Question 53
The value of $$\sin^2 64^\circ + \cos 64^\circ \sin 26^\circ + 2 \cos 43^\circ \cosec 47^\circ$$ is:
Solution
$$\sin^2 64^\circ + \cos 64^\circ \sin 26^\circ + 2 \cos 43^\circ \cosec 47^\circ$$
= $$\sin^2 64^\circ + \cos 64^\circ \sin(90 - 64^\circ) + 2 \cos 43^\circ \cosec(90 - 43^\circ)$$
= $$\sin^2 64^\circ + \cos^2 64^\circ + 2 \cos 43^\circ se43^\circ$$
= $$1 + 2$$ = 3
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