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Question Number 210234 by peter frank last updated on 03/Aug/24

Commented by Pnk2024 last updated on 03/Aug/24

$$weknowthat$$$${sin}^2\theta +{cos}^2\theta =1$$$$\Rightarrow {cos}^2\theta =1-{sin}^2\theta$$$$\Rightarrow cos\theta \times cos\theta =(1+sin\theta )(1-sin\theta )$$$$\Rightarrow \frac{cos\theta }{1-sin\theta }=\frac{1+sin\theta }{cos\theta }$$$$\Rightarrow \frac{cos\theta }{1-sin\theta }=\frac{1+sin\theta -cos\theta }{cos\theta -(1-sin\theta )}bytheoremofequalratio$$$$\Rightarrow \frac{sin\theta -cos\theta +1}{sin\theta +cos\theta -1}=\frac{cos\theta /cos\theta }{(1-sin\theta )/cos\theta }$$$$=\frac{1}{\frac{1}{cos\theta }-\frac{sin\theta }{cos\theta }}$$$$\Rightarrow \frac{sin\theta -cos\theta +1}{sin\theta +cos\theta -1}=\frac{1}{sec\theta -tan\theta }$$$$thisisproved$$

Commented by peter frank last updated on 04/Aug/24

$$\mathrm{thank}\mathrm{you}$$

Answered by efronzo1 last updated on 04/Aug/24

$$\frac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}\stackrel{?}{=}\frac{1}{\sec \theta -\tan \theta }$$$$\underbrace{⋐}$$

Commented by peter frank last updated on 04/Aug/24

$$\mathrm{thank}\mathrm{you}$$

Answered by BaliramKumar last updated on 04/Aug/24

$$\frac{(\mathrm{sinx}-\mathrm{cosx}+1)/\mathrm{cosx}}{(\mathrm{sinx}+\mathrm{cosx}-1)/\mathrm{cosx}}=\frac{\mathrm{tanx}+\mathrm{secx}-1}{\mathrm{tanx}-\mathrm{secx}+1}$$$$\frac{\mathrm{tanx}+\mathrm{secx}-1}{\mathrm{tanx}-\mathrm{secx}+({\sec }^2\mathrm{x}-{\tan }^2\mathrm{x})}$$$$\frac{\mathrm{tanx}+\mathrm{secx}-1}{(\mathrm{tanx}-\mathrm{secx})+(\mathrm{secx}-\mathrm{tanx})(\mathrm{secx}+\mathrm{tanx})}$$$$\frac{\mathrm{tanx}+\mathrm{secx}-1}{(\mathrm{secx}-\mathrm{tanx})(-1+\mathrm{secx}+\mathrm{tanx})}$$$$\frac{\cancel{(\mathrm{tanx}+\mathrm{secx}-1)}}{(\mathrm{secx}-\mathrm{tanx})\cancel{(\mathrm{tanx}+\mathrm{secx}-1)}}$$$$=\frac{1}{\mathrm{secx}-\mathrm{tanx}}$$$$$$$$$$

Commented by peter frank last updated on 04/Aug/24

$$\mathrm{thank}\mathrm{you}$$

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