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Question-68397

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Question Number 68397 by Faradtimmy last updated on 10/Sep/19
Answered by $@ty@m123 last updated on 10/Sep/19
(a) LHS=(√3)cos θ−sin θ =2(((√3)/2)cos θ−(1/2)sin θ) =2(cos 30cos θ−sin 30sin θ) =2cos (30+θ) Pl. check the question. (b) (√((1−sin θ)/(1+sin θ))) = (√(((1−sin θ)/(1+sin θ))×((1−sin θ)/(1−sin θ)))) =(√(((1−sin θ)^2 )/(1−sin^2 θ))) =(√((((1−sin θ)/(cos θ)))^2 )) =(√((sec θ−tan θ)^2 )) =−(sec θ−tan θ), ∵90^o <θ≤180^o Similarly, (√((1+sin θ)/(1−sin θ))) = −(sec θ+tan θ) ∴ the given expression =−(sec θ−tan θ)−(sec θ+tan θ) =−2sec θ
$$\left({a}\right)\:{LHS}=\sqrt{\mathrm{3}}\mathrm{cos}\:\theta−\mathrm{sin}\:\theta \\ $$$$=\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:\theta−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\theta\right) \\ $$$$=\mathrm{2}\left(\mathrm{cos}\:\mathrm{30cos}\:\theta−\mathrm{sin}\:\mathrm{30sin}\:\theta\right) \\ $$$$=\mathrm{2cos}\:\left(\mathrm{30}+\theta\right) \\ $$$${Pl}.\:{check}\:{the}\:{question}. \\ $$$$\left({b}\right)\:\sqrt{\frac{\mathrm{1}−\mathrm{sin}\:\theta}{\mathrm{1}+\mathrm{sin}\:\theta}} \\ $$$$=\:\sqrt{\frac{\mathrm{1}−\mathrm{sin}\:\theta}{\mathrm{1}+\mathrm{sin}\:\theta}×\frac{\mathrm{1}−\mathrm{sin}\:\theta}{\mathrm{1}−\mathrm{sin}\:\theta}} \\ $$$$=\sqrt{\frac{\left(\mathrm{1}−\mathrm{sin}\:\theta\right)^{\mathrm{2}} }{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\theta}} \\ $$$$=\sqrt{\left(\frac{\mathrm{1}−\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}\right)^{\mathrm{2}} } \\ $$$$=\sqrt{\left(\mathrm{sec}\:\theta−\mathrm{tan}\:\theta\right)^{\mathrm{2}} } \\ $$$$=−\left(\mathrm{sec}\:\theta−\mathrm{tan}\:\theta\right),\:\:\because\mathrm{90}^{\mathrm{o}} <\theta\leqslant\mathrm{180}^{\mathrm{o}} \\ $$$${Similarly}, \\ $$$$\sqrt{\frac{\mathrm{1}+\mathrm{sin}\:\theta}{\mathrm{1}−\mathrm{sin}\:\theta}}\:=\:−\left(\mathrm{sec}\:\theta+\mathrm{tan}\:\theta\right) \\ $$$$\therefore\:{the}\:{given}\:{expression} \\ $$$$=−\left(\mathrm{sec}\:\theta−\mathrm{tan}\:\theta\right)−\left(\mathrm{sec}\:\theta+\mathrm{tan}\:\theta\right) \\ $$$$=−\mathrm{2sec}\:\theta \\ $$
Answered by $@ty@m123 last updated on 10/Sep/19
(c) ((sec 8θ−1)/(sec 4θ−1)) =((sec 8θ(1−cos 8θ))/(sec 4θ(1−cos 4θ))) =((cos 4θ.2sin^2 4θ)/(cos 8θ.2sin^2 2θ)) =((2sin 4θcos 4θsin4θ )/(2cos 8θsin^2 2θ)) =((sin 8θ.2sin 2θcos 2θ)/(2cos 8θsin^2 2θ)) =((tan 8θ)/(tan 2θ)) 2.(a) tan 9−tan 27−tan 63+tan 81 =tan 9−tan 27−cot 27+cot 9 =((sin 9)/(cos 9))+((cos 9)/(sin 9))−(((sin 27)/(cos 27))+((cos 27)/(sin 27))) =(1/(cos 9sin 9))−(1/(cos 27sin 27)) =(2/(sin 18))−(2/(sin 54)) =2(((sin 54−sin 18)/(sin 18sin 54))) =((4cos36sin 18 )/(sin 18sin 54)) =((4cos 36)/(sin 54)) =((4cos 36)/(cos 36)) =4
$$\left({c}\right)\:\frac{\mathrm{sec}\:\mathrm{8}\theta−\mathrm{1}}{\mathrm{sec}\:\mathrm{4}\theta−\mathrm{1}} \\ $$$$=\frac{\mathrm{sec}\:\mathrm{8}\theta\left(\mathrm{1}−\mathrm{cos}\:\mathrm{8}\theta\right)}{\mathrm{sec}\:\mathrm{4}\theta\left(\mathrm{1}−\mathrm{cos}\:\mathrm{4}\theta\right)} \\ $$$$=\frac{\mathrm{cos}\:\mathrm{4}\theta.\mathrm{2sin}\:^{\mathrm{2}} \mathrm{4}\theta}{\mathrm{cos}\:\mathrm{8}\theta.\mathrm{2sin}\:^{\mathrm{2}} \mathrm{2}\theta} \\ $$$$=\frac{\mathrm{2sin}\:\mathrm{4}\theta\mathrm{cos}\:\mathrm{4}\theta\mathrm{sin4}\theta\:}{\mathrm{2cos}\:\mathrm{8}\theta\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}\theta} \\ $$$$=\frac{\mathrm{sin}\:\mathrm{8}\theta.\mathrm{2sin}\:\mathrm{2}\theta\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{2cos}\:\mathrm{8}\theta\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}\theta} \\ $$$$=\frac{\mathrm{tan}\:\mathrm{8}\theta}{\mathrm{tan}\:\mathrm{2}\theta} \\ $$$$\mathrm{2}.\left({a}\right)\:\mathrm{tan}\:\mathrm{9}−\mathrm{tan}\:\mathrm{27}−\mathrm{tan}\:\mathrm{63}+\mathrm{tan}\:\mathrm{81} \\ $$$$=\mathrm{tan}\:\mathrm{9}−\mathrm{tan}\:\mathrm{27}−\mathrm{cot}\:\mathrm{27}+\mathrm{cot}\:\mathrm{9} \\ $$$$=\frac{\mathrm{sin}\:\mathrm{9}}{\mathrm{cos}\:\mathrm{9}}+\frac{\mathrm{cos}\:\mathrm{9}}{\mathrm{sin}\:\mathrm{9}}−\left(\frac{\mathrm{sin}\:\mathrm{27}}{\mathrm{cos}\:\mathrm{27}}+\frac{\mathrm{cos}\:\mathrm{27}}{\mathrm{sin}\:\mathrm{27}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{9sin}\:\mathrm{9}}−\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{27sin}\:\mathrm{27}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{sin}\:\mathrm{18}}−\frac{\mathrm{2}}{\mathrm{sin}\:\mathrm{54}} \\ $$$$=\mathrm{2}\left(\frac{\mathrm{sin}\:\mathrm{54}−\mathrm{sin}\:\mathrm{18}}{\mathrm{sin}\:\mathrm{18sin}\:\mathrm{54}}\right) \\ $$$$=\frac{\mathrm{4cos36sin}\:\mathrm{18}\:}{\mathrm{sin}\:\mathrm{18sin}\:\mathrm{54}} \\ $$$$=\frac{\mathrm{4cos}\:\mathrm{36}}{\mathrm{sin}\:\mathrm{54}} \\ $$$$=\frac{\mathrm{4cos}\:\mathrm{36}}{\mathrm{cos}\:\mathrm{36}} \\ $$$$=\mathrm{4} \\ $$

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