Question : Using $\operatorname{cosec}(\alpha+\beta)=\frac{\sec \alpha \times \sec \beta \times \operatorname{cosec} \alpha \times \operatorname{cosec} \beta}{\sec \alpha \times \operatorname{cosec} \beta+\operatorname{cosec} \alpha \times \sec \beta}$, find the value of $\operatorname{cosec} 75°$.
Option 1: $\frac{\sqrt{6}+\sqrt{2}}{4}$
Option 2: $\frac{\sqrt{6}-\sqrt{2}}{4}$
Option 3: $\sqrt{6}-\sqrt{2}$
Option 4: $\sqrt{6}+\sqrt{2}$
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Correct Answer: $\sqrt{6}-\sqrt{2}$
Solution :
Given: $\operatorname{cosec}(\alpha+\beta)=\frac{\sec \alpha \times \sec \beta \times \operatorname{cosec} \alpha \times \operatorname{cosec} \beta}{\sec \alpha \times \operatorname{cosec} \beta+\operatorname{cosec} \alpha \times \sec \beta}$
To find the value of $\operatorname{cosec} 75°$, take $\alpha=45°,\beta=30°$
Now putting their values in the equation we get,
$\operatorname{cosec}(45°+30°) = \frac{\sec 45° \times \sec 30° \times \operatorname{cosec} 45° \times \operatorname{cosec} 30°}{\sec 45° \times \operatorname{cosec} 30°+\operatorname{cosec} 45° \times \sec 30°}$
⇒ $\operatorname{cosec}(45°+30°) = \frac{\sqrt2\times \frac{2}{\sqrt{3}}\times \sqrt2\times 2 }{\sqrt 2×2+\sqrt2×\frac{2}{\sqrt{3}})}$
⇒ $\operatorname{cosec}(45°+30°) = \frac{\frac{2\sqrt2\times 2\sqrt2}{\sqrt{3}}}{2×\sqrt{2}×(1+\frac{1}{\sqrt{3}})}$
⇒ $\operatorname{cosec}(45°+30°) = \frac{2×\sqrt{2}×(\sqrt{3}-1)}{(\sqrt{3}+1)×(\sqrt{3}-1)}$
⇒ $\operatorname{cosec}(45°+30°) = \frac{2×\sqrt{6}-2×\sqrt{2}}{2}$
⇒ $\operatorname{cosec}75°= \sqrt{6}-\sqrt{2}$
Hence, the correct answer is $(\sqrt{6}-\sqrt{2})$.
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