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-

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

\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 .