A) 0
B) 3/2
C) 7/4
D) 5/4
Correct Answer: D
Solution :
\[{{m}_{1}}=\tan \alpha \]and \[{{m}_{2}}=\tan \beta \] \[\Rightarrow \cot \alpha =\frac{1}{{{m}_{1}}}\]and \[\cot \beta =\frac{1}{{{m}_{2}}}\] Hence, \[{{\cot }^{2}}\alpha +{{\cot }^{2}}\beta =\frac{1}{m_{1}^{2}}+\frac{1}{m_{2}^{2}}=\frac{m_{1}^{2}+m_{2}^{2}}{{{({{m}_{1}}{{m}_{2}})}^{2}}}\] \[=\frac{{{({{m}_{1}}+{{m}_{2}})}^{2}}-2{{m}_{1}}{{m}_{2}}}{{{({{m}_{1}}{{m}_{2}})}^{2}}}=\frac{{{(3)}^{2}}-2(2)}{{{(2)}^{2}}}=\frac{5}{4}\] .
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