# ISI2017-DCG-17

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If  $\cos ^{2}x+ \cos ^{4} x=1$, then $\tan ^{2} x+ \tan ^{4} x$ is equal to

1. $1$
2. $0$
3. $2$
4. none of these

recategorized

$cos^2x+cos^4x=1$

$\implies cos^4x=1-cos^2x$

$\implies cos^4x=sin^2x\ldots(1)$

$tan^2x+tan^4x=\frac{sin^2x}{cos^2x}+\frac{sin^4x}{cos^4x}$

$=\frac{cos^4x}{cos^2x}+\frac{sin^4x}{sin^2x}\ldots from(1)$

$=cos^2x+sin^2x=1$

Answer $A$

Given: $$\cos^2x + \cos^4x = 1$$

$$\cos^2x + \cos^4x = 1 \implies \cos^4x = 1 - \cos^2x = \sin^2x$$

$$\implies \cos^4x = \sin^2x\dots \qquad \to (1)$$

To find: $$\tan^2x + \tan^4x$$

$$\implies \tan^2x + \tan^4x = \frac{\sin^2x}{\cos^2x} + \frac{\sin^4x}{\cos^4x} \qquad \to (1)$$

$$\frac{\sin^2x}{\cos^2x} + \frac{\sin^4x}{\cos^4x} = \frac{\cos^4x}{\cos^2x} + \frac{\sin^4x}{\sin^2x} = \cos^2x + \sin^2x = 1$$

$\therefore A$ is the correct option.

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