Question : If $\frac{1}{\operatorname{cosec} \theta+1}+\frac{1}{\operatorname{cosec} \theta-1}=2 \sec \theta, 0^{\circ}<\theta<90^{\circ}$, then the value of $\frac{\tan \theta+2 \sec \theta}{\operatorname{cosec} \theta}$ is:
Option 1: $\frac{4+\sqrt{2}}{2}$
Option 2: $\frac{2+\sqrt{3}}{2}$
Option 3: $\frac{4+\sqrt{3}}{2}$
Option 4: $\frac{2+\sqrt{2}}{2}$
Correct Answer: $\frac{4+\sqrt{2}}{2}$
Solution :
Given: $\frac{1}{\operatorname{cosec} \theta+1}+\frac{1}{\operatorname{cosec} \theta-1}=2 \sec \theta$
⇒ $\frac{\operatorname{cosec} \theta - 1 + \operatorname{cosec} \theta + 1}{\operatorname{cosec}^2 \theta - 1} = 2 \sec \theta$
⇒ $\frac{2 \operatorname{cosec} \theta}{\operatorname{cosec}^2 \theta - 1} = 2 \sec \theta$
⇒ $\operatorname{cosec} \theta = \sec \theta (\operatorname{cosec}^2 \theta - 1)$
We know that,
$\operatorname{cosec}^2 \theta = 1 + \cot^2 \theta$ and $\sec^2 \theta = 1 + \tan^2 \theta$.
Putting the values, we get:
⇒ $\operatorname{cosec} \theta = \sec \theta (\cot^2 \theta)$
⇒ $ \cot \theta =1$
Since $0^{\circ}<\theta<90^{\circ}$, the only solution is $\theta = 45^{\circ}$.
Now, Putting $\theta = 45^{\circ}$, we get:
$\frac{\tan \theta+2 \sec \theta}{\operatorname{cosec} \theta}$
$=\frac{\tan 45^{\circ}+2 \sec 45^{\circ}}{\operatorname{cosec} 45^{\circ}} = \frac{1+2\sqrt{2}}{\sqrt{2}} = \frac{\sqrt2 +4}{\sqrt{2}} =\frac {4 + \sqrt{2}}{2}$
Hence, the correct answer is $\frac {4 + \sqrt{2}}{2}$.
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