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

$\large 4 sinA + \frac{1}{cosA} = 0 $

$\large \frac{4sinA cosA + 1}{cosA} = 0 $

4 sinA cosA + 1 = 0

2(2 sinA cosA ) = -1

2( sin2A ) = -1

⇒ sin2A = -1/2

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

tanA = – 2 ± √3

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