Question : If $a-\frac{1}{a-5}=10$, then the value of $(a-5)^3-\frac{1}{(a-5)^3}$ is:
Option 1: 140
Option 2: 70
Option 3: 100
Option 4: 120
Correct Answer: 140
Solution : $a-\frac{1}{a-5}=10$ .............(1) Subtract 5 in equation (1) on both sides $(a-5)-\frac{1}{a-5}=10-5$ ⇒ $(a-5)-\frac{1}{a-5}=5$ Now, take $(a - 5) = x$ So, the equation becomes ⇒ $x-\frac{1}{x}=5$ Cubing both sides ⇒ $x^3-\frac{1}{x^3}-3(x-\frac{1}{x}) = 125$ ⇒ $x^3-\frac{1}{x^3} = 125+3(5)$ ⇒ $x^3-\frac{1}{x^3} = 140$ Hence, the correct answer is 140.
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Question : If $A+\frac{1}{1+\frac{1}{2+\frac{1}{3}}}=\frac{9}{10}$, then the value of A is:
Option 1: $\frac{1}{5}$
Option 2: $\frac{3}{10}$
Option 3: $\frac{2}{5}$
Option 4: $\frac{1}{10}$
Question : If $2A=3B$, then what is the value of $\frac{A+B}{A}$?
Option 1: $\frac{5}{4}$
Option 2: $\frac{2}{3}$
Option 3: $\frac{5}{2}$
Option 4: $\frac{5}{3}$
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: $3 \sqrt{5}$
Question : If $\tan A=\frac{4}{3}, 0 \leq A \leq 90^{\circ}$, then find the value of $\sin A$.
Option 1: $\frac{3}{5}$
Option 2: $1$
Option 3: $\frac{3}{4}$
Option 4: $\frac{4}{5}$
Question : If $\tan A=\frac{2}{5}$, then find the value of $\frac{\sec ^2 A}{\operatorname{cosec}^2 A}$.
Option 1: $\frac{2}{5}$
Option 2: $\frac{4}{25}$
Option 3: $\frac{9}{25}$
Option 4: $\frac{3}{5}$
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