A) \[{{\cot }^{2}}A\]
B) \[{{\cot }^{2}}B\]
C) \[-ta{{n}^{2}}A\]
D) \[-{{\cot }^{2}}A\]
Correct Answer: B
Solution :
We have, \[A+B={{90}^{o}}\,\Rightarrow A={{90}^{o}}-B\] ?(i) Now, \[\frac{\tan A\tan B+\tan A\cot B}{\sin A\sec B}-\frac{{{\sin }^{2}}B}{{{\cos }^{2}}A}\] \[=\frac{\tan ({{90}^{o}}-B)\tan B+\tan ({{90}^{o}}-B)\cot B}{\sin ({{90}^{o}}-B)\sec B}\] \[-\frac{{{\sin }^{2}}B}{{{\left[ \cos ({{90}^{o}}-B) \right]}^{2}}}\] [using (i)] \[=\frac{\cot B.\tan B+\cot B.\cot B}{\cos B.\sec B}-\frac{{{\sin }^{2}}B}{{{\sin }^{2}}B}={{\cot }^{2}}B\]
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