| 1. |
TanA + secA =4, then what is the value of sinA |
Answer» Answer15/17 ExplanationTANA + secA = 4 We can WRITE it as, sinA/cosA + 1/cosA = 4 → (sinA + 1)/cosA = 4 → sinA + 1 = 4cosA Now squaring both sides, (sinA + 1)² = (16cosA)² → sin²A + 1 + 2sinA = 16cos²A Replace cos²A as 1 - sin²A (SINCE, cos²A + sin²A = 1) → sin²A + 1 + 2sinA = 16(1 - sin²A) → sin²A + 1 + 2sinA = 16 - 16sin²A → sin²A + 16sin²A + 1 - 16 + 2sinA = 0 → 17sin²A + 2sinA - 15 = 0 Now, splitting the middle term → 17sin²A + 17sinA - 15sinA - 15 = 0 → 17sinA(sinA + 1) - 15(sinA + 1) = 0 → (17sinA - 15)(sinA + 1) = 0 → 17sinA - 15 = 0 or sinA + 1 = 0 → sinA = 15/17 or sinA = -1 But, for sinA = -1, tanA becomes undefined. Hence, we will neglect this VALUE. |
|