Question : If $b \cos \theta=a$, then $\operatorname{cosec} \theta+\cot \theta=$_______.
Option 1: $\sqrt{\frac{1}{b+a}}$
Option 2: $\sqrt{\frac{b-a}{b+a}}$
Option 3: $\sqrt{\frac{b+a}{b-a}}$
Option 4: $\sqrt{\frac{1}{b-a}}$
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Correct Answer: $\sqrt{\frac{b+a}{b-a}}$
Solution :
Given, $b \cos \theta=a$
⇒ $\cos \theta=\frac{a}{b}$
We know that $\sin^2 \theta + \cos^2 \theta = 1$
So, $\sin^2 \theta + (\frac{a}{b})^2 = 1$
⇒ $\sin^2 \theta = 1-(\frac{a}{b})^2$
⇒ $\sin \theta = \frac{\sqrt{b^2-a^2}}{b}$
Now, $\operatorname{cosec} \theta+\cot \theta=\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}$
⇒ $\operatorname{cosec} \theta+\cot \theta=\frac{b}{\sqrt{b^2-a^2}}+\frac{\frac{a}{b}}{\frac{\sqrt{b^2-a^2}}{b}}$
⇒ $\operatorname{cosec} \theta+\cot \theta=\frac{b}{\sqrt{b^2-a^2}}+\frac{a}{\sqrt{b^2-a^2}}$
⇒ $\operatorname{cosec} \theta+\cot \theta=\frac{a+b}{\sqrt{(b-a)(b+a)}}$
$\therefore \operatorname{cosec} \theta+\cot \theta=\sqrt{\frac{b+a}{b-a}}$
Hence, the correct answer is $\sqrt{\frac{b+a}{b-a}}$.
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