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Question

# If $cotA=\sqrt{ac},cotB=\sqrt{\left(\frac{c}{a}\right)},cotC=\sqrt{\left(\frac{{a}^{3}}{c}\right)},c={a}^{2}+a+1$, then

A

$A+B=C$

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B

$B+C=A$

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C

$C+A=B$

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D

None of these

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Solution

## The correct option is A $A+B=C$ Given that, $cotA=\sqrt{ac},cotB=\sqrt{\left(\frac{c}{a}\right)},cotC=\sqrt{\left(\frac{{a}^{3}}{c}\right)},c={a}^{2}+a+1$As we know, $\begin{array}{rcl}\mathbit{c}\mathbit{o}\mathbit{t}\mathbf{\left(}\mathbit{A}\mathbf{}\mathbf{+}\mathbf{}\mathbit{B}\mathbf{\right)}\mathbf{}& \mathbf{=}& \mathbf{}\frac{\mathbf{c}\mathbf{o}\mathbf{t}\mathbf{}\mathbf{A}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{t}\mathbf{}\mathbf{B}\mathbf{}\mathbf{–}\mathbf{}\mathbf{1}}{\mathbf{c}\mathbf{o}\mathbf{t}\mathbf{}\mathbf{A}\mathbf{}\mathbf{+}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{t}\mathbf{}\mathbf{B}}\end{array}$ Put the given values of $cotA,cotB,cotC$ in above formula, we get$\begin{array}{rcl}cot\left(A+B\right)& =& \frac{\sqrt{ac}\sqrt{\left(\frac{c}{a}\right)}–1}{\sqrt{ac}+\sqrt{\left(\frac{c}{a}\right)}}\\ & =& \frac{\left(c–1\right)}{\left[\frac{\left(a\sqrt{c}+\sqrt{c}\right)}{\sqrt{a}}\right]}\\ & =& \frac{\left(c–1\right)\sqrt{a}}{\left(a+1\right)\sqrt{c}}\\ & =& \frac{\left({a}^{2}+a\right)\sqrt{a}}{\left(a+1\right)\sqrt{c}}\\ & =& \frac{\left(a+1\right)a\sqrt{a}}{\left(a+1\right)\sqrt{c}}\\ & =& \frac{a\sqrt{a}}{\sqrt{c}}\\ & =& \sqrt{\left(\frac{{a}^{3}}{c}\right)}\\ & =& cotC\end{array}$$\mathbf{\therefore }\mathbit{A}\mathbf{}\mathbf{+}\mathbf{}\mathbit{B}\mathbf{}\mathbf{=}\mathbf{}\mathbit{C}$Hence, Option ‘A’ is Correct.

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