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A^T A^-1 or A^-1 B for matrices like [1, tan t; -tan t, 1]

This type of question, asked 5 times: 10 May 2023 S1, 13 May 2024 S2, 20 Apr 2025 S1, 22 Apr 2025 S1, 17 Apr 2026 S2. All 5 questions below, oldest first.

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  1. Question 1

    10 May 2023 S1

    115
    If A+B=(1tan⁡θ2−tan⁡θ21)A+B=\begin{pmatrix}1 & \tan\frac{\theta}{2}\\-\tan\frac{\theta}{2} & 1\end{pmatrix}, where A is symmetric and B is skew-symmetric matrix, then the matrix A−1B+AB−1A^{-1}B+AB^{-1} at θ=π6\theta=\frac{\pi}{6} is given by

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  2. Question 2

    13 May 2024 S2

    126
    If A=(1tan⁡x−tan⁡x1)A=\begin{pmatrix}1 & \tan x\\ -\tan x & 1\end{pmatrix}, then ATA−1=A^T A^{-1}=

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  3. Question 3

    20 Apr 2025 S1

    148
    If A=[1tan⁡x−tan⁡x1]A =\begin{bmatrix} 1 & \tan x \\ - \tan x & 1 \end{bmatrix}, then ATA−1=A^{T}A^{- 1}=

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  4. Question 4

    22 Apr 2025 S1

    102
    If A=[1cot⁡θ2−cot⁡θ21]A =\begin{bmatrix} 1 & \cot\frac{\theta}{2} \\ - \cot\frac{\theta}{2} & 1 \end{bmatrix} then A−1=A^{- 1}=

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  5. Question 5

    17 Apr 2026 S2

    62
    If A=[1−tan⁡θ2tan⁡θ21]A = \begin{bmatrix} 1 & -\tan\dfrac{\theta}{2} \\ \tan\dfrac{\theta}{2} & 1 \end{bmatrix} and B=[1tan⁡θ2−tan⁡θ21]B = \begin{bmatrix} 1 & \tan\dfrac{\theta}{2} \\ -\tan\dfrac{\theta}{2} & 1 \end{bmatrix} then A−1BA^{-1}B is equal to

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