Question 55

If $$\tan^4 x - \tan^2 x = 1$$, then the value of $$\sin^4 x + \sin^2 x$$ is:

Solution

$$\tan^4 x - \tan^2 x = 1$$

$$=$$> $$\tan^4x=1+\tan^2x$$

$$=$$> $$\tan^4x=\sec^2x$$

$$=$$> $$\tan^2x=\sec x$$

$$=$$> $$\frac{\sin^2x}{\cos^2x}=\frac{1}{\cos x}$$

$$=$$> $$\sin^2x=\cos x$$

$$\therefore\ \sin^4x+\sin^2x=\left(\sin^2x\right)^2+\sin^2x$$

$$=\left(\cos x\right)^2+\sin^2x$$

$$=\cos^2x+\sin^2x$$

$$=1$$

Hence, the correct answer is Option A

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