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))


Question and Answers Forum

All Questions Topic List
Trigonometry Questions
Previous in All Question Next in All Question
Previous in Trigonometry Next in Trigonometry
Question Number 19097 by Tinkutara last updated on 04/Aug/17

$If \tan (\frac{π}{4} + x) = \tan^{3} (\frac{π}{4} + α) then$ $prove that cosec 2 x = \frac{1 + 3 \sin^{2} 2 α}{3 \sin 2 α + \sin^{3} 2 α}$
	Answered by 951172235v last updated on 01/Feb/19
	$((cos x+sin x)/(cos x−sin x)) = (((cos α+sin α)/(cos α−sin α)))^3 →(1) {(1)+1}{(1)−1} ((4cos xsin x)/((cos x−sin x)^2 )) = ((4(cos^3 α+3cos αsin^2 α)(sin^3 α+3cos^2 αsin α))/((cos α−sin α)^6 )) ((sin 2x)/(1−sin 2x)) =((2cos αsin α(1+2sin^2 α)(1+2cos^2 α))/((1−2αsin 2α)^3 )) ((1 )/(cosec 2x−1)) =((sin 2α(3+sin^2 2α))/((1−sin 2α)^3 )) cosec 2x =1+(((1−sin 2α)^3 )/((3sin 2α+sin^3 2α))) cosec 2x =((3sin 2α+sin^3 2α+1−3sin 2α+3sin^2 2α−sin^3 2α)/(3sin 2α+sin^3 α)) = ((1+3sin^2 2α)/(3sin2α+sin^3 2α )) ans.$
	$\frac{\cos x + \sin x}{\cos x - \sin x} = {(\frac{\cos α + \sin α}{\cos α - \sin α})}^{3} \to (1)$ ${(1) + 1} {(1) - 1}$ $\frac{4 \cos xsin x}{{(\cos x - \sin x)}^{2}} = \frac{4 (\cos^{3} α + 3 \cos α \sin^{2} α) (\sin^{3} α + 3 \cos^{2} α \sin α)}{{(\cos α - \sin α)}^{6}}$ $\frac{\sin 2 x}{1 - \sin 2 x} = \frac{2 \cos α \sin α (1 + 2 \sin^{2} α) (1 + 2 \cos^{2} α)}{{(1 - 2 α \sin 2 α)}^{3}}$ $\frac{1}{cosec 2 x - 1} = \frac{\sin 2 α (3 + \sin^{2} 2 α)}{{(1 - \sin 2 α)}^{3}}$ $cosec 2 x = 1 + \frac{{(1 - \sin 2 α)}^{3}}{(3 \sin 2 α + \sin^{3} 2 α)}$ $cosec 2 x = \frac{3 \sin 2 α + \sin^{3} 2 α + 1 - 3 \sin 2 α + 3 \sin^{2} 2 α - \sin^{3} 2 α}{3 \sin 2 α + \sin^{3} α}$ $= \frac{1 + 3 \sin^{2} 2 α}{3 \sin 2 α + \sin^{3} 2 α} ans .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum