$$\frac{2+ an^2(2A)+\cot^2(2A)}{\sec(2A)\cdot\cosec(2A)}$$

Question

$$\frac{2+	an^2(2A)+\cot^2(2A)}{\sec(2A)\cdot\cosec(2A)}$$

Accepted Answer

**Solution:**
$$\frac{2+	an^2(2A)+\cot^2(2A)}{\sec(2A)\cdot\cosec(2A)}$$
Note: $2 + 	an^2	heta + \cot^2	heta = 	an^2	heta + \cot^2	heta + 1 + 1 = (	an	heta + \cot	heta)^2$
So the numerator becomes $(	an 2A + \cot 2A)^2$
Now, $	an 2A + \cot 2A = \frac{\sin 2A}{\cos 2A} + \frac{\cos 2A}{\sin 2A} = \frac{\sin^2 2A + \cos^2 2A}{\sin 2A \cos 2A} = \frac{1}{\sin 2A \cos 2A} = \sec 2A \cdot \cosec 2A$
So the expression becomes:
$$\frac{(\sec 2A \cdot \cosec 2A)^2}{\sec 2A \cdot \cosec 2A} = \sec 2A \cdot \cosec 2A$$

Answer

Answer

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