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Byju's Answer
Standard XII
Mathematics
Adjoint of a Matrix
The value of ...
Question

The value of ⎡⎢ ⎢ ⎢⎣1−tanθ4tanθ41⎤⎥ ⎥ ⎥⎦⎡⎢ ⎢ ⎢⎣1tanθ4−tanθ41⎤⎥ ⎥ ⎥⎦−1 is

A
⎡⎢ ⎢ ⎢⎣cosθ2sinθ2−sinθ2cosθ2⎤⎥ ⎥ ⎥⎦
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B
⎡⎢ ⎢ ⎢⎣cosθ2−sinθ2sinθ2cosθ2⎤⎥ ⎥ ⎥⎦
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C
⎡⎢ ⎢ ⎢⎣sinθ2cosθ2−cosθ2sinθ2⎤⎥ ⎥ ⎥⎦
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D
⎡⎢ ⎢ ⎢⎣sinθ2−cosθ2cosθ2sinθ2⎤⎥ ⎥ ⎥⎦
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Solution

The correct option is A ⎡⎢ ⎢ ⎢⎣cosθ2−sinθ2sinθ2cosθ2⎤⎥ ⎥ ⎥⎦
=⎡⎢ ⎢ ⎢⎣1−tanθ4tanθ41⎤⎥ ⎥ ⎥⎦1(1+tan2θ4)⎡⎢ ⎢ ⎢⎣1−tanθ4tanθ41⎤⎥ ⎥ ⎥⎦

=1sec2θ4⎡⎢ ⎢ ⎢⎣1−tanθ4tanθ41⎤⎥ ⎥ ⎥⎦⎡⎢ ⎢ ⎢⎣1−tanθ4tanθ41⎤⎥ ⎥ ⎥⎦

=cos2θ4⎡⎢ ⎢ ⎢⎣1−tan2θ4−2tanθ42tanθ41−tan2θ4⎤⎥ ⎥ ⎥⎦

=cos2θ4⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣cos2θ4−sin2θ4cos2θ4−2sinθ4cosθ42sinθ4cosθ4cos2θ4−sin2θ4cos2θ4⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦

=⎡⎢ ⎢ ⎢⎣cosθ2−sinθ2sinθ2cosθ2⎤⎥ ⎥ ⎥⎦

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