If $A$ and $B$ are acute angle such that $A + B$ and $A - B$ satisfy the equation $4 {sin}^{2} θ - 2 \sqrt{6} sin θ + 1 = 0$ , then-

A = \frac{π}{4}

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B = \frac{π}{6}

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A = \frac{π}{6}

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B = \frac{π}{4}

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Solution

The correct options are
A $B = \frac{π}{6}$
B $A = \frac{π}{4}$
On solving the equation
$sin θ = \frac{2 \sqrt{6} - 2 \sqrt{2}}{8}$
$sin θ = \frac{\sqrt{3} - 1}{2 \sqrt{2}}$
$A - B = \frac{π}{15}$
$A + B = \frac{14 π}{15}$
$A = \frac{π}{4}$
$B = \frac{π}{6}$
Option A and B

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If A and B are acute angle such that A+B and A−B satisfy the equation 4sin2θ−2√6sinθ+1=0, then-

The correct options are A B=π6 B A=π4On solving the equation sinθ=2√6−2√28 sinθ=√3−12√2 A−B=π15 A+B=14π15 A=π4 B=π6 Option A and B

If $A$ and $B$ are acute angle such that $A + B$ and $A - B$ satisfy the equation $4 {sin}^{2} θ - 2 \sqrt{6} sin θ + 1 = 0$ , then-

The correct options are
A $B = \frac{π}{6}$
B $A = \frac{π}{4}$
On solving the equation
$sin θ = \frac{2 \sqrt{6} - 2 \sqrt{2}}{8}$
$sin θ = \frac{\sqrt{3} - 1}{2 \sqrt{2}}$
$A - B = \frac{π}{15}$
$A + B = \frac{14 π}{15}$
$A = \frac{π}{4}$
$B = \frac{π}{6}$
Option A and B