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Question 47
The value of $$\frac{2(\sin^6 \theta +Â \cos^6 \theta) - 3(\sin^4 \theta + \cos^4 \theta)}{\cos^4 \theta - \sin^4 \theta - 2 \cos^2 \theta}$$ is:
Solution
$$\frac{2(\sin^6 \theta + \cos^6 \theta) - 3(\sin^4 \theta + \cos^4 \theta)}{\cos^4 \theta - \sin^4 \theta - 2 \cos^2 \theta}$$
Put the $$\theta = 90\degree$$,
$$\frac{2(\sin^6 90\degree + \cos^690\degree) - 3(\sin^4 90\degree + \cos^4 90\degree)}{\cos^4 90\degree - \sin^4 90\degree - 2 \cos^2 90\degree}$$
=Â $$\frac{2(1 + 1) - 3(1 + 1)}{1 - 1 - 2 }$$
= $$\frac{4 - 6}{-2} = 1$$
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