(a)OR(b)

Prove that : $\displaystyle \frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \operatorname{cosec} \theta$

\[\begin{aligned} \mathrm{LHS} & =\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta} \\ & =\frac{\dfrac{\sin \theta}{\cos \theta}}{1-\dfrac{\cos \theta}{\sin \theta}}+\frac{\dfrac{\cos \theta}{\sin \theta}}{1-\dfrac{\sin \theta}{\cos \theta}} \\ & =\frac{\sin ^{2} \theta}{\cos \theta(\sin \theta-\cos \theta)}-\frac{\cos ^{2} \theta}{\sin \theta(\sin \theta-\cos \theta)} \\ & =\frac{1}{(\sin \theta-\cos \theta)}\left[\frac{\sin ^{3} \theta-\cos ^{3} \theta}{\sin \theta \cos \theta}\right] \\ & =\frac{(\sin \theta-\cos \theta)\left(\sin ^{2} \theta+\sin \theta \cos \theta+\cos ^{2} \theta\right)}{(\sin \theta-\cos \theta) \sin \theta \cos \theta} \\ & =\frac{(1+\sin \theta \cos \theta)}{\sin \theta \cos \theta} \\ & =1+\sec \theta \operatorname{cosec} \theta=\mathrm{RHS} \end{aligned} \]

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