Question : If $p^2-4 p-1=0$, then the value of $p^2+3 p+\frac{1}{p^2}-\frac{3}{p}$ is:
Option 1: 40
Option 2: 50
Option 3: 65
Option 4: 30
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Correct Answer: 30
Solution : Given: $p^2-4 p-1=0$ ⇒ $p-\frac{1}{p}=4$ Squaring both sides, we get: $⇒(p-\frac{1}{p})^2=(4)^2$ $⇒ (p^2+\frac{1}{p^2}-2×p×\frac{1}{p})=16$ $⇒ (p^2+\frac{1}{p^2})=16+2$ $⇒ (p^2+\frac{1}{p^2})=18$ Now, $p^2+3 p+\frac{1}{p^2}-\frac{3}{p}$ $=p^2+\frac{1}{p^2}+3 p-\frac{3}{p}$ $=p^2+\frac{1}{p^2}+3(p-\frac{1}{p})$ Putting the values, we get: $=18+3×4$ $=30$ Hence, the correct answer is 30.
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Question : If $\sec A=\frac{9}{4}$, then what is the value of $\cot A$?
Option 1: $\frac{4}{\sqrt{65}}$
Option 2: $\frac{9}{\sqrt{65}}$
Option 3: $\frac{\sqrt{65}}{9}$
Option 4: $\frac{\sqrt{65}}{4}$
Question : If $\cos^{4}\theta-\sin^{4}\theta=\frac{2}{3}$, then the value of $1-2\sin^{2}\theta$ is:
Option 1: $\frac{2}{3}$
Option 2: $\frac{3}{2}$
Option 3: $1$
Option 4: $0$
Question : If $2x+\frac{1}{4x}=1$, then the value of $x^{2}+\frac{1}{64x^{2}}$ is:
Option 1: $0$
Option 2: $1$
Option 3: $\frac{1}{4}$
Option 4: $2$
Question : If $\frac{\sin \theta-\cos \theta}{\sin \theta+\cos \theta}=\frac{4}{5}$, then the value of $\frac{\operatorname{cosec}^2 \theta}{2-\operatorname{cosec}^2 \theta}$ is:
Option 1: $\frac{16}{25}$
Option 2: $\frac{40}{41}$
Option 3: $\frac{41}{40}$
Option 4: $\frac{31}{30}$
Question : If $a+\frac{1}{a}=2$ and $b+\frac{1}{b}=-2$, then what is the value of $a^2+\frac{1}{a^2}+b^2+\frac{1}{b^2} ?$
Option 1: 0
Option 2: 2
Option 3: 8
Option 4: 4
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