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If $3 cot\ A = 4$, check whether $\frac{1−tan^2A}{1+tan^2A} = cos^2 A – sin^2 A$ 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 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, $\frac{1−tan^2A}{1+tan^2A} = cos^2 A – sin^2 A$.