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.