If A = ⎡ ⎣ 1 t a n θ 2 − t a n θ 2 1 ⎤ ⎦ and AB = I, then B = Options: A . &nbsp c o s 2 θ 2 . A B . &nbsp c o s 2 θ 2 . A T C . &nbsp c o s 2 θ 2 . I D . &nbsp None 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 =

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Question -

If $A = ⎡ ⎣ \begin{matrix} 1 & t a n \frac{θ}{2} - t a n \frac{θ}{2} & 1 \end{matrix} ⎤ ⎦$ and AB = I, then B =

Options:

A .

c o s^{2} \frac{θ}{2} . A

B .

c o s^{2} \frac{θ}{2} . A^{T}

C .

c o s^{2} \frac{θ}{2} . 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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