{2}A}{1+{\cot }^{2}A}\right)$

$⟹\left(\frac{1+{\tan }^{2}A}{1+\frac{1}{{\tan }^{2}A}}\right)$

$(\cot \theta =\frac{1}{\tan \theta })$

$⟹\frac{{\tan }^{2}A+1}{\left(\frac{1+{\tan }^{2}A}{{\tan }^{2}A}\right)}$

$⟹\frac{1}{\frac{1}{{\tan }^{2}A}}$

$⟹{\tan }^{2}A\cdots \left(1\right)$

Considering the term ${\left(\frac{1-{\tan }^{2}A}{1-{\cot }^{2}A}\right)}^2$

$⟹{\left(\frac{1-{\tan }^{2}A}{1-\frac{1}{{\tan }^{2}A}}\right)}^2$

$(\cot \theta =\frac{1}{\tan \theta })$

$⟹{\left(\frac{1-{\tan }^{2}A}{\frac{{\tan }^{2}A-1}{{\tan }^{2}A}}\right)}^2$

$⟹{\left(\frac{-1}{\frac{1}{{\tan }^{2}A}}\right)}^2$

$⟹{\tan }^{4}A\cdots \left(2\right)$

$\therefore$ From $\left(1\right),\left(2\right)$

The equations are not true.