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Maths-

General

Easy

Question

If A( $theta$ ) = $open square brackets table row 1 cell tan invisible function application theta end cell row cell negative tan invisible function application theta end cell 1 end table close square brackets$ and AB = I, then (sec² $theta$ ) B is equal to-

A( $θ$ )
A(– $θ$ )
A( $θ$ /2)
A(– $θ$ /2)

The correct answer is: A(– $theta$ )

As AB = I, we get B = A^–1
$therefore B equals fraction numerator 1 over denominator 1 plus tan squared invisible function application theta end fraction open square brackets table row 1 cell negative tan invisible function application theta end cell row cell tan invisible function application theta end cell 1 end table close square brackets$
$rightwards double arrow$ (sec^{2 $theta$}) B = $open square brackets table row 1 cell negative tan invisible function application theta end cell row cell tan invisible function application theta end cell 1 end table close square brackets equals A left parenthesis negative theta right parenthesis$

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