A) \[\left( \frac{1+2{{\cos }^{2}}A}{sinA\,\cos A} \right)\]
B) \[\frac{(sinA\,cosA)}{(1-2co{{s}^{2}}A)}\]
C) \[\frac{(sinA\,cosA)}{(1+2co{{s}^{2}}A)}\]
D) \[\frac{(1-2co{{s}^{2}}A)}{\sin A\cos A}\]
Correct Answer: D
Solution :
\[x=\tan A-\cot A\] \[=\frac{\sin A}{\cos A}-\frac{\cos A}{\sin \,A}\] \[=\frac{{{\sin }^{2}}A-{{\cos }^{2}}A}{\sin A.\cos A}\] \[=\frac{1-{{\cos }^{2}}A-{{\cos }^{2}}A}{\sin A.\cos A}\] \[=\frac{1-2{{\cos }^{2}}A}{\sin \,A.cosA}\]
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