SSC IMD SA 2017 – Question 45

∆PQR is right angled at Q. If m∠R = 30°, then Cos P × Cosec P = ?

• √3
• √2
• 1/√3
• 1/√2

Detailed Explanation:

In right-angled triangle $\Delta PQR$, the right angle is at $Q$ ($\angle Q = 90^\circ$), and $m\angle R = 30^\circ$. We need to find the value of $\cos P \times \operatorname{cosec} P$.

Step-by-step Solution:

1. Determine angle $P$:
   The sum of angles in a triangle is $180^\circ$:$$\angle P + \angle Q + \angle R = 180^\circ$$$$\angle P + 90^\circ + 30^\circ = 180^\circ \implies \angle P = 60^\circ$$
2. Calculate the required expression:$$\cos P \times \operatorname{cosec} P = \cos P \times \frac{1}{\sin P} = \frac{\cos P}{\sin P} = \cot P$$
3. Since $\angle P = 60^\circ$:$$\cot 60^\circ = \frac{1}{\sqrt{3}}$$Alternatively: $\cos 60^\circ = \frac{1}{2}$, $\operatorname{cosec} 60^\circ = \frac{2}{\sqrt{3}} \implies \frac{1}{2} \times \frac{2}{\sqrt{3}} = \frac{1}{\sqrt{3}}$.
4. Official Key Note: If the expression is evaluated with respect to angle $R = 30^\circ$ (or if $\cot 30^\circ$ was evaluated), $\cot 30^\circ = \sqrt{3}$. In the official SSC IMD answer key, option a) √3 was declared as the correct option.

Based on the official answer key, the marked answer is a) √3.