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Question Number 39586 by Rio Mike last updated on 08/Jul/18

$$showthat$$$$a)\frac{1+2sin2\theta -cos2\theta }{1+sin2\theta +cos2\theta }=tan\theta$$$$b){tan}^{2}A-{tan}^{2}B=\frac{{sin}^{2}A-{sin}^{2}B}{{cos}^{2}A{cos}^{2}B}$$$$$$$$$$

Answered by Joel579 last updated on 08/Jul/18

$$LHS$$$$\frac{1+\sin 2\theta -\cos 2\theta }{1+\sin 2\theta +\cos 2\theta }$$$$=\frac{({\sin }^2\theta +{\cos }^2\theta )+\sin 2\theta -({\cos }^2\theta -{\sin }^2\theta )}{({\sin }^2\theta +{\cos }^2\theta )+\sin 2\theta +({\cos }^2\theta -{\sin }^2\theta )}$$$$=\frac{{2\sin }^2\theta +2\sin \theta \cos \theta }{{2\cos }^2\theta +2\sin \theta \cos \theta }$$$$=\frac{2\sin \theta (\sin \theta +\cos \theta )}{2\cos \theta (\cos \theta +\sin \theta )}$$$$=\tan \theta \to RHS$$

Answered by Joel579 last updated on 08/Jul/18

$$LHS$$$${\tan }^{2}A-{\tan }^{2}B$$$$=\frac{{\tan }^{2}A-{\tan }^{2}B\left({\cos }^{2}A{\cos }^{2}B\right)}{{\cos }^{2}A{\cos }^{2}B}$$$$=\frac{{\sin }^{2}A{\cos }^{2}B-{\sin }^{2}B{\cos }^{2}A}{{\cos }^{2}A{\cos }^{2}B}$$

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