Question

(i) If $tan\theta +sin\theta =m$ and $tan\theta -sin\theta =n$, show that $m^2-n^2=4\sqrt{mn}$
(ii) If $\frac{x}{a}sin\theta +\frac{y}{b}cos\theta =1$ and $\frac{x}{a}cos\theta -\frac{y}{b}sin\theta =1$ prove that $\frac{x^2}{a^2}+\frac{y^2}{b^2}=2$

Answer 1

$\left(i\right)\tan \theta +\sin \theta =m$

Squaring both sides we get

${\tan }^2\theta +{\sin }^2\theta +2\tan \theta \sin \theta =m^2$ ........$\left(1\right)$

$\tan \theta -\sin \theta =n$

Squaring both sides we get

${\tan }^2\theta +{\sin }^2\theta -2\tan \theta \sin \theta =n^2$ .........$\left(2\right)$

Subtract $\left(2\right)$ from $\left(1\right)$ we get

$m^2-n^2={\tan }^2\theta +{\sin }^2\theta +2\tan \theta \sin \theta -{\tan }^2\theta -{\sin }^2\theta +2\tan \theta \sin \theta =4\tan \theta \sin \theta$

$\Rightarrow m^2-n^2=4\sqrt{{\tan }^2\theta {\sin }^2\theta }$

$\Rightarrow m^2-n^2=4\sqrt{\frac{{\sin }^2\theta }{{\cos }^2\theta }(1-{\cos }^2\theta )}$

$\Rightarrow m^2-n^2=4\sqrt{\frac{{\sin }^2\theta }{{\cos }^2\theta }-\frac{{\sin }^2\theta }{{\cos }^2\theta }{\cos }^2\theta }$

$\Rightarrow m^2-n^2=4\sqrt{{\tan }^2\theta -{\sin }^2\theta }$

$\Rightarrow m^2-n^2=4\sqrt{(\tan \theta +\sin \theta )(\tan \theta -\sin \theta )}$

$\therefore m^2-n^2=4\sqrt{mn}$

Hence proved.

$\left(ii\right)$ Consider $\frac{x}{a}\sin \theta +\frac{y}{b}\cos \theta =1$

Squaring both sides, we get

${(\frac{x}{a}\sin \theta +\frac{y}{b}\cos \theta )}^2=1$

$\Rightarrow \frac{x^2}{a^2}{\sin }^2\theta +\frac{y^2}{b^2}{\cos }^2\theta +2\frac{x}{a}\sin \theta \times \frac{y}{b}\cos \theta =1$

$\Rightarrow \frac{x^2}{a^2}{\sin }^2\theta +\frac{y^2}{b^2}{\cos }^2\theta +2\frac{xy}{ab}\sin \theta \cos \theta =1$ .........$\left(1\right)$

Consider $\frac{x}{a}\sin \theta -\frac{y}{b}\cos \theta =1$

Squaring both sides, we get

${(\frac{x}{a}\sin \theta -\frac{y}{b}\cos \theta )}^2=1$

$\Rightarrow \frac{x^2}{a^2}{\sin }^2\theta +\frac{y^2}{b^2}{\cos }^2\theta -2\frac{x}{a}\sin \theta \times \frac{y}{b}\cos \theta =1$

$\Rightarrow \frac{x^2}{a^2}{\sin }^2\theta +\frac{y^2}{b^2}{\cos }^2\theta -2\frac{xy}{ab}\sin \theta \cos \theta =1$ .........$\left(2\right)$

Adding $\left(1\right)$ and $\left(2\right)$ we get

$\Rightarrow \frac{x^2}{a^2}{\sin }^2\theta +\frac{y^2}{b^2}{\cos }^2\theta +2\frac{xy}{ab}\sin \theta \cos \theta +\frac{x^2}{a^2}{\sin }^2\theta +\frac{y^2}{b^2}{\cos }^2\theta -2\frac{xy}{ab}\sin \theta \cos \theta =1+1$

$\Rightarrow \frac{x^2}{a^2}({\sin }^2\theta +{\cos }^2\theta )+\frac{y^2}{b^2}({\sin }^2\theta +{\cos }^2\theta )=2$

We know that ${\sin }^2\theta +{\cos }^2\theta =1$

$\therefore \frac{x^2}{a^2}+\frac{y^2}{b^2}=2$

Hence proved.