If $ A(\theta)=\left[\begin{array}{lr}1 & \tan \theta \ -\tan \theta & 1\end{array}\right] $ and $ A B=I $, then $ \left(\sec ^{2} \theta\right) B $ is equal to

Question

If $ A(\theta)=\left[\begin{array}{lr}1 & \tan \theta \ -\tan \theta & 1\end{array}\right] $ and $ A B=I $, then $ \left(\sec ^{2} \theta\right) B $ is equal to, $ A(\theta) $, $ A\left(\frac{\theta}{2}\right) $, $ A(-\theta) $, $ A\left(-\frac{\theta}{2}\right) $

MARKS App · Accepted Answer

A B = I ⇒ B = A - 1 = 1 1 + tan 2 ⁡ θ 1 - tan ⁡ θ tan ⁡ θ 1 = 1 sec 2 ⁡ θ 1 - tan ⁡ θ tan ⁡ θ 1 ⇒ ( sec 2 ⁡ θ ) B = 1 - tan ⁡ θ tan ⁡ θ 1 = A ( - θ )

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