Question : If $a+b+c+d=4$, then find the value of $\frac{1}{(1-a)(1-b)(1-c)}+\frac{1}{(1-b)(1-c)(1-d)}+\frac{1}{(1-c)(1-d)(1-a)}+\frac{1}{(1-d)(1-a)(1-b)}$?
Option 1: 0
Option 2: 5
Option 3: 1
Option 4: 4
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Correct Answer: 0
Solution : Given: $a+b+c+d=4$ $\frac{1}{(1-a)(1-b)(1-c)}+\frac{1}{(1-b)(1-c)(1-d)}+\frac{1}{(1-c)(1-d)(1-a)}+\frac{1}{(1-d)(1-a)(1-b)}$ = $\frac{1-d+1-a+1-b+1-c}{(1-a)(1-b)(1-c)(1-d)}$ = $\frac{4-(a+b+c+d)}{(1-a)(1-b)(1-c)(1-d)}$ = $\frac{4-4}{(1-a)(1-b)(1-c)(1-d)}$ = $\frac{0}{(1-a)(1-b)(1-c)(1-d)}$ = $0$ Hence, the correct answer is 0.
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Question : If $\small c+\frac{1}{c}=3$, then the value of $\left (c-3 \right )^{7}+\frac{1}{c^{7}}$ is:
Option 1: 2
Option 2: 0
Option 3: 3
Option 4: 1
Question : If $x+\frac{1}{x}=0$, then the value of $x^{5}+\frac{1}{x^{5}}$ is:
Option 2: –1
Option 4: 0
Question : If 3 cot A = 4 and A is an acute angle, then find the value of sec A.
Option 1: $\frac{3}{4}$
Option 2: $\frac{5}{4}$
Option 3: $\frac{4}{5}$
Option 4: $\frac{5}{3}$
Question : The value of the expression $\frac{(a-b)^{2}}{(b-c)(c-a)}+\frac{(b-c)^{2}}{(a-b)(c-a)}+\frac{(c-a)^{2}}{(a-b)(b-c)}$ is:
Option 1: $0$
Option 2: $3$
Option 3: $\frac{1}{3}$
Option 4: $2$
Question : If $a^3+b^3=9$ and $a+b=3$, then the value of $\frac{1}a+\frac{1}b$ is:
Option 1: $\frac{1}2$
Option 2: $\frac{3}2$
Option 3: $\frac{5}2$
Option 4: $–1$
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