Question

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

Answer 1

Given: $3\cot A=4$

$⟹\cot A=\frac{4}{3}$

$\tan A=\frac{3}{4}=\frac{BC}{AB}$

By Pythgoras theorem, $AC=5$ $[\because (AC{)}^2=(AB{)}^2+(BC{)}^2]$

Hence $\cos A=\frac{4}{5}$

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

LHS $=\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}$

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

$=\frac{\frac{7}{16}}{\frac{25}{16}}$

$=\frac{7}{25}$......(1)

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

$=\frac{16}{25}-\frac{9}{25}=\frac{7}{25}$.....(2)

From equation (1) and (2), we get

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