Question : If $\cos^2x+\cos^4x=1$, then $\tan^2x+\tan^4x$?
Option 1: $0$
Option 2: $1$
Option 3: $2 \tan^2x$
Option 4: $2\tan^4x$
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Correct Answer: $1$
Solution : $\cos^2x+\cos^4x=1$ ⇒ $\cos^4x = 1 - \cos^2x$ ⇒ $\cos^4x = \sin^2x$ [Using $1-\cos^2x=\sin^2x$] ⇒ $\cos^2x\times \cos^2x=\sin^2x$ ⇒ $\tan^2x=\cos^2x$ --------(i) On squaring, $\tan^4x=\cos^4x$ ------(ii) ⇒ $\tan^2x+\tan^4x = \cos^2x+\cos^4x$ ⇒ $\tan^2x+\tan^4x=1$ Hence, the correct answer is $1$.
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Question : The least value of $\tan^2x+\cot^2x$ is:
Option 1: 3
Option 2: 2
Option 3: 0
Option 4: 1
Question : If $\tan^4\theta + \tan^2\theta=1$, what is the value of $11(\cos^4\theta+\cos^2\theta)$?
Option 1: – 11
Option 2: 8
Option 4: 11
Question : If $5 \sin^2 A+3 \cos^2 A=4$, $0<A<90°$, then what is the value of $\tan A$?
Option 1: 0
Option 2: 3
Option 3: 1
Option 4: 2
Question : If $\alpha$ and $\beta$ are the roots of equation $x^{2}-2x+4=0$, then what is the equation whose roots are $\frac{\alpha ^{3}}{\beta ^{2}}$ and $\frac{\beta ^{3}}{\alpha ^{2}}?$
Option 1: $x^{2}-4x+8=0$
Option 2: $x^{2}-32x+4=0$
Option 3: $x^{2}-2x+4=0$
Option 4: $x^{2}-16x+4=0$
Question : If $\cos \theta=\frac{\sqrt{3}}{2}$, then $\tan ^2 \theta \cos ^2 \theta=?$
Option 1: $\frac{1}{\sqrt{3}}$
Option 2: $\frac{1}{4}$
Option 3: $\frac{1}{2}$
Option 4: $\sqrt{3}$
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