A) \[\frac{1}{A{{B}^{2}}}-\frac{1}{C{{A}^{2}}}\]
B) \[\frac{1}{A{{B}^{2}}}-\frac{1}{C{{B}^{2}}}\]
C) \[\frac{1}{B{{C}^{2}}}+\frac{1}{C{{A}^{2}}}\]
D) \[\frac{1}{B{{C}^{2}}}-\frac{1}{C{{A}^{2}}},\]if \[CA>CB\]
Correct Answer: D
Solution :
According to question, |
In \[\Delta ABC,\] |
\[\angle C=90{}^\circ ;CD\bot AB\] |
In\[\Delta ABC,\frac{1}{2}\times CD\times AB=\frac{1}{2}\times AC\times BC\] ?(i) |
\[A{{B}^{2}}=B{{C}^{2}}+C{{A}^{2}}\] ?(ii) (by Pythagoras theorem) |
\[AB\times CD=AC\times BC\](given) |
\[\left\{ \begin{align} & A{{C}^{2}}=A{{D}^{2}}+C{{D}^{2}} \\ & B{{C}^{2}}=C{{D}^{2}}+B{{D}^{2}} \\ \end{align} \right\}\] |
\[\sqrt{(B{{C}^{2}}+C{{A}^{2}})}\times CD=AC\times BC\] |
\[(B{{C}^{2}}+C{{A}^{2}})\times C{{D}^{2}}=A{{C}^{2}}\times B{{C}^{2}}\] |
\[\frac{1}{C{{D}^{2}}}=\frac{B{{C}^{2}}+C{{A}^{2}}}{A{{C}^{2}}\times B{{C}^{2}}}\] |
\[\frac{1}{C{{D}^{2}}}=\frac{1}{A{{C}^{2}}}+\frac{1}{B{{C}^{2}}}\] |
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