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Question Number 192466 by Engr_Jidda last updated on 18/May/23

Commented by Frix last updated on 19/May/23

$$({\sin }^4\theta -{\cos }^4\theta ){\csc }^2\theta =2$$$$\frac{{2\sin }^2\theta -1}{{\sin }^2\theta }=2$$$${2\sin }^2\theta -1={2\sin }^2\theta$$$$-1=0$$$$\mathrm{Wrong}.\nexists \theta :({\sin }^4\theta -{\cos }^4\theta ){\csc }^2\theta =2$$

Answered by cortano12 last updated on 19/May/23

$$\Rightarrow (\sin {}^2\theta -\cos {}^2\theta ).1.\frac{1}{\sin {}^2\theta }$$$$\Rightarrow (\sin {}^2\theta -(1-\sin {}^2\theta )).\frac{1}{\sin {}^2\theta }$$$$\Rightarrow (2\sin {}^2\theta -1).\frac{1}{\sin {}^2\theta }$$$$\Rightarrow 2-\csc {}^2\theta$$

Answered by Spillover last updated on 19/May/23

$$(\sin {}^2\theta +\cos {}^2\theta )(\sin {}^2\theta -\cos {}^2\theta )]\mathrm{cosec}{}^2\theta$$$$(\sin {}^2\theta -\cos {}^2\theta )]\mathrm{cosec}{}^2\theta$$$$1-\cot {}^2\theta$$$$pleasecheck$$

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