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Question Number 159466 by abdullah_ff last updated on 17/Nov/21

$$\mathrm{if}tanA=\frac{a}{b}$$$$\mathrm{then}\mathrm{prove}\mathrm{that}sinA=\pm \frac{a}{\sqrt{a^{}+b^{}}}$$$$\mathrm{please}\mathrm{help}..$$

Answered by Ar Brandon last updated on 17/Nov/21

$$\tan A=\frac{a}{b}\Rightarrow \cot A=\frac{b}{a}$$$${\cot }^{2}A+1={\mathrm{cosec}}^{2}A$$$$\frac{b^2}{a^2}+1={\mathrm{cosec}}^{2}A=\frac{a^2+b^2}{a^2}$$$$\Rightarrow \sin A=\pm \frac{a}{\sqrt{a^2+b^2}}$$

Answered by MJS_new last updated on 17/Nov/21

$$\tan \alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{\sin \alpha }{\sqrt{1-{\sin }^2\alpha }}$$$$\Rightarrow$$$${\tan }^2\alpha =\frac{{\sin }^2\alpha }{1-{\sin }^2\alpha }$$$$\Rightarrow$$$$\sin \alpha =\pm \frac{\tan \alpha }{\sqrt{1+{\tan }^2\alpha }}$$$$\tan \alpha =\frac{a}{b}\Rightarrow \sin \alpha =\pm \frac{\frac{a}{b}}{\sqrt{1+\frac{a^2}{b^2}}}=\pm \frac{a}{\sqrt{a^2+b^2}}$$

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