If 4 sinA + secA = 0 then tanA equals to

Q: If 4 sinA + secA = 0 then tanA equals to

(A) 4 + √2

(B) – 2 + √3

(C) 2 + 4√3

(D) –2 –√3

Sol.  4sinA + secA = 0

⇒ sin2A = -1/2

$\large \frac{2 tanA}{1+ tan^2A} = – \frac{1}{2}$

tanA = – 2 ± √3

Hence (B), (D) are the correct answers.