A) \[\frac{1}{2}\]
B) \[\frac{1}{4}\]
C) 2
D) 1
Correct Answer: D
Solution :
\[\frac{1}{1+{{\tan }^{2}}\theta }+\frac{1}{1+{{\cot }^{2}}\theta }\] \[=\frac{1}{1+{{\tan }^{2}}\theta }+\frac{{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }\] \[=\frac{1+{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }\] \[=1\]
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