Question

If $\alpha$ is the measure of acute angle and $3sin\alpha =2cos\alpha$, then prove that $⟮\frac{1-ta{n}^2\alpha }{1+ta{n}^2\alpha }{⟯}^2+⟮\frac{2tan\alpha }{1+ta{n}^2\alpha }{⟯}^2=1$.

Answer 1

$3\sin \alpha =2\cos \alpha \Rightarrow \tan \alpha =\frac{2}{3}$

Given question $={\left(\frac{1-\frac{4}{9}}{1+\frac{4}{9}}\right)}^2+{\left(\frac{2\times \frac{2}{3}}{1+\frac{4}{9}}\right)}^2$

$=\frac{5^2}{{13}^2}+\frac{{12}^2}{{13}^2}=\frac{{13}^2}{{13}^2}$

$=1$

Hence proved