Question : Value of $\sec^2\theta-\frac{\sin^2\theta-2\sin^4\theta}{2\cos^4\theta-\cos^2\theta}$ is:
Option 1: 1
Option 2: 2
Option 3: –1
Option 4: 0
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Correct Answer: 1
Solution :
$\sec^2\theta-\frac{\sin^2\theta-2\sin^4\theta}{2\cos^4\theta-\cos^2\theta}$
$=\sec^2\theta-\frac{\sin^2\theta(1-2\sin^2\theta)}{\cos^2\theta(2\cos^2\theta-1)}$
$=\sec^2\theta-\frac{\sin^2\theta(1-2\sin^2\theta)}{\cos^2\theta[2(1-\sin^2\theta)-1]}$
$=\sec^2\theta-\frac{\sin^2\theta(1-2\sin^2\theta)}{\cos^2\theta(1-2\sin^2\theta)}$
$=\sec^2\theta-\tan^2\theta$
$=1$
Hence, the correct answer is 1.
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