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If-tan-pi-4-x-tan-3-pi-4-then-prove-that-cosec-2x-1-3-sin-2-2-3-sin-2-sin-3-2-

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Question Number 19097 by Tinkutara last updated on 04/Aug/17
If tan ((π/4) + x) = tan^3 ((π/4) + α) then prove that cosec 2x = ((1 + 3 sin^2 2α)/(3 sin 2α + sin^3 2α))
$$\mathrm{If}\:\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}\:+\:{x}\right)\:=\:\mathrm{tan}^{\mathrm{3}} \:\left(\frac{\pi}{\mathrm{4}}\:+\:\alpha\right)\:\mathrm{then} \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{cosec}\:\mathrm{2}{x}\:=\:\frac{\mathrm{1}\:+\:\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}\alpha}{\mathrm{3}\:\mathrm{sin}\:\mathrm{2}\alpha\:+\:\mathrm{sin}^{\mathrm{3}} \:\mathrm{2}\alpha} \\ $$
Answered by 951172235v last updated on 01/Feb/19
((cos x+sin x)/(cos x−sin x)) = (((cos α+sin α)/(cos α−sin α)))^3 →(1) {(1)+1}{(1)−1} ((4cos xsin x)/((cos x−sin x)^2 )) = ((4(cos^3 α+3cos αsin^2 α)(sin^3 α+3cos^2 αsin α))/((cos α−sin α)^6 )) ((sin 2x)/(1−sin 2x)) =((2cos αsin α(1+2sin^2 α)(1+2cos^2 α))/((1−2αsin 2α)^3 )) ((1 )/(cosec 2x−1)) =((sin 2α(3+sin^2 2α))/((1−sin 2α)^3 )) cosec 2x =1+(((1−sin 2α)^3 )/((3sin 2α+sin^3 2α))) cosec 2x =((3sin 2α+sin^3 2α+1−3sin 2α+3sin^2 2α−sin^3 2α)/(3sin 2α+sin^3 α)) = ((1+3sin^2 2α)/(3sin2α+sin^3 2α )) ans.
$$\frac{\mathrm{cos}\:\mathrm{x}+\mathrm{sin}\:\mathrm{x}}{\mathrm{cos}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}}\:\:=\:\:\left(\frac{\mathrm{cos}\:\alpha+\mathrm{sin}\:\alpha}{\mathrm{cos}\:\alpha−\mathrm{sin}\:\alpha}\right)^{\mathrm{3}} \rightarrow\left(\mathrm{1}\right) \\ $$$$\left\{\left(\mathrm{1}\right)+\mathrm{1}\right\}\left\{\left(\mathrm{1}\right)−\mathrm{1}\right\} \\ $$$$\frac{\mathrm{4cos}\:\mathrm{xsin}\:\mathrm{x}}{\left(\mathrm{cos}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}\right)^{\mathrm{2}} }\:=\:\:\frac{\mathrm{4}\left(\mathrm{cos}\:^{\mathrm{3}} \alpha+\mathrm{3cos}\:\alpha\mathrm{sin}^{\mathrm{2}} \alpha\right)\left(\mathrm{sin}\:^{\mathrm{3}} \alpha+\mathrm{3cos}\:^{\mathrm{2}} \alpha\mathrm{sin}\:\alpha\right)}{\left(\mathrm{cos}\:\alpha−\mathrm{sin}\:\alpha\right)^{\mathrm{6}} } \\ $$$$\frac{\mathrm{sin}\:\mathrm{2x}}{\mathrm{1}−\mathrm{sin}\:\mathrm{2x}}\:\:\:=\frac{\mathrm{2cos}\:\alpha\mathrm{sin}\:\alpha\left(\mathrm{1}+\mathrm{2sin}\:^{\mathrm{2}} \alpha\right)\left(\mathrm{1}+\mathrm{2cos}\:^{\mathrm{2}} \alpha\right)}{\left(\mathrm{1}−\mathrm{2}\alpha\mathrm{sin}\:\mathrm{2}\alpha\right)^{\mathrm{3}} } \\ $$$$\frac{\mathrm{1}\:}{\mathrm{cosec}\:\mathrm{2x}−\mathrm{1}}\:=\frac{\mathrm{sin}\:\mathrm{2}\alpha\left(\mathrm{3}+\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}\alpha\right)}{\left(\mathrm{1}−\mathrm{sin}\:\mathrm{2}\alpha\right)^{\mathrm{3}} } \\ $$$$\mathrm{cosec}\:\mathrm{2x}\:=\mathrm{1}+\frac{\left(\mathrm{1}−\mathrm{sin}\:\mathrm{2}\alpha\right)^{\mathrm{3}} }{\left(\mathrm{3sin}\:\mathrm{2}\alpha+\mathrm{sin}\:^{\mathrm{3}} \mathrm{2}\alpha\right)} \\ $$$$\mathrm{cosec}\:\mathrm{2x}\:=\frac{\mathrm{3sin}\:\mathrm{2}\alpha+\mathrm{sin}\:^{\mathrm{3}} \mathrm{2}\alpha+\mathrm{1}−\mathrm{3sin}\:\mathrm{2}\alpha+\mathrm{3sin}\:^{\mathrm{2}} \mathrm{2}\alpha−\mathrm{sin}\:^{\mathrm{3}} \mathrm{2}\alpha}{\mathrm{3sin}\:\mathrm{2}\alpha+\mathrm{sin}\:^{\mathrm{3}} \alpha} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}+\mathrm{3sin}\:^{\mathrm{2}} \mathrm{2}\alpha}{\mathrm{3sin2}\alpha+\mathrm{sin}^{\mathrm{3}} \mathrm{2}\alpha\:}\:\mathrm{ans}. \\ $$

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