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Fraction with sin^6 + cos^6, sin^4 + cos^4 or tan^6 - sec^6 power identities: simplify

This type of question, asked 2 times: 15 Nov 2020 Tier II, 3 Feb 2022 Tier II. All 3 questions below, oldest first.

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  1. Question 1

    15 Nov 2020 Tier II

    54
    The value of cos⁡6θ+sin⁡6θ+3sin⁡2θcos⁡2θcosec θsec⁡θ(sin⁡θ+cos⁡θ−1)(sin⁡θ+cos⁡θ+1)\frac{\cos^6\theta + \sin^6\theta + 3\sin^2\theta\cos^2\theta}{\text{cosec}\,\theta\sec\theta(\sin\theta + \cos\theta - 1)(\sin\theta + \cos\theta + 1)} is:

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  2. Question 2

    3 Feb 2022 Tier II

    32
    The expression tan⁡6θ−sec⁡6θ+3sec⁡2θtan⁡2θtan⁡2θ+cot⁡2θ+2\frac{\tan^6\theta - \sec^6\theta + 3\sec^2\theta\tan^2\theta}{\tan^2\theta + \cot^2\theta + 2}, 0∘<θ<90∘0^\circ < \theta < 90^\circ, is equal to:

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  3. Question 3

    3 Feb 2022 Tier II

    99
    The expression (1−2sin⁡2θcos⁡2θ)(cot⁡θ+1)cos⁡θ(sin⁡4θ+cos⁡4θ)(1+tan⁡θ)cosec⁡θ−1\frac{(1 - 2\sin^2\theta\cos^2\theta)(\cot\theta + 1)\cos\theta}{(\sin^4\theta + \cos^4\theta)(1 + \tan\theta)\operatorname{cosec}\theta} - 1, 0∘<θ<90∘0^\circ < \theta < 90^\circ, equals:

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