Question

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

Answer 1

Given, $3cotA=4$

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

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

Now, we know that $tanA=\frac{sinA}{cosA}$

$\therefore \frac{sinA}{cosA}=\frac{3}{4}$

Also, $si{n}^{2}A+co{s}^{2}A=1$

$sinA=\frac{3\times cosA}{4}$

$\therefore {\frac{3cosA}{4}}^2+co{s}^{2}A=1$

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

$co{s}^{2}A(1+\frac{9}{16})=1$

$co{s}^{2}A\left(\frac{25}{16}\right)=1$

$co{s}^{2}A=\frac{16}{25}$

$\therefore cosA=\sqrt{\frac{16}{25}}=\frac{4}{5}$

$sinA=tanA\times cosA=\frac{3}{4}\times \frac{4}{5}=\frac{3}{5}$


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

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

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

$=\frac{7}{25}$

$\therefore LHS=RHS$
Hence Proved