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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

we know that sin^2 θ+cos^2 θ=1 ⇒ cos^2 θ=1−sin^2 θ ⇒ cosθ ×cosθ = (1+sinθ)(1−sinθ) ⇒ ((cosθ)/(1−sinθ)) = ((1+sinθ)/(cosθ)) ⇒ ((cosθ)/(1−sinθ))=((1+sinθ−cosθ)/(cosθ−(1−sinθ))) by theorem of equal ratio ⇒ ((sinθ−cosθ+1)/(sinθ+cosθ−1))=((cosθ /cosθ)/((1−sinθ)/cosθ)) =(1/((1/(cosθ))−((sinθ)/(cosθ)))) ⇒ ((sinθ−cosθ+1)/(sinθ+cosθ−1))=(1/(secθ−tanθ)) this is proved

$$\:{we}\:{know}\:{that} \\ $$$$\:\:{sin}^{\mathrm{2}} \theta+{cos}^{\mathrm{2}} \theta=\mathrm{1} \\ $$$$\Rightarrow\:\:{cos}^{\mathrm{2}} \theta=\mathrm{1}−{sin}^{\mathrm{2}} \theta \\ $$$$\Rightarrow\:\:{cos}\theta\:×{cos}\theta\:=\:\left(\mathrm{1}+{sin}\theta\right)\left(\mathrm{1}−{sin}\theta\right) \\ $$$$\Rightarrow\:\frac{{cos}\theta}{\mathrm{1}−{sin}\theta}\:=\:\frac{\mathrm{1}+{sin}\theta}{{cos}\theta} \\ $$$$\Rightarrow\:\frac{{cos}\theta}{\mathrm{1}−{sin}\theta}=\frac{\mathrm{1}+{sin}\theta−{cos}\theta}{{cos}\theta−\left(\mathrm{1}−{sin}\theta\right)}\:{by}\:{theorem}\:{of}\:{equal}\:{ratio} \\ $$$$\Rightarrow\:\frac{{sin}\theta−{cos}\theta+\mathrm{1}}{{sin}\theta+{cos}\theta−\mathrm{1}}=\frac{{cos}\theta\:/{cos}\theta}{\left(\mathrm{1}−{sin}\theta\right)/{cos}\theta}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\frac{\mathrm{1}}{{cos}\theta}−\frac{{sin}\theta}{{cos}\theta}} \\ $$$$\Rightarrow\:\frac{{sin}\theta−{cos}\theta+\mathrm{1}}{{sin}\theta+{cos}\theta−\mathrm{1}}=\frac{\mathrm{1}}{{sec}\theta−{tan}\theta} \\ $$$$\:\:\:\:\:{this}\:{is}\:{proved}\: \\ $$

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

thank you

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

Answered by efronzo1 last updated on 04/Aug/24

((sin θ−cos θ+1)/(sin θ+cos θ−1)) =^? (1/(sec θ−tan θ)) ⋐

$$\:\:\frac{\mathrm{sin}\:\theta−\mathrm{cos}\:\theta+\mathrm{1}}{\mathrm{sin}\:\theta+\mathrm{cos}\:\theta−\mathrm{1}}\:\overset{?} {=}\:\frac{\mathrm{1}}{\mathrm{sec}\:\theta−\mathrm{tan}\:\theta} \\ $$$$\:\:\:\underbrace{\Subset} \\ $$

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

thank you

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

Answered by BaliramKumar last updated on 04/Aug/24

(((sinx − cosx + 1)/cosx)/((sinx + cosx − 1)/cosx)) = ((tanx + secx − 1)/(tanx − secx + 1)) ((tanx + secx − 1)/(tanx − secx + (sec^2 x − tan^2 x))) ((tanx + secx − 1)/((tanx − secx) + (secx − tanx)(secx + tanx))) ((tanx + secx − 1)/((secx − tanx)(−1 + secx + tanx))) (((tanx + secx − 1))/((secx − tanx)(tanx + secx − 1))) = (1/(secx − tanx))

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

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

thank you

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

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