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Question Number 159845 by abdullah_ff last updated on 21/Nov/21

if q = 1−sinθ; then prove that (secθ − tanθ)^2 = (1/q)

$$\mathrm{if}\:{q}\:=\:\mathrm{1}−{sin}\theta;\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\left({sec}\theta\:−\:{tan}\theta\right)^{\mathrm{2}} \:=\:\frac{\mathrm{1}}{{q}} \\ $$

Commented by tounghoungko last updated on 21/Nov/21

sin θ = 1−q cos θ =(√(1−(1−2q+q^2 )))=(√(2q−q^2 )) sec θ=(1/( (√(2q−q^2 )))) ; tan θ = ((1−q)/( (√(2q−q^2 )))) sec θ−tan θ = (q/( (√(2q−q^2 )))) (sec θ−tan θ)^2 = (q^2 /(2q−q^2 ))

$$\mathrm{sin}\:\theta\:=\:\mathrm{1}−{q}\: \\ $$$$\:\mathrm{cos}\:\theta\:=\sqrt{\mathrm{1}−\left(\mathrm{1}−\mathrm{2}{q}+{q}^{\mathrm{2}} \right)}=\sqrt{\mathrm{2}{q}−{q}^{\mathrm{2}} } \\ $$$$\:\mathrm{sec}\:\theta=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{q}−{q}^{\mathrm{2}} }}\:;\:\mathrm{tan}\:\theta\:=\:\frac{\mathrm{1}−{q}}{\:\sqrt{\mathrm{2}{q}−{q}^{\mathrm{2}} }} \\ $$$$\:\mathrm{sec}\:\theta−\mathrm{tan}\:\theta\:=\:\frac{{q}}{\:\sqrt{\mathrm{2}{q}−{q}^{\mathrm{2}} }} \\ $$$$\left(\mathrm{sec}\:\theta−\mathrm{tan}\:\theta\right)^{\mathrm{2}} \:=\:\frac{{q}^{\mathrm{2}} }{\mathrm{2}{q}−{q}^{\mathrm{2}} }\: \\ $$

Answered by JDamian last updated on 21/Nov/21

(((1−sinθ)/(cosθ)))^2 =(q^2 /(cos^2 θ))=(q^2 /(1−sin^2 θ))=(q^2 /(1−(1−q)^2 ))= (q^2 /(1−1−q^2 +2q))=(q/(2−q)) ■

$$\left(\frac{\mathrm{1}−{sin}\theta}{{cos}\theta}\right)^{\mathrm{2}} =\frac{{q}^{\mathrm{2}} }{{cos}^{\mathrm{2}} \theta}=\frac{{q}^{\mathrm{2}} }{\mathrm{1}−{sin}^{\mathrm{2}} \theta}=\frac{{q}^{\mathrm{2}} }{\mathrm{1}−\left(\mathrm{1}−{q}\right)^{\mathrm{2}} }= \\ $$$$\frac{{q}^{\mathrm{2}} }{\mathrm{1}−\mathrm{1}−{q}^{\mathrm{2}} +\mathrm{2}{q}}=\frac{{q}}{\mathrm{2}−{q}}\:\:\:\blacksquare \\ $$

Answered by som(math1967) last updated on 21/Nov/21

(secθ−tanθ)^2 =(((1−sinθ)/(cosθ)))^2 =(((1−sinθ)^2 )/(cos^2 θ)) =(((1−sinθ)^2 )/((1+sinθ)(1−sinθ))) =((1−sinθ)/(1+sinθ))≠(1/q)

$$\left({sec}\theta−{tan}\theta\right)^{\mathrm{2}} \\ $$$$=\left(\frac{\mathrm{1}−{sin}\theta}{{cos}\theta}\right)^{\mathrm{2}} \\ $$$$=\frac{\left(\mathrm{1}−{sin}\theta\right)^{\mathrm{2}} }{{cos}^{\mathrm{2}} \theta} \\ $$$$=\frac{\left(\mathrm{1}−{sin}\theta\right)^{\mathrm{2}} }{\left(\mathrm{1}+{sin}\theta\right)\left(\mathrm{1}−{sin}\theta\right)} \\ $$$$=\frac{\mathrm{1}−{sin}\theta}{\mathrm{1}+{sin}\theta}\neq\frac{\mathrm{1}}{{q}} \\ $$

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