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Question

Member of solutions of $tan invisible function application open parentheses fraction numerator pi over denominator 2 end fraction sin invisible function application theta close parentheses equals cot invisible function application open parentheses fraction numerator pi over denominator 2 end fraction cos invisible function application theta close parentheses comma blank theta element of open square brackets 0 comma blank 6 pi close square brackets comma blank$ is

5
7
4
5

The correct answer is: 7

$tan invisible function application open parentheses fraction numerator pi over denominator 2 end fraction sin invisible function application theta close parentheses equals cot invisible function application open parentheses fraction numerator pi over denominator 2 end fraction cos invisible function application theta close parentheses$
$therefore tan invisible function application open parentheses fraction numerator pi over denominator 2 end fraction sin invisible function application theta close parentheses equals tan invisible function application open parentheses fraction numerator pi over denominator 2 end fraction minus fraction numerator pi over denominator 2 end fraction cos invisible function application theta close parentheses$
$therefore fraction numerator pi over denominator 2 end fraction sin invisible function application theta equals n pi plus fraction numerator pi over denominator 2 end fraction minus fraction numerator pi over denominator 2 end fraction cos invisible function application theta blank$
$sin invisible function application theta plus cos invisible function application theta equals 2 n plus 1$
$rightwards double arrow sin invisible function application theta plus cos invisible function application theta equals 1 plus-or-minus 1$
$rightwards double arrow 1 plus sin invisible function application 2 theta equals 1$
$rightwards double arrow sin invisible function application 2 theta equals 0$
$rightwards double arrow theta equals n pi$

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