Question

If tan theta = M/N then show that Msin theta - Ncos theta / Msin theta + Ncos theta = M^2 - N^2 / M^2 + N^2

Answer 1

plz tell ...if you don't understand last Line...........

Answer 2

$\dfrac{m\sin \theta - n\cos \theta}{m\sin \theta+n\cos \theta}=\dfrac{m^2-n^2}{m^2+n^2}$

Step-by-step explanation:

We have,

$\tan \theta=\dfrac{m}{n}$     ...... (1)

To prove that, $\dfrac{m\sin \theta - n\cos \theta}{m\sin \theta+n\cos \theta}=\dfrac{m^2-n^2}{m^2+n^2}$.

L.H.S.$=\dfrac{m\sin \theta - n\cos \theta}{m\sin \theta+n\cos \theta}$

Dividng nemerator and denominator by $\cos \theta$, we get

$=\dfrac{m\dfrac{\sin \theta }{\cos \theta}-n\dfrac{\cos \theta}{\cos \theta}}{m\dfrac{\sin \theta }{\cos \theta}+n\dfrac{\cos \theta }{\cos \theta}}$

$=\dfrac{m\tan \theta-n}{m\tan \theta+n}$

Using (1), we get

$=\dfrac{m(\dfrac{m}{n})-n}{m(\dfrac{m}{n})+n}$

$=\dfrac{m^2-n^2}{m^2+n^2}$

= R.H.S., proved.

Hence, $\dfrac{m\sin \theta - n\cos \theta}{m\sin \theta+n\cos \theta}=\dfrac{m^2-n^2}{m^2+n^2}$.