A) \[\frac{{{c}^{2}}}{ab}\]
B) \[a+b\]
C) \[\frac{{{a}^{2}}}{bc}\]
D) \[\frac{{{b}^{2}}}{ac}\]
Correct Answer: A
Solution :
\[\cot A+\cot B\] \[=\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}\] \[=\frac{\cos A\sin B+\cos B\sin A}{\sin A\sin B}\] \[=\frac{\sin (A+B)}{\sin A\sin B}\] \[=\frac{\sin (\pi -C)}{\sin A\sin B}\] \[(\because \,\,A+B+C=\pi )\] \[=\frac{\sin C}{\sin A\sin B}\] \[=\frac{1}{\sin A\sin B}\] \[(\because \,\,C={{90}^{o}})\] ? (i) Applying sine rule, \[\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\] \[\Rightarrow \] \[\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{1}{c}\] \[(\because \,\,\sin {{90}^{o}}=1)\] \[\Rightarrow \] \[\sin A=\frac{a}{c},\,\,\sin B=\frac{b}{c}\] From Eq. (i), \[\cot A+\cot B=\frac{1}{\frac{a}{c}\cdot \frac{b}{c}}\] \[=\frac{{{c}^{2}}}{ab}\]
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