Question

tanA + secA =4, then what is the value of sinA

Answer 1

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Answer 2

Answer

15/17

Explanation

tanA + 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.