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Q.
If $l \sin \theta-m \cos \theta=\sqrt{l^2+m^2}$ and $\frac{\cos ^2 \theta}{a^2}+\frac{\sin ^2 \theta}{b^2}=\frac{1}{l^2+m^2}$ then which of the following options is correct?
Trigonometry
Solution:
$l \sin \theta-m \cos \theta=\sqrt{l^2+m^2}$
Divide both sides by $\sqrt{l^2+m^2}$
$ \left(\frac{1}{\sqrt{l^2+m^2}}\right) \sin \theta+\left(\frac{1-m}{\sqrt{l^2+m^2}}\right) \cos \theta=1 $
$\Rightarrow \sin ^2 \theta+\cos ^2 \theta=1$
$\Rightarrow \sin \theta \cdot \sin \theta+\cos \theta \cdot \cos \theta=1 $
$ \Rightarrow\left\{\because \sin \theta=\frac{l}{\sqrt{l^2+n^2}} \text { and } \cos \theta=\frac{-m}{\sqrt{l^2+m^2}}\right\} $
$ \Rightarrow \frac{\cos ^2 \theta}{a^2}+\frac{\sin ^2 \theta}{b^2}=\frac{1}{l^2+m^2}$
Putting the value of $\sin \theta$ and $\cos \theta$, we get
$\frac{m^2}{\left(l^2+m^2\right) a^2}+\frac{l^2}{\left(l^2+m^2\right) b^2}=\frac{1}{l^2+a^2}$
Cancel the same terms on both sides.
$ \frac{1}{l^2+m^2}\left(\frac{m^2}{a^2}+\frac{l^2}{b^2}\right)=\frac{1}{l^2+a^2} $
$ \therefore \frac{m^2}{a^2}+\frac{l^2}{b^2}=1$