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Question Number 198423 by pascal889 last updated on 19/Oct/23

$$\mathrm{show}\mathrm{that}$$$$\frac{1+\sin \varnothing }{\cos \varnothing }=\frac{\cos \varnothing }{1-\sin \varnothing }$$$$\mathrm{Pls}\mathrm{help}$$$$$$

Answered by BaliramKumar last updated on 19/Oct/23

$$\frac{(1+\mathrm{sinx})}{\mathrm{cosx}}\times \frac{(1-\mathrm{sinx})}{(1-\mathrm{sinx})}=\frac{1^2-{\sin }^2\mathrm{x}}{\mathrm{cosx}(1-\mathrm{sinx})}$$$$=\frac{{\cos }^2\mathrm{x}}{\mathrm{cosx}(1-\mathrm{sinx})}=\frac{\mathrm{cosx}}{1-\mathrm{sinx}}$$

Answered by Rasheed.Sindhi last updated on 19/Oct/23

$${\sin }^2\theta +{\cos }^2\theta =1\left[Anidentitythatweknow\right]$$$$\Rightarrow 1-{\sin }^2\theta ={\cos }^2\theta$$$$\Rightarrow (1-\sin \theta )(1+\sin \theta )=\left(\cos \theta \right)\left(\cos \theta \right)$$$$\Rightarrow \frac{1+\sin \theta }{\cos \theta }=\frac{\cos \theta }{1-\sin \theta }$$

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