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(1 + tan a tan b)^2 + (tan a - tan b)^2 = sec^2 a sec^2 b

The same question, asked 2 times: CDS 1 2022, CDS 2 2024. Every wording below, oldest first.

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  1. Question 1

    CDS 1 2022

    55
    What is (sec⁡2α+tan⁡α⋅tan⁡β−tan⁡2α)2+(tan⁡α−tan⁡β)2−sec⁡2α⋅sec⁡2β(\sec^2\alpha + \tan\alpha \cdot \tan\beta - \tan^2\alpha)^2 + (\tan\alpha - \tan\beta)^2 - \sec^2\alpha \cdot \sec^2\beta equal to?

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  2. Question 2

    CDS 2 2024

    12
    What is (1+tan⁡α tan⁡β)2+(tan⁡α−tan⁡β)2(1 + \tan\alpha\,\tan\beta)^2 + (\tan\alpha - \tan\beta)^2 equal to ?

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