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The value of sin²1° + sin²5° + sin²9° + .... + sin²89° is
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11 1 2 - 11√2
- 11
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11 √2
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Correct Option: A
No. of terms in 1 + 5 + 9 + ... + 89 = n
∴ a + (n –1) d = t 2
⇒ 1 + (n – 1) 4 = 89
⇒ (n– 1) 4 = 89 – 1 = 88
⇒ n – 1 = 22
⇒ n = 23
Now, sin²1° + sin²89° + sin²5° + sin²85° + ... + to 22 terms + sin²45°
= (sin²1° + cos²1°) + (sin²5° + cos²5°) + ... + to 11 terms +
 | 1 |  | ² = 11 × 1 + (1 / 2) | |
| √2 |
| = 11 + | = 11 | 2 | 2 |
