Question : If $-\sin \theta+\operatorname{cosec} \theta=6$, then what is the value of $\sin \theta+\operatorname{cosec} \theta$ ?
Option 1: $6$
Option 2: $\sqrt{40}$
Option 3: $\sqrt{34}$
Option 4: $\sqrt{38}$
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Correct Answer: $\sqrt{40}$
Solution :
$-\sin \theta + \operatorname{cosec} \theta = 6$
$⇒-\sin \theta + \frac{1}{\sin \theta} = 6$
$⇒ 1 - \sin^2 \theta = 6 \sin \theta$
$⇒ \sin^2\theta + 6\sin \theta - 1 = 0$
Applying the Sridhar Acharya formula:
$x = \frac{(-b \pm \sqrt{b^2 - 4ac)}}{2a}$
Here $x = \sin \theta, a = 1, b = 6, c = -1$
$⇒ \sin \theta = \frac{-6 \pm \sqrt{6^2 - 4 \times (-1)}}{2}$
$⇒ \sin \theta = \frac{-6 \pm \sqrt{40}}{2}$
$⇒ \sin \theta =-3 \pm \sqrt{10}$
$⇒ \sin \theta + \operatorname{cosec} \theta = \sin \theta + 6 +\sin \theta$ (since, $\operatorname{cosec} \theta = 6 +\sin \theta$ )
$⇒ \sin \theta + \operatorname{cosec} \theta = 2\sin \theta +6$
$⇒ \sin \theta + \operatorname{cosec} \theta = 2(-3 \pm \sqrt{10}) + 6$
$⇒ \sin \theta + \operatorname{cosec} \theta = -6 \pm 2\sqrt{10} + 6$
$⇒ \sin \theta + \operatorname{cosec} \theta = \pm 2\sqrt{10}$
$⇒ \sin \theta + \operatorname{cosec} \theta = \pm \sqrt{40}$
Hence, the correct answer is $+ \sqrt{40}$.
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