A) 0
B) 1
C) 2
D) \[\frac{1}{2}\]
Correct Answer: C
Solution :
\[{{\tan }^{2}}C+{{\tan }^{2}}A\] |
\[={{\left( \frac{BD}{CD} \right)}^{2}}+{{\left( \frac{BD}{AD} \right)}^{2}}\] |
\[={{\left( \frac{4}{4} \right)}^{2}}+{{\left( \frac{4}{4} \right)}^{2}}=1+1=2\] |
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