Question:
If \(\sin p \times \cos(90^\circ – p) = \dfrac{1}{4}\), where \(0^\circ < p < 90^\circ\), then find the value of \(\cot^2 2p + \cos^2 3p\).
यदि \(\sin p \times \cos(90^\circ – p) = \dfrac{1}{4}\), जहाँ \(0^\circ < p < 90^\circ\) है, तो \(\cot^2 2p + \cos^2 3p\) का मान ज्ञात कीजिए।
Options:
- \(\dfrac{1}{5}\)
- \(\dfrac{1}{3}\)
- \(\dfrac{1}{4}\)
- \(\dfrac{1}{7}\)
- \(\dfrac{1}{5}\)
- \(\dfrac{1}{3}\)
- \(\dfrac{1}{4}\)
- \(\dfrac{1}{7}\)
Correct Answer: b)
Explanation:
We know \(\cos(90^\circ – p) = \sin p\).
So, \(\sin p \times \sin p = \sin^2 p = \dfrac{1}{4} \implies \sin p = \dfrac{1}{2} \implies p = 30^\circ\).
Then \(2p = 60^\circ\) and \(3p = 90^\circ\).
\(\cot^2 2p + \cos^2 3p = \cot^2 60^\circ + \cos^2 90^\circ = \left(\dfrac{1}{\sqrt{3}}\right)^2 + 0 = \dfrac{1}{3}\).
हम जानते हैं कि \(\cos(90^\circ – p) = \sin p\)।
अतः, \(\sin^2 p = \dfrac{1}{4} \implies \sin p = \dfrac{1}{2} \implies p = 30^\circ\)।
तब \(2p = 60^\circ\) और \(3p = 90^\circ\)।
\(\cot^2 2p + \cos^2 3p = \cot^2 60^\circ + \cos^2 90^\circ = \left(\dfrac{1}{\sqrt{3}}\right)^2 + 0 = \dfrac{1}{3}\)।