Question:
If \(\text{cosec}^2\alpha = 9\), then find the value of \(\cos^2\alpha + \sec(90^\circ – \alpha)\) where \(0^\circ < \alpha < 90^\circ\).
यदि \(\text{cosec}^2\alpha = 9\) है, तो \(\cos^2\alpha + \sec(90^\circ – \alpha)\) का मान ज्ञात कीजिए, जहाँ \(0^\circ < \alpha < 90^\circ\) है।
Options:
- 9/8
- 17/8
- 35/9
- 21/8
- 9/8
- 17/8
- 35/9
- 21/8
Correct Answer: c)
Explanation:
\(\text{cosec}^2\alpha = 9 \implies \sin^2\alpha = \dfrac{1}{9}\).
\(\cos^2\alpha = 1 – \sin^2\alpha = 1 – \dfrac{1}{9} = \dfrac{8}{9}\).
Also, \(\sec(90^\circ – \alpha) = \text{cosec}\alpha = \sqrt{9} = 3\).
\(\cos^2\alpha + \sec(90^\circ – \alpha) = \dfrac{8}{9} + 3 = \dfrac{8 + 27}{9} = \dfrac{35}{9}\).
\(\text{cosec}^2\alpha = 9 \implies \sin^2\alpha = 1/9\)।
\(\cos^2\alpha = 1 – 1/9 = 8/9\)।
तथा \(\sec(90^\circ – \alpha) = \text{cosec}\alpha = 3\)।
अतः मान \(= 8/9 + 3 = 35/9\) है।