(a) If tan θ+ sin θ= m and tan θ- sin θ= n, then prove that m^2- n^2=4 √ mn.

\[\begin{aligned} \mathrm{LHS} & =\mathrm{m}^{2}-\mathrm{n}^{2} \\ & =(\tan \theta+\sin \theta)^{2}-(\tan \theta-\sin \theta)^{2} \\ & =4 \tan \theta \sin \theta \\ & =4 \sqrt{\tan ^{2} \theta \cdot \sin ^{2} \theta} \\ & =4 \sqrt{\tan ^{2} \theta\left(1-\cos ^{2} \theta\right)} \\ & =4 \sqrt{\tan ^{2} \theta-\sin ^{2} \theta} \\ & =4 \sqrt{(\tan \theta+\sin \theta)(\tan \theta-\sin \theta)} \\ & =4 \sqrt{\mathrm{mn}}=\mathrm{RHS} \end{aligned} \]

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

\(\displaystyle =\frac{\cot \mathrm{A}}{1+\tan \mathrm{A}}=\mathrm{RHS}\)