In right-angled tríngle {\ABC}, \angle C=90^{\circ} and {\CD} is perpendicular to A B. If { {\AB}} \times {\CD}= {\CA} \times {\CB}, then \frac{1}{ {\CD}^2} is equal to

\frac{1}{A B^2}-\frac{1}{C A^2}

\frac{1}{A B^2}-\frac{1}{C B^2}

\frac{1}{ {\BC}^2}+ {1}/{ {\C} {\A}^2}

\frac{1}{ {\BC}^2}-\frac{1}{ {\CA}^2} if {\CA}> {\CB}

UPSC 2013 CDS 1 Math Paper

Question : 92 of 100

Marks: +1, -0

In right-angled tríngle ABC,∠C=90∘ and CD is perpendicular to AB. If AB×CD=CA×CB, then ‌

CD2

is equal to

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