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Question Number 69871 by Shamim last updated on 28/Sep/19

$$\mathrm{Here},{\mathrm{m}}^2-{\mathrm{n}}^2=4\sqrt{\mathrm{mn}}\mathrm{and}\tan \theta +\sin \theta =\mathrm{m}$$$$\mathrm{then}\mathrm{prove}\mathrm{that},\tan \theta -\sin \theta =\mathrm{n}.$$

Answered by MJS last updated on 29/Sep/19

$$\mathrm{first}\mathrm{how}\mathrm{to}\mathrm{solve}$$$$m^2-n^2=4\sqrt{mn}$$$$\mathrm{for}n$$$$\mathrm{squaring}\mathrm{is}\mathrm{a}\mathrm{bad}\mathrm{idea},\mathrm{it}\mathrm{leads}\mathrm{to}$$$$n^4-2m^2{n}^2-16mn+m^4=0$$$$\mathrm{and}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{hard}\mathrm{to}\mathrm{solve}$$$$\mathrm{instead}\mathrm{put}m=p+q,n=p-q$$$$4pq=4\sqrt{p^2-q^2}$$$$p^2{q}^2=p^2-q^2$$$$\Rightarrow q=\pm \frac{p}{\sqrt{p^2+1}}$$$$\Rightarrow m=p\pm \frac{p}{\sqrt{p^2+1}},n=p\mp \frac{p}{\sqrt{p^2+1}}$$$$\mathrm{but}m=\tan \theta +\sin \theta$$$$\mathrm{with}p=\tan \theta \mathrm{we}\mathrm{get}\frac{p}{\sqrt{p^2+1}}=\pm \sin \theta$$$$\Rightarrow n=\tan \theta -\sin \theta$$

Commented by $@ty@m123 last updated on 29/Sep/19

$$Wonderful!$$

Commented by Rasheed.Sindhi last updated on 29/Sep/19

$$\mathcal{X}cellent\mathcal{S}ir!$$

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