1.If A=[1tanαtanα21]A = \begin{bmatrix}1&\tan\alpha\\\tan\frac{\alpha}{2}&1\end{bmatrix}A=[1tan2αtanα1] and AB=IAB=IAB=I, then B=B =B=a.cos2α2⋅A\cos^2\frac{\alpha}{2}\cdot Acos22α⋅Ab.cos2α2⋅I\cos^2\frac{\alpha}{2}\cdot Icos22α⋅Ic.sin2α2⋅A\sin^2\frac{\alpha}{2}\cdot Asin22α⋅Ad.cos2α2⋅AT\cos^2\frac{\alpha}{2}\cdot A^Tcos22α⋅ATLogin to continueOnly logged in users canattempt or see the solution.
