Question

If (tan θ + sin θ) = m and (tan θ − sin θ) = n, prove that (m² − n²)² = 16mn.

Answer 1

$\begin{array}{l}\\ \text{We have}(\tan \theta +\sin \theta )=m\text{and}(\tan \theta -\sin \theta )=n\\ \\ \text{Now, LHS=}{(m^2-n^2)}^2\\ \\ \text{=}{[{(\tan \theta +\sin \theta )}^2-{(\tan \theta -\sin \theta )}^2]}^2\\ ={[({\tan }^2\theta +{\sin }^2\theta +2\tan \theta \sin \theta )-({\tan }^2\theta +{\sin }^2\theta -2\tan \theta \sin \theta )]}^2\\ ={\left[({\tan }^2\theta +{\sin }^2\theta +2\tan \theta \sin \theta -{\tan }^2\theta -{\sin }^2\theta +2\tan \theta \sin \theta )\right]}^2\\ ={\left(4\tan \theta \sin \theta \right)}^2\\ =16{\tan }^2\theta {\sin }^2\theta \\ =16\frac{{\sin }^2\theta }{{\cos }^2\theta }{\sin }^2\theta \\ =16\frac{(1-{\cos }^2\theta ){\sin }^2\theta }{{\cos }^2\theta }\\ =16\left[{\tan }^2\theta (1-{\cos }^2\theta )\right]\\ =16({\tan }^2\theta -{\tan }^2\theta {\cos }^2\theta )\\ =16({\tan }^2\theta -\frac{{\sin }^2\theta }{{\cos }^2\theta }\times {\cos }^2\theta )\\ =16({\tan }^2\theta -{\sin }^2\theta )\\ =16(\tan \theta +\sin \theta )(\tan \theta -\sin \theta )\\ =16mn[(\tan \theta +\sin \theta )(\tan \theta -\sin \theta )=mn]\\ \therefore (m^2-n^2){(m^2-n^2)}^2=16mn\end{array}$