Prove that : \[\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\tan \theta+\cot \theta . \]

\[\begin{aligned} & \text { LHS }=\left(\frac{\dfrac{\sin \theta}{\cos \theta}}{1-\dfrac{\cos \theta}{\sin \theta}}\right)+\left(\frac{\dfrac{\cos \theta}{\sin \theta}}{1-\dfrac{\sin \theta}{\cos \theta}}\right) \\ & =\frac{\sin ^{2} \theta}{\cos \theta(\sin \theta-\cos \theta)}+\frac{\cos ^{2} \theta}{\sin \theta(\cos \theta-\sin \theta)} \\ & =\frac{\sin ^{3} \theta-\cos ^{3} \theta}{\sin \theta \cos \theta(\sin \theta-\cos \theta)} \\ & =\frac{(\sin \theta-\cos \theta)\left(\sin ^{2} \theta+\cos ^{2} \theta+\sin \theta \cos \theta\right)}{\sin \theta \cos \theta(\sin \theta-\cos \theta)} \\ & =\frac{\sin ^{2} \theta}{\sin \theta \cos \theta}+\frac{\cos ^{2} \theta}{\sin \theta \cos \theta}+\frac{\sin \theta \cos \theta}{\sin \theta \cos \theta} \\ & =\tan \theta+\cot \theta+1=\mathrm{RHS} \end{aligned} \] Sol. Let centre be \(\displaystyle \mathrm{O}(2,1) \Rightarrow \mathrm{OA}=\mathrm{OB}\) \[\begin{aligned} & \sqrt{(5-2)^{2}+(6-1)^{2}}=\sqrt{(-3-2)^{2}+(\mathrm{K}-1)^{2}} \\ & \Rightarrow 9=(\mathrm{K}-1)^{2} \\ & \Rightarrow \mathrm{~K}=-2,4 \end{aligned} \] For \(\displaystyle \mathrm{K}=-2, \mathrm{AB}=\sqrt{128}\) or \(\displaystyle 8 \sqrt{2}\) For \(\displaystyle \mathrm{K}=4, \mathrm{AB}=\sqrt{68}\) or \(\displaystyle 2 \sqrt{17}\) $\displaystyle 3$ : $\displaystyle 2$ \(\displaystyle \mathrm{AP}: \mathrm{PB}=3: 2\) Substituting \(\displaystyle \mathrm{x}=2\) and \(\displaystyle \mathrm{y}=1\) in the given equation L.H.S. \(\displaystyle =\mathrm{x}-3 \mathrm{y}\) \[=2-3(1) \] = -$\displaystyle 1$ = R.H.S. \(\displaystyle \therefore P\) lies on the given line \(\displaystyle \mathrm{PA}=\sqrt{(2+1)^{2}+(1-7)^{2}}=\sqrt{45}\) or \(\displaystyle 3 \sqrt{5}\) \(\displaystyle \mathrm{PB}=\sqrt{(2-4)^{2}+(1+3)^{2}}=\sqrt{20}\) or \(\displaystyle 2 \sqrt{5}\)