Question:
If $\sin \alpha = \frac{5}{13}$, then find the value of $\cos \alpha \cdot \csc \alpha \cdot \cot \alpha$.
यदि $\sin \alpha = \frac{5}{13}$ है, तो $\cos \alpha \cdot \csc \alpha \cdot \cot \alpha$ का मान ज्ञात कीजिए।
Options:
- a) 12/5
- b) 25/144
- c) 144/25
- d) 5/12
- a) 12/5
- b) 25/144
- c) 144/25
- d) 5/12
correct answer : c)
Explanation:
Let’s first simplify the expression:
$\cos \alpha \cdot \csc \alpha \cdot \cot \alpha = \cos \alpha \cdot \frac{1}{\sin \alpha} \cdot \cot \alpha$
$= \cot \alpha \cdot \cot \alpha = \cot^2 \alpha$
Given $\sin \alpha = \frac{5}{13}$:
In a right-angled triangle, opposite = 5, hypotenuse = 13.
By Pythagorean theorem, adjacent $= \sqrt{13^2 – 5^2} = \sqrt{144} = 12$.
Therefore, $\cot \alpha = \frac{\text{adjacent}}{\text{opposite}} = \frac{12}{5}$.
So, $\cot^2 \alpha = \left(\frac{12}{5}\right)^2 = \frac{144}{25}$.
Hence, option (c) is the correct answer.
पहले व्यंजक को सरल करें:
$\cos \alpha \cdot \csc \alpha \cdot \cot \alpha = \cos \alpha \cdot \frac{1}{\sin \alpha} \cdot \cot \alpha$
$= \cot \alpha \cdot \cot \alpha = \cot^2 \alpha$
दिया गया है $\sin \alpha = \frac{5}{13}$:
एक समकोण त्रिभुज में, सम्मुख भुजा (opposite) = 5, कर्ण (hypotenuse) = 13।
पाइथागोरस प्रमेय द्वारा, आसन्न भुजा (adjacent) $= \sqrt{13^2 – 5^2} = \sqrt{144} = 12$।
इसलिए, $\cot \alpha = \frac{\text{आसन्न}}{\text{सम्मुख}} = \frac{12}{5}$।
तो, $\cot^2 \alpha = \left(\frac{12}{5}\right)^2 = \frac{144}{25}$।
अतः, विकल्प (c) सही उत्तर है।