A) \[{{\sec }^{2}}A\]
B) -1
C) \[{{\cot }^{2}}A\]
D) \[{{\tan }^{2}}A\]
Correct Answer: D
Solution :
We have, \[\frac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\frac{1+{{\tan }^{2}}A}{1+\frac{1}{{{\tan }^{2}}A}}\] \[=\frac{1+{{\tan }^{2}}A}{\frac{{{\tan }^{2}}A+1}{{{\tan }^{2}}A}}\] \[=\left( 1+{{\tan }^{2}}A \right)\frac{\left( {{\tan }^{2}}A \right)}{\left( 1+{{\tan }^{2}}A \right)}\] \[={{\tan }^{2}}a\]
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