If A=≤ft[{array}{cc}1 & - θ/2 \\ θ/2 & 1{array}] and B=≤ft[{array}{cc}1 & θ/2 \\ - θ/2 & 1{array}] then A^{-1} B is equal to

Correct Answer is : [cos⁡θsin⁡θ−sin⁡θcos⁡θ]≤ft[{array}{cc} θ & θ \\ - θ & θ{array}][cosθ−sinθsinθcosθ]

≤ft[{array}{cc} θ & θ \\ - θ & θ{array}]

≤ft[{array}{cc} θ & - θ \\ θ & θ{array}]

≤ft[{array}{cc} θ & θ \\ θ & θ{array}]

Correct Answer is : [cos⁡θsin⁡θ−sin⁡θcos⁡θ]≤ft[{array}{cc} θ & θ \\ - θ & θ{array}][cosθ−sinθsinθcosθ]

Test Index

MHTCET 17 Apr 2026 Shift 2 Paper

Question : 112 of 150

Marks: +1, -0

A=\left[\begin{array}{cc}1 & -\tan \frac{\theta}{2} \\ \tan \frac{\theta}{2} & 1\end{array}\right]

B=\left[\begin{array}{cc}1 & \tan \frac{\theta}{2} \\ -\tan \frac{\theta}{2} & 1\end{array}\right]

A^{-1} B

is equal to

$\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$
$\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]$
$\left[\begin{array}{cc}\sin \theta & -\cos \theta \\ \cos \theta & \sin \theta\end{array}\right]$
$\left[\begin{array}{cc}\cos \theta & \sin \theta \\ \sin \theta & \cos \theta\end{array}\right]$

Solution:

Concept:

Use the half-angle substitution

t=\tan \frac{\theta}{2}

to compute

A^{-1}

and then multiply by

B

Explanation:

Let

t=\tan \frac{\theta}{2}

Then

A=\begin{bmatrix}1 & -t \\ t & 1\end{bmatrix}

and

B=\begin{bmatrix}1 & t \\ -t & 1\end{bmatrix}

For a

2\times 2

matrix

\begin{bmatrix}a & b \\ c & d\end{bmatrix}

, the inverse is

\frac{1}{ad-bc}\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}

Here

|A|=1(1)-(-t)(t)=1+t^{2}

A^{-1}=\frac{1}{1+t^{2}}\begin{bmatrix}1 & t \\ -t & 1\end{bmatrix}=\frac{1}{1+t^{2}}B

Therefore,

A^{-1}B=\frac{1}{1+t^{2}}B\cdot B

Since

B\cdot B=\begin{bmatrix}1 & t \\ -t & 1\end{bmatrix}\begin{bmatrix}1 & t \\ -t & 1\end{bmatrix}=\begin{bmatrix}1-t^{2} & 2t \\ -2t & 1-t^{2}\end{bmatrix}

we get

A^{-1}B=\frac{1}{1+t^{2}}\begin{bmatrix}1-t^{2} & 2t \\ -2t & 1-t^{2}\end{bmatrix}

Use the identities

\cos \theta=\frac{1-t^{2}}{1+t^{2}}

and

\sin \theta=\frac{2t}{1+t^{2}}

Thus,

A^{-1}B=\begin{bmatrix}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{bmatrix}

Answer:

\begin{bmatrix}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{bmatrix}

, which is Option A.

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