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Question Number 10955 by Joel576 last updated on 04/Mar/17
If ((cos θ)/(1 − sin θ)) = a a ≠ (π/2) + 2kπ So, tan (θ/2) = ... (A) (a/(a + 1)) (D) ((a + 1)/(a − 1)) (B) (1/(a + 1)) (E) (a/(a − 1)) (C) ((a − 1)/(a + 1))
$$\mathrm{If}\:\:\frac{\mathrm{cos}\:\theta}{\mathrm{1}\:−\:\mathrm{sin}\:\theta}\:=\:{a}\:\:\:\:\:\:\:\:\:\:\:{a}\:\neq\:\frac{\pi}{\mathrm{2}}\:+\:\mathrm{2}{k}\pi \\ $$$$\mathrm{So},\:\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\:=\:… \\ $$$$\left(\mathrm{A}\right)\:\:\frac{{a}}{{a}\:+\:\mathrm{1}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\:\frac{{a}\:+\:\mathrm{1}}{{a}\:−\:\mathrm{1}} \\ $$$$\left(\mathrm{B}\right)\:\:\frac{\mathrm{1}}{{a}\:+\:\mathrm{1}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{E}\right)\:\:\frac{{a}}{{a}\:−\:\mathrm{1}} \\ $$$$\left(\mathrm{C}\right)\:\:\frac{{a}\:−\:\mathrm{1}}{{a}\:+\:\mathrm{1}} \\ $$
Answered by ajfour last updated on 04/Mar/17
((sin ((π/2)−θ))/(1−cos ((π/2)−θ)))=((2sin ((π/4)−(θ/2))cos((π/4)−(θ/2)) )/(2sin^2 ((π/4)−(θ/2)))) =cot ((π/4)−(θ/2)) = a tan ((π/4)−(θ/2)) =(1/a) ; let tan ((θ/2))= p ((1−p)/(1+p)) =(1/a) ⇒ a−ap =1+p p =((a−1)/(a+1)) .
$$ \\ $$$$\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\theta\right)}{\mathrm{1}−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}−\theta\right)}=\frac{\mathrm{2sin}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right)\mathrm{cos}\left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right)\:}{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right)} \\ $$$$=\mathrm{cot}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right)\:=\:{a} \\ $$$$\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right)\:=\frac{\mathrm{1}}{{a}}\:\:;\:{let}\:\mathrm{tan}\:\left(\frac{\theta}{\mathrm{2}}\right)=\:{p} \\ $$$$\frac{\mathrm{1}−{p}}{\mathrm{1}+{p}}\:=\frac{\mathrm{1}}{{a}}\:\Rightarrow\:{a}−{ap}\:=\mathrm{1}+{p} \\ $$$${p}\:=\frac{{a}−\mathrm{1}}{{a}+\mathrm{1}}\:. \\ $$
Commented by Joel576 last updated on 05/Mar/17
thank you very much
$${thank}\:{you}\:{very}\:{much} \\ $$
Answered by ridwan balatif last updated on 04/Mar/17
another way (((cosθ)/(1−sinθ)))^2 =(a)^2 ((cos^2 θ)/((1−sinθ)^2 ))=a^2 (((1−sin^2 θ))/((1−sinθ)^2 ))=a^2 (((1−sinθ)(1+sinθ))/((1−sinθ)(1−sinθ)))=a^2 ((1+sinθ)/(1−sinθ))=a^2 sinθ=((a^2 −1)/(a^2 +1)) cosθ=((2a)/(a^2 +1)) remember: tan(θ/2)=((sinθ)/(1+cosθ)) tan(θ/2)=(((a^2 −1)/(a^2 +1))/(1+((2a)/(a^2 +1)))) tan(θ/2)=(((a^2 −1)/(a^2 +1))/(((a^2 +2a+1)/(a^2 +1)) )) tan(θ/2)=((a^2 −1)/(a^2 +2a+1)) tan(θ/2)=(((a−1)(a+1))/((a+1)^2 )) tan(θ/2)=((a−1)/(a+1)) Answer: (C)
$$\mathrm{another}\:\mathrm{way} \\ $$$$\left(\frac{\mathrm{cos}\theta}{\mathrm{1}−\mathrm{sin}\theta}\right)^{\mathrm{2}} =\left(\mathrm{a}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{cos}^{\mathrm{2}} \theta}{\left(\mathrm{1}−\mathrm{sin}\theta\right)^{\mathrm{2}} }=\mathrm{a}^{\mathrm{2}} \\ $$$$\frac{\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta\right)}{\left(\mathrm{1}−\mathrm{sin}\theta\right)^{\mathrm{2}} }=\mathrm{a}^{\mathrm{2}} \\ $$$$\frac{\left(\mathrm{1}−\mathrm{sin}\theta\right)\left(\mathrm{1}+\mathrm{sin}\theta\right)}{\left(\mathrm{1}−\mathrm{sin}\theta\right)\left(\mathrm{1}−\mathrm{sin}\theta\right)}=\mathrm{a}^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}+\mathrm{sin}\theta}{\mathrm{1}−\mathrm{sin}\theta}=\mathrm{a}^{\mathrm{2}} \\ $$$$\mathrm{sin}\theta=\frac{\mathrm{a}^{\mathrm{2}} −\mathrm{1}}{\mathrm{a}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{cos}\theta=\frac{\mathrm{2a}}{\mathrm{a}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{remember}:\:\mathrm{tan}\frac{\theta}{\mathrm{2}}=\frac{\mathrm{sin}\theta}{\mathrm{1}+\mathrm{cos}\theta} \\ $$$$\mathrm{tan}\frac{\theta}{\mathrm{2}}=\frac{\frac{\mathrm{a}^{\mathrm{2}} −\mathrm{1}}{\mathrm{a}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{1}+\frac{\mathrm{2a}}{\mathrm{a}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\mathrm{tan}\frac{\theta}{\mathrm{2}}=\frac{\frac{\mathrm{a}^{\mathrm{2}} −\mathrm{1}}{\mathrm{a}^{\mathrm{2}} +\mathrm{1}}}{\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{2a}+\mathrm{1}}{\mathrm{a}^{\mathrm{2}} +\mathrm{1}}\:\:} \\ $$$$\mathrm{tan}\frac{\theta}{\mathrm{2}}=\frac{\mathrm{a}^{\mathrm{2}} −\mathrm{1}}{\mathrm{a}^{\mathrm{2}} +\mathrm{2a}+\mathrm{1}} \\ $$$$\mathrm{tan}\frac{\theta}{\mathrm{2}}=\frac{\left(\mathrm{a}−\mathrm{1}\right)\left(\mathrm{a}+\mathrm{1}\right)}{\left(\mathrm{a}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{tan}\frac{\theta}{\mathrm{2}}=\frac{\mathrm{a}−\mathrm{1}}{\mathrm{a}+\mathrm{1}} \\ $$$$\mathrm{Answer}:\:\left(\mathrm{C}\right) \\ $$
Commented by Joel576 last updated on 05/Mar/17
thank you very much
$${thank}\:{you}\:{very}\:{much} \\ $$

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