Solve the following.
$\frac{\sin 40^{\circ}}{\cos 50^{\circ}}+\frac{{cosec}~50^{\circ}}{\sec 40^{\circ}}-4 \cos 50^{\circ} {cosec}~40^{\circ}$
Answer & explanation
Correct answer: option 2
\(\frac{sin40º}{cos50º}\) + \(\frac{cosec50º}{sec40º}\) - 4 cos50º . cosec40º
{ using , sinA = cosB Iff A + B = 90 º
& secA = cosecB Iff A + B = 90º }
= \(\frac{sin40º}{sin40º}\) + \(\frac{cosec50º}{cosec50º}\) - 4 cos50º . sec50º
= 1 + 1 - 4
= -2