If $3\ cot\ A = 4$, check whether $\frac{1−tan^2\ A}{1+tan^2\ A} = cos^2\ A - sin^2\ A$ or not.

Given:

$3\ cot\ A = 4$.

To do:

We have to check whether $\frac{1−tan^2A}{1+tan^2A} = cos^2 A – sin^2 A$ or not.

Solution:  

Let, in a triangle $ABC$ right-angled at $B$, $3\ cot\ A = 4$.

This implies,

$cot\ A = \frac{4}{3}$

We know that,

In a right-angled triangle $ABC$ with a right angle at $B$,

By Pythagoras theorem,

$AC^2=AB^2+BC^2$

By trigonometric ratios definitions,

$sin\ A=\frac{Opposite}{Hypotenuse}=\frac{BC}{AC}$

$cos\ A=\frac{Adjacent}{Hypotenuse}=\frac{AB}{AC}$

$tan\ A=\frac{Opposite}{Adjacent}=\frac{BC}{AB}$

$cot\ A=\frac{Adjacent}{Opposite}=\frac{AB}{BC}$

Here,

$AC^2=AB^2+BC^2$

$\Rightarrow AC^2=(4)^2+(3)^2$

$\Rightarrow AC^2=16+9$

$\Rightarrow AC=\sqrt{25}=5$

Therefore,

$sin\ A=\frac{BC}{AC}=\frac{3}{5}$

$cos\ A=\frac{AB}{AC}=\frac{4}{5}$

 $tan\ A=\frac{BC}{AB}=\frac{3}{4}$

Let us consider LHS,

$\frac{1−tan^2A}{1+tan^2A} =\frac{1-(\frac{3}{4})^2}{1+(\frac{3}{4})^2}$

$=\frac{1-\frac{9}{16}}{1+\frac{9}{16}}$

$=\frac{\frac{16-9}{16}}{\frac{16+9}{16}}$

$=\frac{7}{25}$

Now, consider RHS

$cos^2 A – sin^2 A=(\frac{4}{5})^2-(\frac{3}{5})^2$

$=\frac{16}{25}-\frac{9}{25}$

$=\frac{16-9}{25}$

$=\frac{7}{25}$

LHS $=$ RHS

Therefore, $\frac{1−tan^2A}{1+tan^2A} = cos^2 A – sin^2 A$.