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

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

$=\frac{{\cos }^{2}A-si{n}^{2}A}{{\cos }^{2}A+si{n}^{2}A}$

$={\cos }^{2}A-{\sin }^{2}A$

$\therefore \frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}$ is always equal to

${\cos }^{2}A-{\sin }^{2}A$ for all values of $\cot A$.