Question

If 3 cot A = 4, evaluate 1−tan2A/1+tan2A

Answer 1

Answer:

The value of $\frac{1-tan2A}{1+tan2A}$ is $\frac{-31}{17}$

Step-by-step explanation:

Given:

$CotA=\frac{3}{4}$ or $TanA=\frac{4}{3}$

We need to find the value of $\frac{1-tan2A}{1+tan2A}$

Let us find $Tan2A$ first

The formula for $Tan 2A$ is

$Tan2A=\frac{2TanA}{1-Tan^2A}$

By substituting the value we get :

$Tan2A=\frac{2(\frac{4}{3} )}{1-(\frac{4}{3} )^2}\\\\Tan2A=\frac{\frac{8}{3} }{1-\frac{16}{9} }$

$Tan2A=\frac{\frac{8}{3} }{-\frac{7}{9} } \\\\Tan2A=\frac{8}{3} *\frac{-9}{7}\\\\Tan2A=\frac{-24}{7}$

Substituting the value we get:

$\frac{1-tan2A}{1+tan2A}=\frac{1-(\frac{-24}{7} ) }{1+(\frac{-24}{7} )}$

$=\frac{(\frac{7+24}{7} ) }{(\frac{7-24}{7} )}\\\\=\frac{(\frac{31}{7} ) }{(\frac{-17}{7} )}\\\\=\frac{-31}{17}$