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Question

$tan invisible function application 3 theta plus tan invisible function application theta equals 2 tan invisible function application 2 theta$ , then $blank theta$ is equal to $open parentheses n element of Z close parentheses$

$n π$
$\frac{n π}{4}$
$2 n π$
None of these

The correct answer is: $n pi$

$tan invisible function application 3 theta plus tan invisible function application theta equals 2 tan invisible function application 2 theta$
$rightwards double arrow tan invisible function application 3 theta minus tan invisible function application 2 theta equals tan invisible function application 2 theta minus tan invisible function application theta$
$rightwards double arrow fraction numerator sin invisible function application open parentheses 3 theta minus 2 theta close parentheses over denominator cos invisible function application 3 theta cos invisible function application 2 theta end fraction equals fraction numerator sin invisible function application open parentheses 2 theta minus theta close parentheses over denominator cos invisible function application 2 theta cos invisible function application theta end fraction$
$rightwards double arrow sin invisible function application theta open parentheses 2 sin invisible function application theta sin invisible function application 2 theta close parentheses equals 0$
$rightwards double arrow sin invisible function application theta equals 0$ or $sin invisible function application 2 theta equals 0$
$rightwards double arrow theta equals n pi blank$ or $blank 2 theta equals n pi comma blank n element of Z$
But $theta equals n pi divided by 2$ is rejected as when $n$ is odd, $tan invisible function application theta$ is not defined and when $n$ is even, i.e., $2 r$ , then $blank theta equals r pi$
Then $theta equals n pi comma blank n element of I$ is the only solution

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e value of $open parentheses 1 plus cos invisible function application fraction numerator pi over denominator 8 end fraction close parentheses open parentheses 1 plus cos invisible function application fraction numerator 3 pi over denominator 8 end fraction close parentheses open parentheses 1 plus cos invisible function application fraction numerator 5 pi over denominator 8 end fraction close parentheses open parentheses 1 plus cos invisible function application fraction numerator 7 pi over denominator 8 end fraction close parentheses$ is

parallel

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parallel

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