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JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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If   2sin⁡  α1+cos⁡  α+sin⁡  α=x,    then   1−cos⁡  α−sin⁡  αcos⁡  α  \;\frac{{2\sin \,\,\alpha }}{{1 + \cos \,\,\alpha + \sin \,\,\alpha }} = x,\,\,\;\mathrm{then}\;\,\frac{{1 - \cos \,\,\alpha - \sin \,\,\alpha }}{{\cos \,\,\alpha }}\; is equal to

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Correct option: D
None of these
Option D (None of these) is correct because the given expression simplifies to −x-x, which is not among the provided options.

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If tan⁡α=pq\tan \alpha=\frac{p}{q}, where α=6β\alpha=6 \beta and α\alpha being an acute angle, then 12[pcosec⁡2β−qsec⁡2β]\frac{1}{2}[p \operatorname{cosec} 2 \beta-q \sec 2 \beta] is equal to The value of cos⁡12∘+cos⁡84∘+cos⁡156∘+cos⁡132∘\cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ} is If the △ABC\triangle A B C is acute angled at CC, then The value of the expression sin⁡6  θ+cos⁡6  θ+3sin⁡2  θ  cos⁡2  θ{\sin ^6}\,\,\theta + {\cos ^6}\,\,\theta + 3{\sin ^2}\,\,\theta \,\,{\cos ^2}\,\,\theta equals sin⁡2A2+sin⁡2B2+sin⁡2C2=\sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}+\sin ^2 \frac{C}{2}= sin⁡2(A/2)+sin⁡2(B/2)−sin⁡2(C/2)=\sin ^2(A / 2)+\sin ^2(B / 2)-\sin ^2(C / 2)=
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