Question

State true or false

If $3cotA=4\text{, then}\frac{1-ta{n}^{2}A}{1+ta{n}^{2}A}=co{s}^{2}A-si{n}^{2}A$

• A. True
• B. False

Answer 1

The correct option is A True

Given $3cotA=4$

$\Rightarrow cotA=\frac{4}{3}$

$tanA=\frac{1}{cotA}=\frac{3}{4}$

Therefore, $sinA=\frac{3}{5}$ and $cosA=\frac{4}{5}$ ..... (3-4-5 Pythagoras Triplet)

Now, LHS = $\frac{1-ta{n}^{2}A}{1+ta{n}^{2}A}$

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

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

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

$=\frac{7}{25}$

RHS = $co{s}^{2}A-si{n}^{2}A$

$={\left(\frac{4}{5}\right)}^2-{\left(\frac{3}{5}\right)}^2$

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

$=\frac{7}{25}$

$\therefore LHS=RHS$