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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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