Question

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

Answer 1

$\angle B={90}^o$

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

Let $AB=4x,BC=3x$.

$A{C}^2=AB+B{C}^2$

$A{C}^2=16k^2+9x^2$

$A{C}^2=25x^2$

$AC=5x$

Now, $\tan A=\frac{3}{4}$

LHS : $\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}=\frac{1-\frac{9}{16}}{1+\frac{9}{16}}=\frac{7}{25}$

RHS : ${\cos }^{2}A-{\sin }^{2}A={\left(\frac{4}{5}\right)}^2-{\left(\frac{3}{5}\right)}^2=\frac{16}{25}-\frac{9}{25}=\frac{7}{25}$

$\Rightarrow LHS=RHS$

$\therefore \frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}={\cos }^{2}A-{\sin }^{2}A$