Question

If $x=a\sin \theta ,y=b\tan \theta ,\mathrm{then}\mathrm{prove}\mathrm{that},\frac{a^2}{x^2}-\frac{b^2}{y^2}=1$.

Answer 1

$$\begin{gathered} \mathrm{Given}:x=a\sin \theta ;y=b\tan \theta \\ \mathrm{LHS}:\frac{a^2}{x^2}-\frac{b^2}{y^2} \\ =\frac{a^2}{a^2{\sin }^2\theta }-\frac{b^2}{b^2{\tan }^2\theta } \\ =\frac{1}{{\sin }^2\theta }-\frac{1}{{\tan }^2\theta } \\ =\frac{1}{{\sin }^2\theta }-\frac{{\cos }^2\theta }{{\sin }^2\theta } \\ =\frac{1-{\cos }^2\theta }{\sin {}^2\theta } \\ =\frac{{\sin }^2\theta }{{\sin }^2\theta }=1=\mathrm{LHS} \\ \mathrm{Hence},\mathrm{proved}. \end{gathered}$$