Question

If $\tan \theta +\sin \theta =mand\tan \theta -\sin \theta =n,showthat{m}^2+n^2=4\sqrt{mn}.$

Answer 1

DISCLAIMER: In the question, there is a misprint. The LHS should be $m^2-n^2\mathrm{in}\mathrm{stead}\mathrm{of}{m}^2+n^2$.

$$\begin{gathered} \mathrm{Given}:\tan \theta +\sin \theta =m\mathrm{and}\tan \theta -\sin \theta =n \\ \mathrm{LHS}:m^2-n^2 \\ ={\left(\tan \theta +\sin \theta \right)}^2-{\left(\tan \theta -\sin \theta \right)}^2 \\ ={\tan }^2\theta +{\sin }^2\theta +2\tan \theta \sin \theta -{\tan }^2\theta -{\sin }^2\theta +2\tan \theta \sin \theta \\ =4\tan \theta \sin \theta \\ \mathrm{RHS}:4\sqrt{mn} \\ =4\sqrt{\left(\tan \theta +\sin \theta \right)\left(\tan \theta -\sin \theta \right)} \\ =4\sqrt{{\tan }^2\theta -{\sin }^2\theta } \\ =4\sqrt{\frac{{\sin }^2\theta -{\sin }^2\theta {\cos }^2\theta }{{\cos }^2\theta }} \\ =\frac{4\sin \theta }{\cos \theta }\sqrt{\frac{{\sin }^2\theta }{1}} \\ =4\tan \theta \sin \theta \\ \therefore \mathrm{LHS}=\mathrm{RHS} \end{gathered}$$