A) \[\tan A\]
B) \[\sin A\]
C) \[\cot A\]
D) \[\operatorname{cosec}A\]
Correct Answer: A
Solution :
If \[A+B=90{}^\circ \]then \[\Rightarrow \,\sqrt{\frac{\tan A.\,\tan \,B+\tan A.\,\cot \,B}{\sin \,A\,secB}-\frac{{{\sin }^{2}}B}{{{\cos }^{2}}A}}\] \[=\sqrt{\frac{\tan \,A.\cot A+\tan A.\tan A}{\frac{\sin A}{\cos B}}-\frac{{{\cos }^{2}}A}{{{\cos }^{2}}A}}\] \[=\sqrt{\frac{1+{{\tan }^{2}}A}{\frac{\sin A}{\sin A}}-1}=\sqrt{1+{{\tan }^{2}}A-1}\] \[=\sqrt{{{\tan }^{2}}A}=\mathbf{tanA}\]
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