If - and - are values of - satisfying the equation -3-1sin-3-1cos-4-2- where -0-2- then the value of - is

nbsp;√(3) 1sinθ+√(3)+1cosθ=4√(2) ⇒√(6) √(2)4sinθ+√(6)+√(2)4cosθ=2 ⇒sinπ12sinθ+ cosπ12cosθ=2 ⇒sin(π12)cosθ+cos(π12)sinθ=sin 2θ ⇒sin(θ +π12)=sin 2θ ⇒θ +π12=nπ+ ...

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Byju's Answer

Standard XII

Mathematics

Equations of the Form acosθ+bsinθ = c

If α and β ar...

Question

If $α$ and $β$ are values of $θ$ satisfying the equation $\frac{\sqrt{3} - 1}{sin θ} + \frac{\sqrt{3} + 1}{cos θ} = 4 \sqrt{2},$ where $θ \in (0, \frac{π}{2}),$ then the value of $| α - β |$ is

\frac{2 π}{3}

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\frac{2 π}{5}

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\frac{2 π}{7}

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\frac{2 π}{9}

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Solution

The correct option is D $\frac{2 π}{9}$
$\frac{\sqrt{3} - 1}{sin θ} + \frac{\sqrt{3} + 1}{cos θ} = 4 \sqrt{2}$
$\Rightarrow \frac{\sqrt{6} - \sqrt{2}}{4 sin θ} + \frac{\sqrt{6} + \sqrt{2}}{4 cos θ} = 2$
$\Rightarrow \frac{sin \frac{π}{12}}{sin θ} + \frac{cos \frac{π}{12}}{cos θ} = 2$
$\Rightarrow sin (\frac{π}{12}) cos θ + cos (\frac{π}{12}) sin θ = sin 2 θ$
$\Rightarrow sin (θ + \frac{π}{12}) = sin 2 θ$
$\Rightarrow θ + \frac{π}{12} = n π + (- 1)^{n} (2 θ),$ $n \in Z$
$\Rightarrow θ = \frac{π}{12}, \frac{11 π}{36}$

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If α and β are values of θ satisfying the equation √3−1sinθ+√3+1cosθ=4√2, where θ∈(0,π2), then the value of |α−β| is

The correct option is D 2π9 √3−1sinθ+√3+1cosθ=4√2 ⇒√6−√24sinθ+√6+√24cosθ=2 ⇒sinπ12sinθ+cosπ12cosθ=2 ⇒sin(π12)cosθ+cos(π12)sinθ=sin2θ ⇒sin(θ+π12)=sin2θ ⇒θ+π12=nπ+(−1)n(2θ), n∈Z ⇒θ=π12,11π36

If $α$ and $β$ are values of $θ$ satisfying the equation $\frac{\sqrt{3} - 1}{sin θ} + \frac{\sqrt{3} + 1}{cos θ} = 4 \sqrt{2},$ where $θ \in (0, \frac{π}{2}),$ then the value of $| α - β |$ is