Question : If $a-\frac{1}{a}=4$, then the value of $a+\frac{1}{a}$ is:
Option 1: $5 \sqrt{5}$
Option 2: $4 \sqrt{5}$
Option 3: $2 \sqrt{5}$
Option 4: $3 \sqrt{5}$
Correct Answer: $2 \sqrt{5}$
Solution : Given: $a-\frac{1}{a}=4$ Squaring both sides, we get, ⇒ $a^2+\frac{1}{a^2}-2=16$ ⇒ $a^2+\frac{1}{a^2}=16+2$ ⇒ $a^2+\frac{1}{a^2}=18$ Adding 2 on both sides, we get, ⇒ $a^2+\frac{1}{a^2}+2=18+2$ ⇒ $(a+\frac{1}{a})^2=20$ $\therefore a+\frac{1}{a}=2\sqrt5$ Hence, the correct answer is $2\sqrt5$.
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Question : If $\sin A=\frac{1}{2}$, then the value of $(\tan A+\cos A)$ is:
Option 1: $\frac{2}{3 \sqrt{3}}$
Option 2: $\frac{3}{2 \sqrt{3}}$
Option 3: $\frac{5}{2 \sqrt{3}}$
Option 4: $\frac{5}{3 \sqrt{3}}$
Question : If $\sin A=\frac{2}{3}$, then find the value of (7 – tan A)(3 + cos A).
Option 1: $\frac{61}{3}+\frac{17}{3 \sqrt{5}}$
Option 2: $\frac{61}{3 \sqrt{5}}+\frac{17}{3}$
Option 3: $\frac{61}{3}+\frac{17}{\sqrt{5}}$
Option 4: $\frac{61}{3}-\frac{17}{3 \sqrt{5}}$
Question : If $x^2+\frac{1}{x^2}=\frac{7}{4}$ for $x>0$, what is the value of $(x^3+\frac{1}{x^3})$?
Option 1: $\frac{3\sqrt{3}}{5}$
Option 2: $\frac{3\sqrt{15}}{5}$
Option 3: $\frac{3\sqrt{15}}{8}$
Option 4: $\frac{3\sqrt{5}}{8}$
Question : If $x^2+\frac{1}{x^2}=\frac{7}{4}$ for $x>0$; then what is the value of $x+\frac{1}{x}$?
Option 1: $2$
Option 2: $\frac{\sqrt{15}}{2}$
Option 3: $\sqrt{5}$
Option 4: $\sqrt{3}$
Question : In $\triangle ABC, \angle B=90^{\circ}$ and AB : BC = 1 : 2. The value of $\cos A+\tan C$ is:
Option 1: $\frac{5+\sqrt{5}}{2 \sqrt{5}}$
Option 2: $\frac{1+\sqrt{5}}{2 \sqrt{5}}$
Option 3: $\frac{2 \sqrt{5}}{2+\sqrt{5}}$
Option 4: $\frac{2+\sqrt{5}}{2 \sqrt{5}}$
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