(A)OR(B)

If $\displaystyle x=h+a \cos \theta, y=k+b \sin \theta$, then prove that : \[\left(\frac{x-h}{a}\right)^{2}+\left(\frac{y-k}{b}\right)^{2}=1 \]

\[\begin{aligned} & x=h+a \cos \theta \Rightarrow \frac{x-h}{a}=\cos \theta \\ & y=k+b \cos \theta \Rightarrow \frac{y-k}{b}=\sin \theta \\ & \therefore \text { LHS }=\left(\frac{x-h}{a}\right)^{2}+\left(\frac{y-k}{b}\right)^{2}=\cos ^{2} \theta+\sin ^{2} \theta=1=\mathrm{RHS} \end{aligned} \]

\[\begin{aligned} & \mathrm{LHS}=\frac{\tan \mathrm{A}}{1+\sec \mathrm{A}}-\frac{\tan \mathrm{A}}{1-\sec \mathrm{A}}=\frac{\dfrac{\sin \mathrm{A}}{\cos \mathrm{~A}}}{1+\dfrac{1}{\cos \mathrm{~A}}}-\frac{\dfrac{\sin \mathrm{A}}{\cos \mathrm{~A}}}{1-\dfrac{1}{\cos \mathrm{~A}}} \\ & =\frac{\sin \mathrm{A}}{\cos \mathrm{~A}+1}-\frac{\sin \mathrm{A}}{\cos \mathrm{~A}-1} \\ & =\sin \mathrm{A}\left(\frac{-2}{-\sin ^{2} \mathrm{~A}}\right) \\ & =\frac{2}{\sin \mathrm{~A}}=2 \operatorname{cosec} \mathrm{~A}=\mathrm{RHS} \end{aligned} \]