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Question Number 206970 by MATHEMATICSAM last updated on 02/May/24

$$\mathrm{If}\sin \theta =\frac{m^2+2mn}{m^2+2mn+2n^2}\mathrm{then}\mathrm{prove}$$$$\mathrm{that}\tan \theta =\frac{m^2+2mn}{2mn+2n^2}.$$

Answered by Rasheed.Sindhi last updated on 02/May/24

$$\sin \theta =\frac{m^2+2mn}{m^2+2mn+2n^2}$$$${\sin }^2\theta ={\left(\frac{m^2+2mn}{m^2+2mn+2n^2}\right)}^2$$$$1-{\sin }^2\theta =1-\frac{{(m^2+2mn)}^2}{{(m^2+2mn+2n^2)}^2}$$$${\cos }^2\theta =\frac{{(m^2+2mn+2n^2)}^2-{(m^2+2mn)}^2}{{(m^2+2mn+2n^2)}^2}$$$${\tan }^2\theta =\frac{{\sin }^2\theta }{{\cos }^2}=\frac{{(m^2+2mn)}^2}{{(m^2+2mn+2n^2)}^2}\times \frac{{(m^2+2mn+2n^2)}^2}{{(m^2+2mn+2n^2)}^2-{(m^2+2mn)}^2}$$$${\tan }^2\theta =\frac{{(m^2+2mn)}^2}{\{(m^2+2mn+2n^2)-(m^2+2mn)\}\{(m^2+2mn+2n^2)+(m^2+2mn)\}}$$$$=\frac{{(m^2+2mn)}^2}{2n^2(2m^2+4mn+2n^2)}$$$$=\frac{{(m^2+2mn)}^2}{4n^2(m^2+2mn+n^2)}$$$$=\frac{{(m^2+2mn)}^2}{4n^2{(m+n)}^2}$$$$\tan \theta =\pm \frac{m(m+2n)}{2n(m+n)}$$

Answered by Frix last updated on 02/May/24

$$t=\frac{s}{\sqrt{1-s^2}}$$$$\stackrel{}{s}=\frac{u}{v}\Rightarrow t=\frac{u}{\sqrt{v^2-u^2}}$$$$u=m^2+2mn$$$$v=m^2+2mn+2n^2$$$$t=\frac{m(m+2n)}{2\mid (m+n)n\mid }$$

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