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Question 68
Solve the following.
$$\left(\frac{\sin 27^\circ}{\cos 63^\circ}\right) - \left(\frac{\cos 27^\circ}{\sin 63^\circ}\right)^{2}$$
Solution
$$\left(\frac{\sin 27^\circ}{\cos 63^\circ}\right) - \left(\frac{\cos 27^\circ}{\sin 63^\circ}\right)^{2}$$
$$\left(\frac{\sin(90^\circ - 63^\circ)}{\cos 63^\circ}\right) - \left(\frac{\cos(90^\circ - 63^\circ)}{\sin 63^\circ}\right)^{2}$$
=Â $$\left(\frac{\\cos 63^\circ}{\cos 63^\circ}\right) - \left(\frac{\sin 63^\circ}{\sin 63^\circ}\right)^{2}$$
= 1 - 1 = 0
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