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Question 53
Solve the following.
$$\frac{\sin 40^\circ}{\cos 50^\circ} + \frac{\cosec 50^\circ}{\sec 40^\circ} - 4 \cos 50^\circ \cosec 40^\circ$$
Solution
$$\frac{\sin 40^\circ}{\cos 50^\circ} + \frac{\cosec 50^\circ}{\sec 40^\circ} - 4 \cos 50^\circ \cosec 40^\circ$$
=$$\frac{\sin 40^\circ}{\cos(90^\circ -Â 40^\circ)} + \frac{\cosec(90^\circ - 40^\circ)}{\sec 40^\circ} - 4 \cos(90^\circ - 40^\circ) \cosec 40^\circ$$
=$$\frac{\sin 40^\circ}{\sin 40^\circ} + \frac{\sec 40^\circ}{\sec 40^\circ} - 4 \sin40^\circ \cosec 40^\circ$$
= 1 +Â 1 - 4 = -2
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