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Maths-

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Question

Let A = $open square brackets table row cell cos to the power of 2 end exponent invisible function application theta end cell cell sin invisible function application theta cos invisible function application theta end cell row cell cos invisible function application theta sin invisible function application theta end cell cell sin to the power of 2 end exponent invisible function application theta end cell end table close square brackets$ and B = $open square brackets table row cell cos to the power of 2 end exponent invisible function application phi end cell cell sin invisible function application phi cos invisible function application phi end cell row cell cos invisible function application phi sin invisible function application phi end cell cell sin to the power of 2 end exponent invisible function application phi end cell end table close square brackets$ then AB = 0, if-

$θ = n ϕ,$ n = 0, 1, 2, …….
$θ = ϕ$ = n $π$ n = 0, 1, 2, ……
$θ = ϕ$ + (2n +1) $\frac{π}{2}$ , n = 0, 1, 2, ……
$θ = ϕ$ + n $\frac{π}{2}$ , n = 0, 1, 2, …..

The correct answer is: $theta equals ϕ$ + (2n +1) $fraction numerator pi over denominator 2 end fraction$ , n = 0, 1, 2, ……

AB = $open square brackets table row cell cos to the power of 2 end exponent invisible function application theta cos to the power of 2 end exponent invisible function application phi plus sin invisible function application theta cos invisible function application theta cos invisible function application phi sin invisible function application phi end cell cell cos to the power of 2 end exponent invisible function application theta sin invisible function application phi cos invisible function application phi plus sin invisible function application theta cos invisible function application theta sin to the power of 2 end exponent invisible function application phi end cell row cell cos invisible function application theta sin invisible function application theta cos to the power of 2 end exponent invisible function application phi plus sin to the power of 2 end exponent invisible function application theta cos invisible function application phi sin invisible function application phi end cell cell cos invisible function application theta sin invisible function application theta sin invisible function application phi cos invisible function application phi plus sin to the power of 2 end exponent invisible function application theta sin to the power of 2 end exponent invisible function application phi end cell end table close square brackets$
AB = $open square brackets table row cell cos invisible function application theta cos invisible function application phi cos invisible function application left parenthesis theta minus phi right parenthesis end cell cell sin invisible function application phi cos invisible function application theta cos invisible function application blank left parenthesis theta minus phi right parenthesis end cell row cell sin invisible function application theta cos invisible function application phi cos invisible function application left parenthesis theta minus phi right parenthesis end cell cell sin invisible function application theta sin invisible function application phi cos invisible function application blank left parenthesis theta minus phi right parenthesis end cell end table close square brackets$
$because$ AB = 0 $rightwards double arrow$ cos $left parenthesis theta minus ϕ right parenthesis$ = 0
$rightwards double arrow$ cos $left parenthesis theta minus ϕ right parenthesis$ = cos (2n +1) $fraction numerator pi over denominator 2 end fraction$ where n = 0, 1, 2, …
 = (2n +1) $fraction numerator pi over denominator 2 end fraction$ + $ϕ$ where n = 0, 1, 2, ……

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