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Simplify a long trig fraction (0 < theta < 90) by converting to sin and cos

This type of question, asked 3 times: 15 Nov 2020 Tier II, 29 Jan 2022 Tier II, 18 Jan 2026 Tier II. All 6 questions below, oldest first.

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  1. Question 1

    15 Nov 2020 Tier II

    6
    The value of sec⁡2θ(2+tan⁡2θ+cot⁡2θ)÷(sin⁡2θ−tan⁡2θ)(cosec2θ+sec⁡2θ)(1+cot⁡2θ)2\frac{\sec^2\theta(2+\tan^2\theta+\cot^2\theta)\div(\sin^2\theta-\tan^2\theta)}{(\text{cosec}^2\theta+\sec^2\theta)(1+\cot^2\theta)^2} is:

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  2. Question 2

    15 Nov 2020 Tier II

    94
    sin⁡θ[(1−tan⁡θ)tan⁡θ+sec⁡2θ](1−sin⁡θ)tan⁡θ(1+tan⁡θ)(sec⁡θ+tan⁡θ)\frac{\sin\theta[(1 - \tan\theta)\tan\theta + \sec^2\theta]}{(1 - \sin\theta)\tan\theta(1 + \tan\theta)(\sec\theta + \tan\theta)} is equal to:

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  3. Question 3

    29 Jan 2022 Tier II

    75
    (1+sec⁡θ cosec θ)2(sec⁡θ−tan⁡θ)2(1+sin⁡θ)(sin⁡θ+sec⁡θ)2+(cos⁡θ+cosec θ)2\frac{(1 + \sec\theta\,\text{cosec}\,\theta)^2(\sec\theta - \tan\theta)^2(1 + \sin\theta)}{(\sin\theta + \sec\theta)^2 + (\cos\theta + \text{cosec}\,\theta)^2}, 0∘<θ<90∘0^\circ < \theta < 90^\circ, is equal to:

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  4. Question 4

    29 Jan 2022 Tier II

    81
    1+cos⁡θ−sin⁡2θsin⁡θ(1+cos⁡θ)×sec⁡2θ+cosec2θtan⁡θ+cot⁡θ\frac{1 + \cos\theta - \sin^2\theta}{\sin\theta(1 + \cos\theta)} \times \frac{\sqrt{\sec^2\theta + \text{cosec}^2\theta}}{\tan\theta + \cot\theta}, 0∘<θ<90∘0^\circ < \theta < 90^\circ, is equal to:

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  5. Question 5

    29 Jan 2022 Tier II

    37
    (tan⁡3θsec⁡2θ+cot⁡3θcosec2θ+2sin⁡θcos⁡θ)÷(1+cosec2θ+tan⁡2θ)\left(\frac{\tan^3\theta}{\sec^2\theta} + \frac{\cot^3\theta}{\text{cosec}^2\theta} + 2\sin\theta\cos\theta\right) \div (1 + \text{cosec}^2\theta + \tan^2\theta), 0∘<θ<90∘0^\circ < \theta < 90^\circ, is equal to:

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  6. Question 6

    18 Jan 2026 Tier II

    17
    Simplify: tan⁡2θ−sin⁡2θ2+tan⁡2θ+cot⁡2θ\frac{\tan^2 \theta - \sin^2 \theta}{2 + \tan^2 \theta + \cot^2 \theta}

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