# ISI2017-DCG-17

37 views

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.

## Related questions

1
38 views
If $\cos x = \dfrac{1}{2}$, the value of the expression $\dfrac{\cos 6x+6 \cos 4x+15 \cos 2x +10}{\cos 5x+5 \cos 3x +10 \cos x}$ is $\frac{1}{2}$ $1$ $\frac{1}{4}$ $0$
If $a,b,c$ are the sides of $\Delta ABC$, then $\tan \frac{B-C}{2} \tan \frac{A}{2}$ is equal to $\frac{b+c}{b-c}$ $\frac{b-c}{b+c}$ $\frac{c-b}{c+b}$ none of these
The angle between the tangents drawn from the point $(-1, 7)$ to the circle $x^2+y^2=25$ is $\tan^{-1} (\frac{1}{2})$ $\tan^{-1} (\frac{2}{3})$ $\frac{\pi}{2}$ $\frac{\pi}{3}$
The value of $\dfrac{1}{\log_2 n}+ \dfrac{1}{\log_3 n}+\dfrac{1}{\log_4 n}+ \dots + \dfrac{1}{\log_{2017} n}\:\:($ where $n=2017!)$ is $1$ $2$ $2017$ none of these