A) 5
B) 13
C) 15
D) 6
Correct Answer: C
Solution :
Let \[{{\tan }^{-1}}2=\alpha \Rightarrow \tan \alpha =2\] and \[{{\cot }^{-1}}3=\beta \Rightarrow \cot \beta =3\] \[{{\sec }^{2}}({{\tan }^{-1}}2)+\text{cose}{{\text{c}}^{\text{2}}}({{\cot }^{-1}}3)\] = \[{{\sec }^{2}}\alpha +\text{cose}{{\text{c}}^{\text{2}}}\alpha \]=\[1+{{\tan }^{2}}\alpha +1+{{\cot }^{2}}\alpha \] = \[2+{{(2)}^{2}}+{{(3)}^{2}}=15\].
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