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JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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If sin⁡4α+4cos⁡4 β+2=42  sin⁡α  cos⁡β ;   α,β  ∈  [0, π],{\sin ^4}\alpha + 4{\cos ^4}\,\beta + 2 = 4\sqrt 2 \,\,\sin \alpha \,\,\cos \beta \,;\,\,\,\alpha ,\beta \,\, \in \,\,\left[ {0,\,\pi } \right], then cos⁡(α+β)−cos⁡(α−β)\cos \left( {\alpha + \beta } \right) - \cos \left( {\alpha - \beta } \right) is equal to
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Correct option: C
−2- \sqrt 2
Option C (−2- \sqrt 2) is correct because by applying AM >= GM inequality, we find the values of sin⁡α\sin \alpha and cos⁡β\cos \beta, and then calculate the expression cos⁡(α+β)−cos⁡(α−β)\cos (\alpha + \beta ) - \cos (\alpha - \beta ).

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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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