Question

If $\tan \theta =\frac{m}{n},$ and sin $\theta =\frac{K}{L}$, then find the value of $\frac{L^2-K^2}{L^2{n}^2}\times (m^2+n^2)$

Answer 1

Let $P=m\alpha$ and $B=n\alpha$

$\therefore \tan \theta =\frac{P}{B}=\frac{m}{n}$
$H^2=P^2+B^2$ ........ [By Pythagoras theorem]
$⟹H^2=m^2{\alpha }^2+n^2{\alpha }^2$
$⟹H=\alpha \sqrt{m^2+n^2}$
$\therefore \sin \theta =\frac{P}{H}=\frac{m\alpha }{\alpha \sqrt{m^2+n^2}}$
$\therefore \sin \theta =\frac{m}{\sqrt{m^2+n^2}}$ ....... $\left(1\right)$

Consider, $\frac{L^2-K^2}{L^2{n}^2}\times (m^2+n^2)$

$=\frac{1-\frac{K^2}{L^2}}{n^2}\times (m^2+n^2)$

$=\frac{(1-(\sin \theta {)}^2)}{n^2}\times (m^2+n^2)$

$=\frac{1-\frac{m^2}{m^2+n^2}}{n^2}\times (m^2+n^2)$

$=\frac{n^2(m^2+n^2)}{(m^2+n^2)n^2}=1$

Hence, the answer is $1$.