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(I + A)(I - A) inverse for a skew-symmetric A

This type of question, asked 2 times: 2021 Paper 3, 23 Jan 2026 S2. All 2 questions below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    2021 Paper 3

    84
    If A=[0−tan⁡(θ2)tan⁡(θ2)0]A=\begin{bmatrix} 0 & - \tan\left( \frac{\theta}{2} \right) \\ \tan\left( \frac{\theta}{2} \right) & 0 \end{bmatrix} and (I2+A)(I2−A)−1\left( I_{2}+A \right)\left( I_{2}-A \right)^{- 1} =[a−bba]=\begin{bmatrix} a & -b \\ b & a \end{bmatrix}, then 13(a2+b2)13\left( a^{2}+b^{2} \right) is equal to

  2. Question 2

    23 Jan 2026 S2

    74
    Let A=[02−3−2013−10]A = \begin{bmatrix} 0 & 2 & - 3 \\ - 2 & 0 & 1 \\ 3 & - 1 & 0 \end{bmatrix} and B be a matrix such that B(I−A)=I+AB(I - A) = I + A. Then the sum of the diagonal elements of BTBB^{T}B is equal to