$$\tan A\operatorname{cosec}^2A=\frac{\ \sin A}{\cos A}\times\frac{\ 1}{\sin^2A}$$
$$=\frac{\ 1}{\cos A\sin A}$$
$$=\ \frac{\ 1}{\cos^2A}\times\frac{\ \cos A}{\sin A}$$
$$=\sec^2A\cot A$$
$$=\left(1+\tan^2A\right)\cot A$$
Hence, Option A is correct answer
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