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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Question : If $\cot \theta=\frac{1}{\sqrt{3}}, 0^{\circ}<\theta<90^{\circ}$, then the value of $\frac{2-\sin ^2 \theta}{1-\cos ^2 \theta}+\left(\operatorname{cosec}^2 \theta-\sec \theta\right)$ is:
Question : If $\frac{\cos \theta}{1-\sin \theta}+\frac{\cos \theta}{1+\sin \theta}=4,0^{\circ}<\theta<90^{\circ}$, then what is the value of $(\sec \theta+\operatorname{cosec} \theta+\cot \theta) ?$
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