If A = ⎡ ⎣ 1 t a n θ 2 − t a n θ 2 1 ⎤ ⎦ and AB = I, then B = Options: A . &nbspcos2θ2.A B . &nbspcos2θ2.AT C . &nbspcos2θ2.I D . &nbspNone of these

Answer: Option B : B | A | = 1 + t a n 2 θ 2 = s e c 2 θ 2 A B = I ⇒ B − I A − 1 [ 1 0 0 1 ] ⎡ ⎢ ⎣ 1 − t a n θ 2 t a n θ 2 1 ⎤ ⎥ ⎦ s e c 2 θ 2 = c o s 2 θ 2 . A T .

If A=⎡⎣1tanθ2−tanθ21⎤⎦ and AB = I

Question

A = ⎡ ⎣ \begin{matrix} 1 & t a n \frac{θ}{2} - t a n \frac{θ}{2} & 1 \end{matrix} ⎤ ⎦

and AB = I, then B =

Options:

A . cos2θ2.A

B . cos2θ2.AT

C . cos2θ2.I

D . None of these

Answer: Option B
:
B

| A | = 1 + t a n^{2} \frac{θ}{2} = s e c^{2} \frac{θ}{2} A B = I \Rightarrow B - I A^{- 1} \frac{[\begin{matrix} 1 & 0 0 & 1 \end{matrix}] ⎡ ⎢ ⎣ \begin{matrix} 1 & - t a n \frac{θ}{2} t a n \frac{θ}{2} & 1 \end{matrix} ⎤ ⎥ ⎦}{s e c^{2} \frac{θ}{2}} = c o s^{2} \frac{θ}{2} . A^{T} .

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