Question Number 198423 by pascal889 last updated on 19/Oct/23
$$\mathrm{show}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}+\mathrm{sin}\varnothing}{\mathrm{cos}\varnothing}\:=\frac{\mathrm{cos}\varnothing}{\mathrm{1}−\mathrm{sin}\varnothing} \\ $$$$\mathrm{Pls}\:\mathrm{help} \\ $$$$ \\ $$
Answered by BaliramKumar last updated on 19/Oct/23
$$\frac{\left(\mathrm{1}+\mathrm{sinx}\right)}{\mathrm{cosx}}×\frac{\left(\mathrm{1}−\mathrm{sinx}\right)}{\left(\mathrm{1}−\mathrm{sinx}\right)}\:=\:\frac{\mathrm{1}^{\mathrm{2}} −\mathrm{sin}^{\mathrm{2}} \mathrm{x}}{\mathrm{cosx}\left(\mathrm{1}−\mathrm{sinx}\right)} \\ $$$$=\frac{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}{\mathrm{cosx}\left(\mathrm{1}−\mathrm{sinx}\right)}\:=\:\frac{\mathrm{cosx}}{\mathrm{1}−\mathrm{sinx}} \\ $$
Answered by Rasheed.Sindhi last updated on 19/Oct/23
![sin^2 θ+cos^2 θ=1 [An identity that we know] ⇒1−sin^2 θ=cos^2 θ ⇒(1−sinθ)(1+sinθ)=(cosθ)(cosθ) ⇒((1+sinθ)/(cosθ )) =((cosθ )/(1−sinθ))](https://www.tinkutara.com/question/Q198429.png)
$$\:\:\mathrm{sin}^{\mathrm{2}} \theta+\mathrm{cos}^{\mathrm{2}} \theta=\mathrm{1}\:\:\left[{An}\:{identity}\:{that}\:{we}\:{know}\right] \\ $$$$\:\:\Rightarrow\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta=\mathrm{cos}^{\mathrm{2}} \theta \\ $$$$\:\:\Rightarrow\left(\mathrm{1}−\mathrm{sin}\theta\right)\left(\mathrm{1}+\mathrm{sin}\theta\right)=\left(\mathrm{cos}\theta\right)\left(\mathrm{cos}\theta\right) \\ $$$$\:\:\Rightarrow\frac{\mathrm{1}+\mathrm{sin}\theta}{\mathrm{cos}\theta\:}\:=\frac{\mathrm{cos}\theta\:}{\mathrm{1}−\mathrm{sin}\theta}\:\:\:\:\: \\ $$