A) \[\tan x+d\cot x+C\]
B) \[\tan x-\cot x+C\]
C) \[\tan x.\cot x+C\]
D) \[\tan x-\cot 2x+C\]
Correct Answer: B
Solution :
\[\int_{{}}^{{}}{\frac{dx}{{{\sin }^{2}}x.\,{{\cos }^{2}}x}}=\int_{{}}^{{}}{\frac{({{\sin }^{2}}x+{{\cos }^{2}}x)}{{{\sin }^{2}}x.\,{{\cos }^{2}}x}}dx\] \[=\int_{{}}^{{}}{\frac{{{\sin }^{2}}x}{{{\sin }^{2}}x.{{\cos }^{2}}x}}dx+\int_{{}}^{{}}{\frac{{{\cos }^{2}}x}{{{\sin }^{2}}x.{{\cos }^{2}}x}}dx\] \[=\int_{{}}^{{}}{\frac{dx}{{{\cos }^{2}}x}dx+\int_{{}}^{{}}{\frac{dx}{{{\sin }^{2}}x}}}\] \[=\int_{{}}^{{}}{{{\sec }^{2}}x\,dx}+\int_{{}}^{{}}{\cos e{{c}^{2}}x\,dx}\] \[=\tan x-\cot x+C\]
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