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. $A+B=C$
• B. $B+C=A$
• C. $C+A=B$
• D. None of these

Answer 1

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,

$\begin{array}{rcl}\mathit{c}\mathit{o}\mathit{t}\mathbf{(}\mathit{A}\mathbf{}\mathbf{+}\mathbf{}\mathit{B}\mathbf{)}\mathbf{}& \mathbf{=}& \mathbf{}\frac{\mathbf{c}\mathbf{o}\mathbf{t}\mathbf{}\mathbf{A}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{t}\mathbf{}\mathbf{B}\mathbf{}\mathbf{–}\mathbf{}\mathbf{1}}{\mathbf{c}\mathbf{o}\mathbf{t}\mathbf{}\mathbf{A}\mathbf{}\mathbf{+}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{t}\mathbf{}\mathbf{B}}\end{array}$

Put the given values of $cotA,cotB,cotC$ in above formula, we get

$\begin{array}{rcl}cot\left(A+B\right)& =& \frac{\sqrt{ac}\sqrt{\left({\displaystyle \frac{c}{a}}\right)}–1}{\sqrt{ac}+\sqrt{\left(\frac{c}{a}\right)}}\\ & =& \frac{(c–1)}{\left[{\displaystyle \frac{(a\sqrt{c}+\sqrt{c})}{\sqrt{a}}}\right]}\\ & =& \frac{(c–1)\sqrt{a}}{(a+1)\sqrt{c}}\\ & =& \frac{(a^2+a)\sqrt{a}}{(a+1)\sqrt{c}}\\ & =& \frac{(a+1)a\sqrt{a}}{(a+1)\sqrt{c}}\\ & =& \frac{a\sqrt{a}}{\sqrt{c}}\\ & =& \sqrt{\left(\frac{a^3}{c}\right)}\\ & =& cotC\end{array}$

$\mathbf{\therefore }\mathit{A}\mathbf{}\mathbf{+}\mathbf{}\mathit{B}\mathbf{}\mathbf{=}\mathbf{}\mathit{C}$

Hence, Option ‘A’ is Correct.