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Question Number 147809 by mathdanisur last updated on 23/Jul/21

$$\frac{{\mathit{s}\mathit{i}\mathit{n}}^6\left(\mathit{\alpha }\right)+{\mathit{c}\mathit{o}\mathit{s}}^6\left(\mathit{\alpha }\right)-1}{{\mathit{s}\mathit{i}\mathit{n}}^4\left(\mathit{\alpha }\right)-{\mathit{s}\mathit{i}\mathit{n}}^2\left(\mathit{\alpha }\right)}=?$$

Answered by gsk2684 last updated on 23/Jul/21

$${\left(\sin {}^2\alpha \right)}^3+{\left(\cos {}^2\alpha \right)}^3$$$$=({\sin }^2\alpha +\cos {}^2\alpha )(\sin {}^4\alpha -\sin {}^2\alpha \cos {}^2\alpha +\cos {}^4\alpha )$$$$=\left(1\right)({(\sin {}^2\alpha +\cos {}^2\alpha )}^2-3\sin {}^2\alpha \cos {}^2\alpha )$$$$=1-3\sin {}^2\alpha \cos {}^2\alpha$$$$answer$$$$\frac{-3\sin {}^2\alpha \cos {}^2\alpha }{\sin {}^2\alpha (\sin {}^2\alpha -1)}=3$$

Commented by mathdanisur last updated on 23/Jul/21

$$ThankyouSir$$$$Answer:3or-3.?$$

Commented by puissant last updated on 23/Jul/21

$$3\mathrm{because}\frac{-{3\sin }^2\alpha {\cos }^2\alpha }{{\sin }^2\alpha ({\sin }^2\alpha -1)}$$$$=\frac{-{3\cos }^2\alpha }{-(1-{\sin }^2\alpha )}=\frac{3(1-{\sin }^2\alpha )}{(1-{\sin }^2\alpha )}=3...$$

Commented by mathdanisur last updated on 23/Jul/21

$$thankyouSir$$

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