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Question Number 35951 by gyugfeet last updated on 26/May/18

$$(cos\theta +cos\beta /sin\theta -sin\beta )=(sin\theta +sin\beta /cos\theta -cos\beta )proveghis$$

Commented by Rasheed.Sindhi last updated on 27/May/18

$$\mathcal{D}\mathrm{o}\mathrm{you}\mathrm{mean}$$$$\cos \theta +\frac{\cos \beta }{\sin \theta }-\sin \beta =\sin \theta +\frac{\sin \beta }{\cos \theta }-\cos \beta$$$$\mathrm{or}$$$$\frac{cos\theta +cos\beta }{sin\theta -sin\beta }=\frac{sin\theta +sin\beta }{cos\theta -cos\beta }?$$

Commented by gyugfeet last updated on 27/May/18

$$thissecondone...$$

Commented by Rasheed.Sindhi last updated on 28/May/18

$$\mathrm{The}\mathrm{identity}\mathrm{is}\mathrm{false}!\mathrm{You}\mathrm{can}\mathrm{verify}$$$$\mathrm{by}\mathrm{various}\mathrm{values}\mathrm{of}\theta \mathrm{\&}\beta .\mathrm{For}\mathrm{example}$$$$\theta =\pi /3\mathrm{\&}\beta =\pi /6$$

Answered by Rasheed.Sindhi last updated on 28/May/18

$$\frac{cos\theta +cos\beta }{sin\theta -sin\beta }=\frac{sin\theta +sin\beta }{cos\theta -cos\beta }$$$$\mathrm{LHS}$$$$\frac{cos\theta +cos\beta }{sin\theta -sin\beta }$$$$=\frac{2\cos \left(\frac{a+b}{2}\right)\cos \left(\frac{a+b}{2}\right)}{2\cos \left(\frac{a+b}{2}\right)\sin \left(\frac{a-b}{2}\right)}=\frac{\cos \left(\frac{a+b}{2}\right)}{\sin \left(\frac{a-b}{2}\right)}$$$$\mathrm{RHS}$$$$\frac{sin\theta +sin\beta }{cos\theta -cos\beta }=\frac{2\sin \left(\frac{a+b}{2}\right)\cos \left(\frac{a-b}{2}\right)}{-2\sin \left(\frac{a+b}{2}\right)\sin \left(\frac{a-b}{2}\right)}$$$$=-\frac{\cos \left(\frac{a-b}{2}\right)}{\sin \left(\frac{a-b}{2}\right)}=-\tan \left(\frac{a-b}{2}\right)$$$$???$$

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