Question 51
The value of $$\frac{1 - 2 \sin^2 \theta \cos^2 \theta}{\sin^4 \theta + \cos^4 \theta} - 1$$ is:
Solution
$$\frac{1 - 2 \sin^2 \theta \cos^2 \theta}{\sin^4 \theta + \cos^4 \theta} - 1$$
= $$\frac{1 - 2 \sin^2 \theta \cos^2 \theta - \sin^4 \theta - \cos^4 \theta}{\sin^4 \theta + \cos^4 \theta}$$
= $$\frac{1 - (sin^2 \theta + \cos^2 \theta)^2 }{\sin^4 \theta + \cos^4 \theta}$$
= $$\frac{1 - 1}{\sin^4 \theta + \cos^4 \theta}$$ = 0
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