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Question Number 107900 by bemath last updated on 13/Aug/20

$$\frac{\mathrm{\Sigma }\mathcal{B}e\mathcal{M}ath\mathrm{\Sigma }}{◻}$$$$Given\tan x-\sec x=\vartheta$$$$then\sin x=?$$

Answered by bemath last updated on 13/Aug/20

$$\frac{\sin x-1}{\cos x}=\upsilon \Rightarrow \sin x-1=\upsilon \cos x...\left(1\right)$$$$\sin x+1=-\frac{1}{\upsilon }\cos x...\left(2\right)$$$$substituteeq\left(2\right)toeq\left(1\right)$$$$\sin x+1=-\frac{1}{\upsilon }\left(\frac{\sin x-1}{\upsilon }\right)$$$${\upsilon }^2\sin x+{\upsilon }^2=-\sin x+1$$$$({\upsilon }^2+1)\sin x=1-{\upsilon }^2$$$$\Rightarrow \sin x=\frac{1-{\upsilon }^2}{1+{\upsilon }^2}$$

Answered by bobhans last updated on 13/Aug/20

$$\frac{\wedge \mathcal{B}\mathrm{obhans}\wedge }{◻}$$$$\tan \mathrm{x}-\sec \mathrm{x}\times \left(\frac{\tan \mathrm{x}+\sec \mathrm{x}}{\tan \mathrm{x}+\sec \mathrm{x}}\right)=\upsilon$$$$\frac{\tan {}^2\mathrm{x}-\sec {}^2\mathrm{x}}{\tan \mathrm{x}+\sec \mathrm{x}}=\upsilon \to \frac{\tan {}^2\mathrm{x}-(1+\tan {}^2\mathrm{x})}{\tan \mathrm{x}+\sec \mathrm{c}}=\upsilon$$$$\tan \mathrm{x}+\sec \mathrm{x}=-\frac{1}{\upsilon }\Rightarrow \frac{\sin \mathrm{x}+1}{\cos \mathrm{x}}=-\frac{1}{\upsilon }$$$$\sin \mathrm{x}+1=-\frac{1}{\upsilon }\cos \mathrm{x}$$

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