If A(θ)=1tan⁡θ−tan⁡θ1 and AB=I, then cos2⁡θB is equal to

As AB=I, we get B=A−1∴ B=11+tan2⁡θ1−tan⁡θtan⁡θ1⇒ cos2⁡θB=1−tan⁡θtan⁡θ1=A(−θ)

If A(θ)=1tan⁡θ−tan⁡θ1 and AB=I, then cos2⁡θB is equal to

As AB=I, we get B=A−1∴ B=11+tan2⁡θ1−tan⁡θtan⁡θ1⇒ cos2⁡θB=1−tan⁡θtan⁡θ1=A(−θ)

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If $A (θ) = (\begin{array}{cc} 1 & \tan θ \\ - \tan θ & 1 \end{array}) and A B = I, then (\cos^{2} θ) B$ is equal to

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$A (θ)$

$A (- θ)$

$A (\frac{θ}{2})$

$A (- \frac{θ}{2})$

answer is B.

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Detailed Solution

$As AB = I, we get B = A^{- 1}$

$∴ B = \frac{1}{1 + \tan^{2} θ} (\begin{array}{cc} 1 & - \tan θ \\ \tan θ & 1 \end{array})$

$\Rightarrow (\cos^{2} θ) B = (\begin{array}{cc} 1 & - \tan θ \\ \tan θ & 1 \end{array}) = A (- θ)$

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If A(θ)=1tan⁡θ−tan⁡θ1 and AB=I, then cos2⁡θB is equal to

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If $A (θ) = (\begin{array}{cc} 1 & \tan θ \\ - \tan θ & 1 \end{array}) and A B = I, then (\cos^{2} θ) B$ is equal to